# Corrigendum to “On subgroup perfect codes in Cayley graphs” [European J.
Combin. 91 (2021) 103228]

Junyang Zhang  School of Mathematical Sciences, Chongqing Normal University,
Chongqing 401331, P. R. China Sanming Zhou  School of Mathematics and
Statistics, The University of Melbourne, Parkville, VIC 3010, Australia

###### Abstract

We correct the statements of two theorems and two corollaries in our paper [On
subgroup perfect codes in Cayley graphs, _European J. Combin._ 91 (2021)
103228]. Proofs of these theorems and three other results are given as well.

††footnotetext: E-mail addresses<EMAIL_ADDRESS>(Junyang Zhang),
<EMAIL_ADDRESS>(Sanming Zhou)

This is a corrigendum to our paper [On subgroup perfect codes in Cayley
graphs, _European J. Combin._ 91 (2021) 103228], which will be referred to as
[1] hereafter. After the paper was published, an error was found in the proof
of Theorem 3.1, thereby rendering the result invalid. Here we will give a
corrected version of this result and its proof. As a consequence of this
change, we need to revise [1, Theorem 3.2] and its proof as well. It turns out
that the subgroup involved in Theorem 3.2 and Corollaries 3.3 and 3.4 in [1]
is required to be a $2$-group or have odd order or odd index. The proofs of
Theorems 3.7, 4.2 and 4.3 in [1] need to be modified due to their reliance on
Corollaries 3.3 and 3.4, but the statements of these results remain true. We
will give new proofs of these three theorems using the corrected Theorem 3.1.

We will use the notation and terminology in [1]. As usual, denote by $|G|$ the
order of a group $G$, $|G:H|$ the index in $G$ of a subgroup $H$ of $G$, and
$|X|$ the cardinality of a set $X$. We will also use the same labels to label
corresponding results from [1] or their revised versions.

The following four lemmas are exactly Lemmas 2.1–2.4 in [1]. We repeat them
here for later convenience.

######  Lemma 2.1.

Let $G$ be a group and $H$ a subgroup of $G$. Then $H$ is a perfect code of
$G$ if and only if it has a Cayley transversal in $G$.

######  Lemma 2.2.

Let $G$ be a group and $H$ a subgroup of $G$. Then $H$ is a perfect code of
$G$ if and only if it is a perfect code of any subgroup of $G$ which contains
$H$.

######  Lemma 2.3.

Let $G$ be a group and $H$ a subgroup of $G$. If $H$ is a perfect code of $G$,
then for any $g\in G$, $g^{-1}Hg$ is a perfect code of $G$. More specifically,
if $H$ is a perfect code in $\hbox{\rm Cay}(G,S)$ for some connection set $S$
of $G$, then $g^{-1}Hg$ is a perfect code in $\hbox{\rm Cay}(G,g^{-1}Sg)$.

######  Lemma 2.4.

([2, Theorem 2.2]) Let $G$ be a group and $H$ a normal subgroup of $G$. Then
$H$ is a perfect code of $G$ if and only if for all $x\in G$, $x^{2}\in H$
implies $(xh)^{2}=1$ for some $h\in H$.

An error occurs in lines 11–12 of the proof of [1, Theorem 3.1], where the
claim that “$xH\neq yH$ for distinct elements $x,y\in X$” is not true as
$z^{-1}H$ may be equal to $wH$ for some $w\in T$. The corrected version of [1,
Theorem 3.1] should read as follows.

######  Theorem 3.1.

Let $G$ be a group and $H$ a subgroup of $G$. Then $H$ is not a perfect code
of $G$ if and only if there exists a double coset $D=HxH$ with $D=D^{-1}$
having an odd number of left cosets of $H$ in $G$ and containing no
involution. In particular, if $H$ is not a perfect code of $G$, then there
exists a $2$-element $x\in G\setminus H$ such that $x^{2}\in H$, $|H:H\cap
xHx^{-1}|$ is odd, and $HxH$ contains no involution.

######  Proof.

We prove the sufficiency first. Assume that there is a double coset $D=HxH$
with $D=D^{-1}$ having $m$ left cosets of $H$ in $G$ for some odd integer $m$
and containing no involution. Set $D=\cup_{i=1}^{m}g_{i}H$, where
$g_{1}H,\ldots,g_{m}H$ are distinct left cosets of $H$ in $G$. Suppose to the
contrary that $H$ is a perfect code of $G$. By Lemma 2.1, $H$ has a Cayley
transversal $T$ in $G$. It follows that for each $1\leq i\leq m$, there exists
$x_{i}\in G$ such that $T\cap g_{i}H=\\{x_{i}\\}$. Set
$X=\\{x_{1},x_{2},\ldots,x_{m}\\}$. Then $X=T\cap D$. Since both $T$ and $D$
are inverse-closed, we have $X^{-1}=(T\cap D)^{-1}=T^{-1}\cap D^{-1}=T\cap
D=X$. Since $m$ is odd, $X$ contains at least one involution. Since $X$ is a
subset of $D$, it follows that $D$ contains at least one involution, but this
contradicts our assumption.

Now we prove the necessity. Assume that $H$ is not a perfect code of $G$. Take
a subset $T$ of $G$ with maximum cardinality such that $1\in T$, $T^{-1}=T$,
$HTH=TH$ and $xH\neq yH$ for all pairs of distinct elements $x,y\in T$. (The
existence of $T$ follows from the fact that there are subsets of $G$, say,
$\\{1\\}$, with all these properties.) Since $H$ is not a perfect code of $G$,
by Lemma 2.1, $T$ is not a left transversal of $H$ in $G$. It follows that
$G\setminus TH\neq\emptyset$ and therefore we can take an element $x\in
G\setminus TH$. Set $D=HxH$. Then $D=Hx^{-1}H$.

Since $HTH=TH$ and $x\notin TH$, we have $D\cap TH=\emptyset$. Furthermore,
since $T=T^{-1}$, we have

$D^{-1}\cap TH=D^{-1}\cap HTH=(D\cap HTH)^{-1}=\emptyset.$

Set $|H:H\cap xHx^{-1}|=\ell$. Since $H\cap x^{-1}Hx=x^{-1}(H\cap xHx^{-1})x$,
we have $|H:H\cap x^{-1}Hx|=\ell$. It is straightforward to verify that
$h_{1}xH=h_{2}xH$ if and only if $h_{1}(H\cap xHx^{-1})=h_{2}(H\cap xHx^{-1})$
for any pair of elements $h_{1},h_{2}\in H$. Therefore, $D$ is the union of
$\ell$ distinct left cosets of $H$ in $G$. So we can express $D$ as the union
of $\ell$ distinct left cosets $x_{1}xH,\ldots,x_{\ell}xH$ for some
$x_{1},\ldots,x_{\ell}\in H$. Similarly, we can write $D^{-1}$ as the union of
$\ell$ distinct left cosets $y_{1}x^{-1}H,\ldots,y_{\ell}x^{-1}H$ for some
$y_{1},\ldots,y_{\ell}\in H$.

We will complete the proof of the necessity by showing that $D=D^{-1}$, $D$
contains no involution, and $\ell$ is odd. If $D\neq D^{-1}$, then we set

$X=T\cup\\{x_{1}xy_{1}^{-1},\ldots,x_{\ell}xy_{\ell}^{-1},y_{1}x^{-1}x_{1}^{-1},\ldots,y_{\ell}x^{-1}x_{\ell}^{-1}\\}.$

If $D$ contains an involution $z$, then $z\in x_{i}xH$ for some $1\leq
i\leq\ell$. We set

$X=T\cup\\{x_{1}x_{i}^{-1}zx_{i}x_{1}^{-1},\ldots,x_{\ell}x_{i}^{-1}zx_{i}x_{\ell}^{-1}\\}.$

If $D=D^{-1}$ and $\ell$ is even, then $x^{-1}H=hxH$ for some $h\in H$. We set

$X=T\cup\left\\{x_{1}xhx_{\frac{\ell}{2}+1}^{-1},\ldots,x_{\frac{\ell}{2}}xhx_{\ell}^{-1},x_{\frac{\ell}{2}+1}h^{-1}x^{-1}x_{1}^{-1},\ldots,x_{\ell}h^{-1}x^{-1}x_{\frac{\ell}{2}}^{-1}\right\\}.$

In each case above, we have defined a subset $X$ of $G$ which contains $T$ as
a proper subset. It can be verified that $X^{-1}=X$, $HXH=XH$ and $xH\neq yH$
for any pair of distinct elements $x,y\in X$, but this contradicts the
maximality of $T$. This contradiction shows that $D=D^{-1}$, $D$ contains no
involution, and $\ell$ is an odd integer.

It remains to prove the last statement in the theorem. Suppose that $H$ is not
a perfect code of $G$. Then by what we have proved above there exists $g\in
G\setminus H$ such that $Hg^{-1}H=HgH$, $|H:H\cap gHg^{-1}|$ is odd, and $HgH$
contains no involution. Since $Hg^{-1}H=HgH$, we have $Hg^{-1}=Hgh$ for some
$h\in H$. Set $y=gh$. Then $y\in G\setminus H$ and $y^{2}\in H$. Let $s$ be
the largest odd divisor of the order of $y$. Set $x=y^{s}$. Then $x$ is a
$2$-element and $x^{2}\in H$. Since $s-1$ is even and $y^{2}\in H$, we have
$y^{s-1}\in H$. Hence $HxH=HyH=HgH$ and
$xHx^{-1}=yHy^{-1}=ghHh^{-1}g^{-1}=gHg^{-1}$. Note that $x\not\in H$ as
$y\not\in H$. Therefore, $x\in G\setminus H$ is a $2$-element such that
$x^{2}\in H$, $|H:H\cap xHx^{-1}|$ is odd, and $HxH$ contains no involution. ∎

The corrected version of [1, Theorem 3.2] should read as follows.

######  Theorem 3.2.

Let $G$ be a group and $H$ a subgroup of $G$. Suppose that either $H$ is a
$2$-group or at least one of $|H|$ and $|G:H|$ is odd. Then $H$ is a perfect
code of $G$ if and only if $H$ is a perfect code of $N_{G}(H)$.

######  Proof.

The necessity follows from Lemma 2.2. To prove the sufficiency, we assume that
$H$ is a perfect code of $N_{G}(H)$. By way of contradiction, suppose that $H$
is not a perfect code of $G$. By Theorem 3.1, there exists a $2$-element $x\in
G\setminus H$ such that $x^{2}\in H$, $|H:H\cap xHx^{-1}|$ is odd, and $HxH$
contains no involution. In particular, $xH$ contains no involution. Since $H$
is a perfect code of $N_{G}(H)$, we have $x\notin N_{G}(H)$ by Lemma 2.4. Set
$L=H\cap xHx^{-1}$. Then $L$ is a proper subgroup of $H$. Since $|H:L|$ is
odd, $H$ cannot be a $2$-group. Since $HxH$ contains no involution, $x$ is not
an involution. Therefore, $x^{2}$ is a non-identity $2$-element. Since
$x^{2}\in H$, it follows that $H$ is of even order. Since

$xLx^{-1}=x(H\cap xHx^{-1})x^{-1}=xHx^{-1}\cap x^{2}Hx^{-2}=xHx^{-1}\cap H=L,$

we have $x\in N_{G}(L)$. It follows that $|N_{G}(L):L|$ is even. Since
$|N_{G}(L):L|$ is a divisor of $|G:L|$, it follows that $|G:L|$ is even. Since
$|G:L|=|G:H||H:L|$ and $|H:L|$ is odd, we conclude that $|G:H|$ is even. Now
we have proved that $H$ is not a $2$-group and both $|H|$ and $|G:H|$ are
even. This contradicts our assumption and therefore $H$ must be a perfect code
of $G$. ∎

The following is a counterexample to both [1, Theorem 3.1] and [1, Theorem
3.2].

######  Example.

The special linear group of degree $2$ over the finite field of order $3$ has
the following representation

$G=\langle a,b,c\mid
a^{4}=c^{3}=1,a^{2}=b^{2},ab=b^{-1}a,ac=cb,bc=cab\rangle.$

Note that $G$ is a split extension of its Sylow $2$-subgroup $\langle
a,b\rangle$ by its Sylow $3$-subgroup $\langle c\rangle$. Note also that
$\langle a,b\rangle$ is isomorphic to the quaternion group, of which $a^{2}$
is the unique involution. Therefore $a^{2}$ is the unique involution of $G$.
Set $H=\langle a^{2},c\rangle$. Then $H$ is a cyclic subgroup of $G$ of order
$6$. Clearly, $a\in G\setminus H$ and $a^{2}\in H$. Furthermore, it is
straightforward to verify that $HaH=Ha^{-1}H=G\setminus H$. Therefore, $HaH$
is an inverse-closed double coset having $3$ left cosets of $H$ in $G$ and
containing no involution. By Theorem 3.1, $H$ is not a perfect code of of $G$.
Since $aH=a^{-1}H$ and $aH$ contains no involution, this shows that the
statement in [1, Theorem 3.1] is not true. Moreover, we have $H=N_{G}(H)$ and
hence $H$ is a perfect code of $N_{G}(H)$. This shows that [1, Theorem 3.2] is
not true without the additional condition that either $H$ is a $2$-group or at
least one of $|H|$ and $|G:H|$ is odd.

Combining Lemma 2.4 and Theorem 3.2, we obtain the following corrected version
of [1, Corollary 3.3].

######  Corollary 3.3.

Let $G$ be a group and $H$ a subgroup of $G$. Suppose that either $H$ is a
$2$-group or at least one of $|H|$ and $|G:H|$ is odd. Then $H$ is a perfect
code of $G$ if and only if for any $x\in N_{G}(H)$, $x^{2}\in H$ implies
$(xh)^{2}=1$ for some $h\in H$.

The corrected version of [1, Corollary 3.4] should read as follows. The proof
of this new version is exactly the same as the proof of [1, Corollary 3.4],
except that Corollary 3.3 should be understood as the corrected Corollary 3.3.

######  Corollary 3.4.

Let $G$ be a group and $H$ a subgroup of $G$. Suppose that either $H$ is a
$2$-group or at least one of $|H|$ and $|G:H|$ is odd. Then $H$ is a perfect
code of $G$ if and only if for any $2$-element $x\in N_{G}(H)$, $x^{2}\in H$
implies $(xh)^{2}=1$ for some $h\in H$.

The proof of [1, Theorem 3.7] relies on [1, Corollary 3.3]. Since the latter
has been modified, we have to revise the proof of the former. Below we will
give a proof of [1, Theorem 3.7] using Lemmas 2.1 and 2.4 and the corrected
Theorem 3.1. Fortunately, both statements in [1, Theorem 3.7] remain true
under the same hypothesis. We repeat this theorem here for convenience.

######  Theorem 3.7.

Let $G$ be a group, $N$ a normal subgroup of $G$, and $H$ a subgroup of $G$
which contains $N$. Then the following hold:

  1. (a)

if $H$ is a perfect code of $G$, then $H/N$ is a perfect code of $G/N$;

  2. (b)

if $N$ and $H/N$ are perfect codes of $G$ and $G/N$, respectively, then $H$ is
a perfect code of $G$.

######  Proof.

(a) Suppose that $H$ is a perfect code of $G$. By Lemma 2.1, there exists a
Cayley transversal $T$ of $H$ in $G$. By the definition of a Cayley
transversal, we then have $T^{-1}=T$, $G=TH$ and $xH\neq yH$ for any pair of
distinct elements $x,y\in T$. Write $T/N=\\{xN:x\in T\\}$. Then
$(T/N)^{-1}=T/N$ and $G/N=(T/N)(H/N)$. Moreover, since $N$ is contained in
$H$, we have $(xN)H/N\neq(yN)H/N$ for any pair of distinct elements $xN,yN\in
T/N$. Therefore, $T/N$ is a Cayley transversal of $H/N$ in $G/N$. By Lemma
2.1, $H/N$ is a perfect code of $G/N$.

(b) Suppose to the contrary that $H$ is not a perfect code of $G$. By Theorem
3.1, there exists a $2$-element $x\in G\setminus H$ such that $x^{2}\in H$,
$|H:H\cap xHx^{-1}|$ is odd, and $HxH$ contains no involution. Since $N$ is a
normal subgroup of $G$ contained in $H$, $xN$ is a $2$-element in
$(G/N)\setminus(H/N)$ and $(xN)^{2}\in H/N$. Therefore, $xN(H/N)=x^{-1}N(H/N)$
and it follows that the double coset $(H/N)(xN)(H/N)$ is inverse-closed.
Moreover, $(H/N)(xN)(H/N)$ contains an odd number of left cosets of $H/N$ in
$G/N$ as this number is equal to $|H/N:(H\cap xHx^{-1})/N|=|H:H\cap
xHx^{-1}|$. Since $H/N$ is a perfect code of $G/N$, by Theorem 3.1,
$(H/N)(xN)(H/N)$ contains an involution. Therefore, $(aNxNbN)^{2}=N$ for some
$aN,bN\in H/N$. That is, $(axb)^{2}N=N$ and hence $(axb)^{2}\in N$. Since $N$
is a perfect code and a normal subgroup of $G$, by Lemma 2.4, there exists
$c\in N$ such that $(axbc)^{2}=1$. Note that $a,bc\in H$. However, $HxH$
contains no involution, a contradiction. ∎

In [1], the proof of Theorem 4.2 relies on Corollaries 3.3 and 3.4. So this
proof needs to be revised, but fortunately the theorem itself remains true
under the same hypothesis. We re-present this theorem below for convenience,
and we give a proof of it using Lemma 2.3 and the corrected Theorem 3.1 and
Corollary 3.4.

######  Theorem 4.2.

Let $G$ be a group and $H$ a subgroup of $G$. If there exists a Sylow
$2$-subgroup of $H$ which is a perfect code of $G$, then $H$ is a perfect code
of $G$.

######  Proof.

Let $g$ be an arbitrary element of $G\setminus H$ such that the double coset
$HgH$ is inverse-closed and contains $m$ left cosets of $H$ in $G$ where $m$
is odd. Then $m=|H:H\cap gHg^{-1}|$. In the following we will prove that $HgH$
contains an involution. Once this is achieved, it then follows from Theorem
3.1 that $H$ is a perfect code of $G$.

Since $Hg^{-1}H=HgH$, we have $g^{-1}=h_{1}gh_{2}$ for some $h_{1},h_{2}\in
H$. Since $g\notin G\setminus H$ by our assumption, we have $gh_{1}\notin
G\setminus H$. Since $(gh_{1})^{2}=h_{2}^{-1}h_{1}\in H$, it follows that
$gh_{1}$ is of even order. Let $s$ be the largest odd divisor of the order of
$gh_{1}$. Set $x=(gh_{1})^{s}$. Then $x$ is a $2$-element in $G\setminus H$
with $x^{2}\in H$. Since $x=gh_{1}(gh_{1})^{s-1}$ and $h_{1}(gh_{1})^{s-1}\in
H$, we have $HxH=HgH$ and $xHx^{-1}=gHg^{-1}$. Set $L=H\cap xHx^{-1}$. Then
$|H:L|=|H:H\cap gHg^{-1}|=m$. Since $xLx^{-1}=xHx^{-1}\cap
x^{2}Hx^{-2}=xHx^{-1}\cap H=L$, $L$ is normal and is of index $2$ in $\langle
x,L\rangle$. Let $P$ be a Sylow $2$-subgroup of $\langle x,L\rangle$ such that
$x\in P$. Set $Q=P\cap L$. Then $x^{2}\in Q$ and $Q$ is of index $2$ in $P$.
It follows that $Q$ is normal in $P$ and is a Sylow $2$-subgroup of $L$. Since
$|H:L|=m$ is odd, $Q$ is also a Sylow $2$-subgroup of $H$. By our assumption,
$H$ has a Sylow $2$-subgroup which is a perfect code of $G$. On the other
hand, by Sylow’s Theorem, any two Sylow $2$-subgroups of $H$ are conjugate in
$H$. Thus, by Lemma 2.3, $Q$ is a perfect code of $G$. Since $Q$ is normal in
$P$ and $x\in P$, we have $x\in N_{G}(Q)$. Hence, by Corollary 3.4, there
exists $b\in Q$ such that $xb$ is an involution. Since $Q$ is a subgroup of
$H$ and $HxH=HgH$, the involution $xb$ is contained in $HgH$, as required. ∎

Finally, although [1, Theorem 4.3] remains true, its proof needs to be revised
due to the reliance on [1, Corollary 3.3]. We now give a proof of this result
using Theorem 4.2 and the corrected Theorem 3.1 and Corollary 3.4. We repeat
this result here for convenience.

######  Theorem 4.3.

Let $G$ be a group. Let $Q$ be a $2$-subgroup of $G$ and $K$ a
$2^{\prime}$-subgroup of $G$. Suppose that all Sylow $2$-subgroups of
$N_{G}(Q)$ are contained in $N_{G}(K)$. Then $QK$ is a perfect code of $G$ if
and only if $Q$ is a perfect code of $G$.

In particular, if $Q$ is a $2$-subgroup of $G$ and $K$ is a normal
$2^{\prime}$-subgroup of $G$, then $QK$ is a perfect code of $G$ if and only
if $Q$ is a perfect code of $G$.

######  Proof.

Since all Sylow $2$-subgroups of $N_{G}(Q)$ are contained in $N_{G}(K)$, $Q$
is contained in $N_{G}(K)$. Hence $H=QK$ is a subgroup of $G$ and $K$ is
normal in $H$.

Since the order of $Q$ is a power of $2$ but the order of $K$ is odd, we have
$Q\cap K=\\{1\\}$. Therefore, $Q$ is a Sylow $2$-subgroup of $H$ and $K$ is a
Hall $2^{\prime}$-subgroup of $H$. Thus, by Theorem 4.2, if $Q$ is a perfect
code of $G$, then $H$ is a perfect code of $G$. This proves the sufficiency.

We now prove the necessity. Assume that $H$ is a perfect code of $G$. Consider
an arbitrary $2$-element $x\in N_{G}(Q)$ with $x^{2}\in Q$. Since $x^{2}\in
H$, we have $(HxH)^{-1}=HxH$. Since all Sylow $2$-subgroups of $N_{G}(Q)$ are
contained in $N_{G}(K)$, we have $x\in N_{G}(K)$. Therefore,
$HxH=QKxH=xQKH=xH$. By Theorem 3.1, $HxH$ contains an involution. Since
$HxH=xQK$, we have $(xab)^{2}=1$ for some $a\in Q$ and $b\in K$. Noting that
$xa\in N_{G}(K)$, we obtain $(xa)^{2}=xab^{-1}a^{-1}x^{-1}b^{-1}\in K$. On the
other hand, since $x\in N_{G}(Q)$, we get $(xa)^{2}=x^{2}x^{-1}axa\in Q$.
Recalling $Q\cap K=\\{1\\}$, it follows that $(xa)^{2}=1$. Thus, by Corollary
3.4, $Q$ is a perfect code of $G$. ∎

Acknowledgements

We would like to thank Dr. Kai Yuan for bringing an error in the proof of [1,
Theorem 3.1] to our attention. We are grateful to the anonymous referees for
their careful reading and valuable suggestions on this paper. The first author
was supported by the Basic Research and Frontier Exploration Project of
Chongqing (cstc2018jcyjAX0010), the Science and Technology Research Program of
Chongqing Municipal Education Commission (KJQN201800512) and the National
Natural Science Foundation of China (No. 11671276). The second author was
supported by the Research Grant Support Scheme of The University of Melbourne.

## References

  * [1] J. Zhang and S. Zhou, On subgroup perfect codes in Cayley graphs, _European J. Combin._ 91 (2021) 103228.
  * [2] H. Huang, B. Xia, and S. Zhou, Perfect codes in Cayley graphs, SIAM J. Discrete Math. 32 (2018) 548–559.


# Exploiting Contextual Structure to Generate Useful Auxiliary Tasks

1st Benedict Quartey  Department of Computer Science
Brown University
<EMAIL_ADDRESS>2nd Ankit Shah  Department of Computer Science
Brown University
<EMAIL_ADDRESS>3rd George Konidaris  Department of Computer Science
Brown University
<EMAIL_ADDRESS>
###### Abstract

Reinforcement learning requires interaction with an environment, which is
expensive for robots. This constraint necessitates approaches that work with
limited environmental interaction by maximizing the reuse of previous
experiences. We propose an approach that maximizes experience reuse while
learning to solve a given task by generating and simultaneously learning
useful auxiliary tasks. To generate these tasks, we construct an abstract
temporal logic representation of the given task and leverage large language
models to generate context-aware object embeddings that facilitate object
replacements. Counterfactual reasoning and off-policy methods allow us to
simultaneously learn these auxiliary tasks while solving the given target
task. We combine these insights into a novel framework for multitask
reinforcement learning and experimentally show that our generated auxiliary
tasks share similar underlying exploration requirements as the given task,
thereby maximizing the utility of directed exploration. Our approach allows
agents to automatically learn additional useful policies without extra
environment interaction.

##  I Introduction

Reinforcement learning (RL) is a general-purpose paradigm that models agents
interacting with environments. It has proven to be a robust approach to
sequential decision-making and has recently seen several exciting successes
[1, 2, 3]. However, to learn valuable behaviors that accomplish tasks, an
agent must explore by repeatedly interacting with its environment, which can
be prohibitively expensive, especially in robotics. Therefore it is imperative
to make efficient use of experience data. One approach is to exploit the fact
that, while exploring to solve any given task, an agent acquires environmental
experiences that could be valuable for learning to solve many closely related
tasks. Consider the task of learning to make tea in an unfamiliar house. To
solve this task, a simple sequence of goals could be: _get to the kitchen,_
_get a cup_ , _get a teabag_ , _get water from the faucet_ , and _get milk
from the fridge_. Once an agent discovers the location of a fridge while
learning to solve this task, it should be able to reuse that experience to
solve a related task, such as _get a drink from the fridge_.

Figure 1: HomeGrid, a deterministic discrete grid-world domain. Agents in this
world complete tasks by visiting grid locations corresponding to objects in
the environment. Tasks are specified with LTL formulae that represent the
sequence/ordering of subgoals necessary for completing a given task.
Satisfying these tasks involves visiting relevant grid cells in an acceptable
order determined by the task specification. As an example, the numbered arrows
in the diagram indicate a policy for satisfying the Goto Fridge task.
00footnotetext: Code: https://github.com/benedictquartey/TaskExplore

Prior experience replay works [4, 5, 6] introduce effective methods of reusing
previous experiences, and off-policy learning algorithms [7, 8, 9] also enable
cross-task learning; the agent can use data generated by a _behavior policy_
to learn to perform a different _target policy_. However, these approaches are
limited in their ability to specify auxiliary tasks that maximally benefit
from counterfactual experience reasoning and off-policy learning, in part
because they do not make any assumptions about task structure.

We posit that task structure, as well as semantic and contextual structure in
object-centric environments, can be exploited to generate auxiliary tasks. We
propose a new method TaskExplore, that given a target task, uses temporal
logic expressions as a means for generating contextually similar auxiliary
tasks, by swapping objects using context-aware embeddings generated by large
language models. In this setting, we can leverage counterfactual reasoning and
off-policy methods to simultaneously learn these auxiliary tasks while
learning the given task, with a behavior policy only conditioned on the given
task. Our approach maximizes the utility of directed exploration experience,
particularly in complex environments that require agents to constrain/direct
their exploration. We show empirically that auxiliary tasks generated by
TaskExplore maximally leverage the directed experience of a single-task
curriculum.

In summary, we present two main contributions in this paper:

  1. 1.

We present a method of using context-aware object embeddings and abstract
temporal logic task representations to generate useful auxiliary tasks that
share underlying exploration requirements with a given target task.

  2. 2.

We demonstrate empirically that this class of generated tasks results in
better transfer than randomly generated tasksets and uniquely benefits from
directed exploration on the primary task.

##  II Background

In this work, we focus on object-centric environments [10] and leverage linear
temporal logic [11] to describe temporally extended tasks involving objects.
These tasks are learned using off-policy learning for linear temporal logic
[9].

###  II-A Object Oriented Markov Decision Process

An Object-oriented Markov decision process (OOMDP) [10] is a Markov decision
process where its state space is structured into objects, and its transition
function focuses on interactions between objects. A given environment in the
OOMDP paradigm augments the state, action, transition function, reward
function and discount factor tuple of regular MDPs with a set of objects
$O=\\{o_{1},...,o_{n}\\}$, where each object is an instance of a class from a
predefined set of classes $C=\\{C_{1},...,C_{n}\\}$. Each class typically also
includes a set of attributes $Att(C)=\\{C.a_{1},...,C.a_{n}\\}$.

###  II-B Linear Temporal Logic

Linear Temporal Logic (LTL) [11], presents an expressive grammar for
specifying temporal behavior. LTL expressions/formulae are composed of atomic
propositions, the logical connectives: negation ($\neg$), conjunction
($\land$), disjunction ($\lor$), implication ($\rightarrow$); and a set of
temporal operators: next ($\bigcirc$), until (U), always ($\square$) and
eventually($\lozenge$). The minimal syntax of an LTL formula is defined below:

$\varphi:=p\ |\ \neg\varphi_{1}\ |\ \varphi_{1}\wedge\varphi_{2}\ |\
\bigcirc\varphi\ |\ \varphi_{1}\text{U}\varphi_{2}$ (1)

where $p$ is an atomic proposition, a boolean literal that captures a property
of the environment; $\varphi_{1}$ and $\varphi_{2}$ are valid LTL formulas.
LTL formulas present an alternative, often more expressive and natural way of
specifying reward objectives in the reinforcement learning setting [12]. They
allow the expression of explicit specifications that characterize the
successful execution of a task. Consider a sequential navigation task of
visiting a kitchen and then visiting a fridge, where _Kitchen_ and _Fridge_
are Boolean atomic propositions that can be observed. The LTL formula below
can be used to specify this task:

$\lozenge(Kitchen\land(\lozenge Fridge))$ (2)

LTL formulas can be progressed given a sequence of truth assignments of
propositions [13] to determine which parts of the formula have been satisfied
by the states seen so far and which parts remain. This is particularly useful
in the reinforcement learning domain as this provides a method of tracking
non-Markovian objectives. In fact, a class of reinforcement learning
algorithms such as geometric-LTL (G-LTL) [12], Q-learning for Reward machines
(Q-RM) [8] and LPOPL [9] leverage this to construct and solve a product MDP
using the environment state space and automaton representation of LTL
specifications.

###  II-C Off-policy Learning with LTL

Off-policy RL methods address the setting where an agent learns a desired
_Target Policy_ while interacting in the environment using a different
_Behavior Policy_. LTL Progression for Off-Policy Learning (LPOPL) [9] adapts
the Q-learning off-policy algorithm [14] to simultaneously learn policies for
multiple LTL tasks during the same environment interaction.

Given a set of LTL task specifications $\phi$, LPOPL first extracts subtasks
from each specification via LTL progression [13], ie. progressing every given
formula over all possible truth assignments of the set of environment
propositions. LPOPL then iteratively performs a series of episodes and learns
a separate Q-value function for each task and extracted subtasks (also
expressed as LTL formulae). At each iteration, a task is selected from $\phi$
and used as the objective of the episode and actions are selected with an
epsilon greedy behavior policy conditioned on that task. However, all Q-value
functions are updated at each step via off-policy updates, allowing the agent
to make progress on tasks that may not be its current objective.

To help describe LPOPL, consider an example with the given set of tasks
$\phi=\\{\lozenge(a\land\lozenge b),\lozenge(c\land\lozenge b)\\}$.
Progressing both tasks would result in extracting the subtask $\lozenge b$.
LPOPL will then initialize three Q-value functions $Q_{\lozenge(a\land\lozenge
b)}$, $Q_{\lozenge(c\land\lozenge b)}$ and $Q_{\lozenge b}$. To run an
episode, a task will be selected from $\phi$, say $\lozenge(a\land\lozenge
b)$, then actions will be selected epsilon greedy on
$Q_{\lozenge(a\land\lozenge b)}$. Once the proposition _a_ becomes _true_
during the episode, the behavior policy becomes epsilon greedy on $Q_{\lozenge
b}$, since progressing $\lozenge(a\land\lozenge b)$ with an assignment of True
for the proposition _a_ transforms the LTL formula to $\lozenge b$. This leads
to a desirable property of LPOPL: task specifications that share progressed
LTL forms can share subtask policies.

We borrow LPOPL’s subtask extraction and Q-value function update strategies to
accelerate off-policy learning of multiple tasks. However, our approach
differs as we do not learn from a curriculum of tasks. We instead use a
behavior policy conditioned on a _single_ given task and reapply experience
and counterfactual reasoning on experiences from this task to simultaneously
solve generated auxiliary tasks.

##  III Related Work

Experience replay methods consider single-task curriculum problems where the
cardinality of the set of tasks used in extracting samples for transfer is one
(1) and includes only the target task [15]. The focus in these works is
discovering optimal methods of organizing and training on the experience
acquired from single tasks. Prioritized experience replay [5] improves on
Experience Replay [4], which uniformly sampled from a replay memory, by
prioritizing important transitions so they are sampled more frequently.
Hindsight experience replay (HER) [6] employed exploration as an implicit
curriculum and introduced learning from alternate realizations from
experiences in the replay memory, by relabelling experiences based on goals
that were actually achieved rather than what the agent was aiming to achieve.
HER’s counterfactual experience reuse is limited to singular goal states, and
so cannot encode expressive temporally extended behaviors such as reaching a
goal state while encountering specific intermediary states [7]. Additionally,
HER’s alternative goals are intermediate samples of the target task and not
distinct alternative tasks.

Other works such as Counterfactual experiences for reward machines (CRM) [7],
Q-learning for Reward Machines [8] and LPOPL [9] introduce algorithms for
applying off-policy updates to simultaneously learn multiple action-value
functions. They address HER’s limitation in applying counterfactual experience
to temporally extended alternative tasks. However, in these works, the
alternative tasks that benefit from counterfactual reasoning are assumed to be
known or given. Additionally, behavior policies that dictate environment
interactions from which counterfactual experiences are generated are
conditioned on curricula consisting of multiple tasks–typically including the
alternative tasks that benefit from these synthetic experiences.

Our work seeks to present a solution to how agents might automatically
generate expressive temporally extended auxiliary tasks––in contrast to
intermediate tasks––that can maximally leverage the directed experience of a
single-task curriculum in object-centric environments. TaskExplore is distinct
from the limited task generation works in curriculum learning literature [16,
17], where tasks are not manually designed. The goal of our approach is not to
generate good intermediate tasks to obtain experience samples, as is the goal
in task generation for curriculum learning [15]. Our focus is rather to
generate distinct auxiliary tasks that maximally leverage directed experience
from single-task curricula, which is more akin to life-long learning contexts,
where agents learn to generalize from very small or constrained datasets [18].

Figure 2: This figure depicts the TaskExplore framework. Given a task
specified in linear temporal logic, we construct an abstract task template
that replaces instance object propositions in the given formula with
embeddings of their descriptions, capturing various relevant attributes of
each object. We then generate auxiliary tasks by selecting objects from the
environment for each proposition node in our abstract task template using the
cosine similarity metric. We initialize policies (Q-value functions) for the
given task and all auxiliary tasks and perform online RL where actions are
selected $\epsilon$-greedy on only the given task, gathering directed
experiences necessary for solving the given task. At each learning step, all
Q-value functions are updated via off-policy Q-learning updates.

##  IV Problem Definition

To define our problem formally we propose Object Oriented Non-Markovian reward
decision process (OO-NMRDP), combining ideas from the Object-Oriented Markov
decision process (OOMDP) [10] and Non-Markovian reward decision process
(NMRDP) [19, 20, 21] formalisms. An OO-NMRDP is an 8-tuple $M=\textlangle
O,C,L,S,A,T,R_{\varphi},\gamma\textrangle$, where O is a set of Boolean
propositions representing objects present in the environment and detectable by
a labeling function $L:S\rightarrow 2^{O}$ that maps states to these Boolean
propositions, specifying which propositions are true in which states. C is the
set of object classes, S is the set of states, A is a set of actions,
$\gamma\in[0,1]$ is the discount factor and $T:S\times A\times
S\rightarrow[0,1]$ represents the transition dynamics of the environment.
Unlike regular MDPs the reward function $R_{\varphi}$ is defined over state
histories, where the agent receives a reward of 1 if and only if the sequence
of seen states in a given episode satisfies the LTL formula $\varphi$.

$R_{\varphi}(\langle
s_{0},...,s_{\rangle})=\begin{cases}1&~{}\mathrm{if}~{}\delta_{0:n-1}\nvDash\varphi~{}\text{and}~{}\delta_{0:n}\models\varphi\\\
0&~{}\mathrm{otherwise},\end{cases}$ (3)

where $\delta_{i:j}=\langle(s_{i}),...,L(s_{j})\rangle$. The learning agent
does not have access to either the set of object classes $C$ or the transition
dynamics of the environment $T$. We express sequential [12] or soft ordering
constraint tasks [22] as LTL formulas over the set of propositions O.

Figure 3: This figure depicts how TaskExplore constructs and leverages
context-aware object embeddings and abstract task representations/templates.
In Figure a, we use an autoregressive LLM to generate detailed descriptions
for the list of objects in our environment and use an encoder language model
to encode these generated descriptions into a 768-dimensional vector for each
object. We then cluster these description embeddings, discovering object
classes that capture the semantic similarity between objects. In Figure b, our
approach constructs a task template by representing proposition nodes in the
abstract syntax graph of a given LTL formula with embeddings of corresponding
objects. With this task template, we can create new contextually similar tasks
by selecting objects from the environment based on their cosine similarity,
balancing selections between highly correlated objects and relevant yet unseen
objects.

##  V Exploiting contextual structure
to generate auxiliary tasks

Humans can exploit the structure inherent in object-centric environments, in
terms of objects and their relationships with each other, and the
compositional structure of tasks. With this structure, they are able to learn
tasks while thinking of alternate ways in which the experience gathered may be
useful for other tasks. This ability allows humans to gain multiple useful
skills when learning any one specific thing. We present how abstract temporal
logic representations of tasks and context-aware embeddings of objects in an
environment could be used to equip RL agents with this ability.

Figure 2 illustrates the high-level steps in our method. Given a sequential
task specified in linear temporal logic $\varphi_{1}$, our framework leverages
the compositional syntax of LTL and semantics of object attributes to develop
a set of auxiliary tasks $Aux_{\varphi 1}$= $\\{a_{1},...,a_{n}\\}$ that
possess similar underlying exploration requirements as $\varphi_{1}$, via
object swaps. We then initialize a policy bank with Q-value functions for
$\varphi_{1}$ and each task in the generated auxiliary task set $Aux_{\varphi
1}$. We simultaneously learn the policy for $\varphi_{1}$, and all the
auxiliary tasks using off-policy updates akin to LPOPL [9]. The agent always
follows an $\epsilon$-greedy policy with respect to $\varphi_{1}$. A key
distinction between our learning approach and LPOPL is the absence of a multi-
task curriculum of target tasks and our behavior policy which is conditioned
on just the given task, constraining exploration in the environment.

In section V-A we present a detailed look into how we construct context-aware
object embeddings from our environment, these embeddings are used to determine
relevant objects for swaps. Section V-B looks at how we use these object
embeddings to construct abstract task templates from which we generate
auxiliary tasks. Figure 3 visualizes these two processes. Finally, in Section
V-C we explain how counterfactual reasoning and off-policy learning are used
to simultaneously learn policies for the generated auxiliary tasks.

###  V-A Exploiting Structure in Object Relationships

Large language models are trained on large text corpora and encode useful
common-sense and context-aware human knowledge, as such act as good priors for
structuring relationships between objects in object-centric environments. Our
method leverages this class of models to generate discriminative context-aware
embeddings of objects. Autoregressive language models such as GPT [23] are
adept at language generation as they are trained to maximize the likelihood of
the next token given previous tokens. Encoder-decoder and encoder-only large
language models such as T5 [24] and BERT [25] on the other hand are adept at
generating compact representations that effectively capture sentence context
and semantics. We use the InstructGPT _text-davinci-003_ model [26] to
generate detailed descriptions for the list of objects in our environment and
use the _Sentence-T5_ encoder model [27] to encode these generated
descriptions into a 768-dimensional vector for each object. We then cluster
these description embeddings using the K-means algorithm [28, 29] with
K-means++ initialization [30].

We investigate Sentence-T5’s ability to generate context-aware embeddings for
clustering based on object name alone, shown in Figure 4a. Figure 4b
alternatively shows the improvement in clustering results using descriptions
generated by the _text-davinci-003_ model, presenting insights into the
benefits of using large language models for context-aware data augmentation
for downstream tasks.

Figure 4: This figure depicts the results of performing k-means clustering on
768-dimensional embedding vectors for each environment object, results are
only visualized in 2D space. Embeddings for each object in Figure (a) are
generated by encoding the shown object name using the Sentence-T5 model.
Conversely, embeddings in Figure (b) are generated by encoding text
descriptions of each object generated by text-davinci-003. The number of
clusters used in the k-means algorithm was four(4) based on the number
distinct exploration zones in HomeGrid. Note that embeddings generated from
LLM object descriptions improved the separation of emergent cluster
boundaries, and desirably increased the distance in cosine space between
similar yet contextually different objects such as Kitchen Cabinet and
Bathroom Cabinet.

###  V-B Exploiting Structure in Task Composition

We leverage the inherent compositionality of LTL to create generalizable
representations or templates of tasks given an instance LTL formula. This task
template allows us to generate related auxiliary tasks. Similar to the
representation format used in prior works [31] we parse a given formula
$\varphi_{1}$ into its abstract syntax tree and represent it as a directed
graph $G_{\varphi_{1}}=(V_{\varphi_{1}},E_{\varphi_{1}})$. Edges
$E_{\varphi_{1}}$ in this graph connect parent operators to their subformulas
and Vertices $V_{\varphi_{1}}$ represent nodes that are either operators or
atomic propositions, as shown in Figure 3b.1. This representation allows us to
efficiently exploit the structure of the given task in constructing auxiliary
tasks.

We create an abstract task template by traversing $G_{\varphi_{1}}$ and
replacing instance proposition nodes––which represent objects in our
environment––with embeddings of their descriptions as shown in Figure 3b.2.
With this abstract task template, we generate $x$ new auxiliary tasks by
swapping relevant objects from the environment at each proposition node. We
select objects whose embedding lies in the same embedding cluster and
prioritize those with the highest cosine similarity with the template
embedding node being considered. We introduce a simple value-dependent object
selection metric based on upper confidence bounds [32] that balances out
selecting high cosine similarity objects and relevant but unseen objects.

Equation 4 governs object selection for each embedding node in a given
template.

$Obj{{}_{t}}^{v_{i}}=\max_{obj}[sim(\textbf{E}_{text}(G_{\varphi{{}_{1}}}^{v_{i}}),\textbf{E}_{text}(obj))+c\sqrt{\frac{\log{N_{t}}}{N_{t}^{obj}}}],$
(4)

where $Obj{{}_{t}}^{v_{i}}$ is the chosen object proposition for template node
$i$ at trial t;
$sim(\textbf{E}_{text}(G_{\varphi{{}_{1}}}^{v_{i}}),\textbf{E}_{text}(obj))$
is the cosine similarity between the embedding of object $obj$ and the
embedding at node $v_{i}$ of the template $G_{\varphi{{}_{1}}}$; c is a
tuneable parameter that balances selecting high-value objects vs relevant but
unseen objects; $N_{t}$ is the total number of object selection trails;
$N_{t}^{obj}$ is the number of trails where object $obj$ was selected.

###  V-C Off-policy Updates via Counterfactual Experience

Concerning off-policy updates, when an episode is run with an epsilon greedy
behavior policy on the given LTL formula, the Q-value function of each
auxiliary task will also be updated as if their corresponding formula was the
current objective. Assuming an action _a_ is taken in state _s_ resulting in a
new state $s^{{}^{\prime}}$, to update a specific $Q_{\varphi}$, the reward
that would have been observed during that transition if the agent’s objective
was $\varphi$ is computed. To achieve this, $\varphi$ is progressed through
the new state $s^{{}^{\prime}}$; if the resulting formula
$\varphi^{{}^{\prime}}$ is _true_ the reward is 1, and 0 otherwise.
$Q_{\varphi}$ is then updated using the following rule:

$Q_{\varphi}(s,a)\leftarrow
Q_{\varphi}(s,a)+\alpha(r+\gamma\max_{a^{{}^{\prime}}}Q_{\varphi^{{}^{\prime}}}(s^{{}^{\prime}},a^{{}^{\prime}})-Q_{\varphi}(s,a).$
(5)

As shown in LPOPL [9] learning is globally optimal as $Q_{\varphi}(s,a)$ is
updated with the maximum of every action $a^{{}^{\prime}}$ from its progressed
subtask $Q_{\varphi^{{}^{\prime}}}(s^{{}^{\prime}},a^{{}^{\prime}})$.

##  VI Experiments

The task specification used in our experiments was a food preparation task
where the agent had to go to the kitchen cabinet, obtain a cooking pot, obtain
seasoning, then go to the fridge, obtain chicken, and finally go to the stove.
In HomeGrid, this task corresponds to visiting the right cells in the correct
order. The LTL formula below represents this task using the atomic
propositions that represent each of the relevant objects:

$\lozenge(C\land\lozenge(P\land\lozenge(I\land\lozenge(F\land\lozenge(H\land\lozenge
Y)))))$ (6)

See Figure 1 for a description of our environment HomeGrid. We find that a
good heuristic for choosing the minimum number of object clusters is the
number of distinct exploration zones in the environment. For HomeGrid, this is
four (4) since there are four distinct useful exploration regions namely
”Kitchen”, ”Bathroom”, ”Living room” and ”Bedroom”. We evaluate our approach
with the following three conditions:

  1. 1.

Ours: Given our food preparation task $\varphi_{1}$ we generate 20 auxiliary
tasks following our approach. We learn these tasks simultaneously with
$\varphi_{1}$ with a behavior policy epsilon greedy on $\varphi_{1}$,
directing exploration towards more relevant experiences for $\varphi_{1}$.

  2. 2.

Baseline 1: To demonstrate that tasks generated by TaskExplore uniquely
leverage the directed experience of a single-task curriculum, we repeat the
approach described above, replacing the behavior policy with a random one that
explores more widely.

  3. 3.

Baseline 2: To show that tasks generated by TaskExplore more relevantly
benefit from directed exploration experience than the general set of possible
tasks, we generate 20 auxiliary tasks by randomly sampling sequential tasks
from the set of propositions in our environment, as typically done in prior
works [31, 22]. These tasks are sampled to have the same length of
propositions as the given task which places them in the same level of
difficulty. We learn these tasks simultaneously with $\varphi_{1}$, using a
behavior policy epsilon greedy on $\varphi_{1}$.

Figure 5: Figure (a) shows the normalized discounted reward obtained by the
agent on the given task as it learns to solve it simultaneously with
TaskExplore generated auxiliary tasks using a random behavior policy
($\pi^{random}$) and an epsilon greedy behavior policy ($\pi^{*}$). Figure (b)
shows the task success rate on auxiliary tasks as learning progresses.
Learning TaskExplore generated tasks while using epsilon greedy ($\pi^{*}$)
behavior policy on the given task significantly outperforms all other
baselines. All results are normalized over 7 different seeded runs

##  VII Results and Discussion

Figure 6: This figure shows the exploration heatmap of the random and e-greedy
behavior policies while learning the given food preparation task with
counterfactual updates on TaskExplore generated auxiliary tasks. The agent
starts each episode in cell (x-axis=14,y-axis=13). This is an ablative diagram
that helps visualize the beneficial effects of directed exploration in complex
environments and how this directed experience can benefit contextually similar
auxiliary tasks that share the same underlying exploration requirements, more
than an exploration strategy that may produce more diverse experiences.

Figure 5 presents the results from our experiments, which highlight several
interesting properties of the auxiliary task set developed by TaskExplore.
Firstly, in Figure 5a, we see that our approach which simultaneously learns to
solve auxiliary tasks with a behavior policy $\epsilon$-greedy on the given
task does not adversely affect performance on the given task. By contrast,
performance on the given task deteriorates when using a random behavior
policy, highlighting the relevance of directed exploration.

Intuitively, learning with a random behavior policy gathers more diverse
experiences which should benefit multiple auxiliary tasks more than the
experience gathered during directed exploration. However, our results in 5b
show that the auxiliary tasks developed by TaskExplore maximally leverage that
constrained experience and actually perform better than when using a random
behavior policy. This is because the generated tasks are contextually similar
to the given task and share similar underlying exploration requirements.
Figure 6 presents further insights into this phenomenon.

Finally, the results in 5b show that developing a curriculum of randomly
sampled tasks from the general distribution space of possible tasks in the
environment cannot leverage directed exploration experience as well as
auxiliary tasks developed by TaskExplore.

##  VIII Conclusion

This paper introduced an approach to how agents might automatically generate
expressive temporally extended auxiliary tasks that can maximally leverage the
directed experience of a single-task curriculum in object-centric
environments. This approach is particularly valuable in the lifelong learning
setting, as agents can generate and solve new tasks from constrained datasets.
In the spirit of reusing computation, a policy bank of these policies can be
saved and reused to accelerate learning future tasks.

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# The sum of EP elements in a ring with involution

Huanyin Chen  and  Marjan Sheibani∗ School of Mathematics
Hangzhou Normal University
Hangzhou, China<EMAIL_ADDRESS>Farzanegan Campus, Semnan University,
Semnan, Iran<EMAIL_ADDRESS>
###### Abstract.

We present a necessary and sufficient condition under which the sum of two EP
elements in a *-ring has core inverse. As an application, we establish the
conditions under which a block complex with EP subblocks has core inverse.

###### Key words and phrases:

core inverse; EP element, additive property; block complex matrix; ring.

###### 2020 Mathematics Subject Classification:

15A09, 16W10, 16U80.

Corresponding author: Marjan Sheibani

##  1\. Introduction

Let $R$ be an associative ring with an involution $*$. Here, $*:R\rightarrow
R$ is an operation satisfying $(x+y)^{*}=x^{*}+y^{*},(xy)^{*}=y^{*}x^{*}$ and
$(x^{*})^{*}=x$ for all $x,y\in R$. A ring $R$ with involution $*$ is called a
*-ring.

The core inverse of a complex matrix was first introduced by Baksalary and G.
Trenkler in [2]. Rakić et al. extended the core inverse of a complex to the
case of an element in a *-ring (see [11]). An element $a\in R$ has core
inverse if there exists some $x\in R$ such that

$axa=a,xR=aR,Rx=Ra^{*}.$

If such $x$ exists, it is unique, and denote it by $a^{\tiny\\#⃝}$. In 2019,
Xu obtained a new characterization of core inverse ( see[15, Theorem 3.1]).
They proved that $a\in R$ has core inverse if and only if there exists some
$x\in R$ such that

$xa^{2}=a,ax^{2}=x,(ax)^{*}=ax.$

In this case, $a^{\tiny\\#⃝}=x$.

An element $a$ in $R$ has group inverse provided that there exists $x\in R$
such that

$xa^{2}=a,ax^{2}=x,ax=xa.$

Such $x$ is unique if exists, denoted by $a^{\\#}$, and called the group
inverse of $a$. An element $a$ in $R$ has Drazin inverse provided that there
exists $x\in R$ such that

$xa^{n+1}=a^{n},ax^{2}=x,ax=xa.$

Such $x$ is unique if exists, denoted by $a^{D}$, and called the Drazin
inverse of $a$. Evidently, $a\in R$ has Drazin inverse if and only $a^{n}\in
R$ has group inverse for some natural number $n$. An element $a\in R$ has
$(1,3)$-inverse provided that there exists some $x\in R$ such that $a=axa$ and
$(ax)^{*}=ax$. We denote $x$ by $a^{(1,3)}$. We list several characterizations
of core inverse in a *-ring.

######  Theorem 1.1.

(see [7, Theorem 3.4], [11, Theorem 2.14] and [15, Theorem 2.6]). Let $R$ be a
*-ring, and let $a\in R$. Then the following are equivalent:

  1. (1)

$a$ has core inverse.

  2. (2)

There exists $x\in R$ such that $axa=a,x=xax,xa^{2}=a,ax^{2}=x,(ax)^{*}=ax$.

  3. (4)

There exists some $p^{2}=p=p^{*}\in R$ (i.e., $p$ is a projection) such that
$pa=0$ and $a+p\in R$ is invertible.

  4. (5)

$a\in R$ has group inverse and $Ra=Ra^{*}a$.

  5. (6)

$a\in R$ has group inverse and $a\in R$ has $(1,3)$-inverse.
In this case, $a^{\tiny\\#⃝}=x$.

The core invertibility in a *-ring is attractive. Many authors have studied
such problems from many different views, e.g., [1, 5, 6, 13, 15, 17].

In [15, Theorem 4.3], Xue, Chen and Zhang proved that $a+b\in R$ has core
inverse under the conditions $ab=0$ and $a^{*}b=0$ for two core invertible
elements $a$ and $b$ in $R$.

In [18, Theorem 4.1], Zhou et al. considered the core inverse of $a+b$ under
the conditions
$a^{2}a^{\tiny\\#⃝}b^{\tiny\\#⃝}b=baa^{\tiny\\#⃝},ab^{\tiny\\#⃝}b=aa^{\tiny\\#⃝}b$
in a Dedekind-finite ring in which $2$ is invertible.

Recall that $a\in R$ is EP (i.e., an EP element) provided that $a$ has group
inverse $a^{\\#}$ and Moore-Penrose inverse $a^{{\dagger}}$ with
$a^{\\#}=a^{{\dagger}}$(see [14]). In 2019, Xu et al. proved that $a\in R$ is
EP if and only if there exists some $x\in R$ such that

$xa^{2}=a,ax^{2}=x,(xa)^{*}=xa.$

In [11], Rakić et al. proved that $a\in R$ is EP if and only if $a\in R$ has
core inverse and $a^{\tiny\\#⃝}=a^{\\#}$. For more results on EP elements in a
ring, we refer [12, 14, 17] to the reader.

In this paper, we present a new additive result for the core inverse in a
*-ring. We give a necessary and sufficient condition under which the sum of
two EP elements has core inverse.

Let $C^{n\times n}$ be a *-ring of $n\times n$ complex matrices, with
conjugate transpose as the involution. As an application, we establish
perturbed conditions under which a block complex matrix with EP subblocks has
core inverse.

Throughout the paper, all *-rings are associative with an identity. Let $a\in
R^{D}$ and $a^{\pi}=1-aa^{D}$. Let $p^{2}=p\in R$, and let $x\in R$. We write
$x=pxp+px(1-p)+(1-p)xp+(1-p)x(1-p),$ and induce a Pierce representation given
by the matrix $x=\left(\begin{array}[]{cc}pxp&px(1-p)\\\
(1-p)xp&(1-p)x(1-p)\end{array}\right)_{p}.$ We use $R^{\\#}$ and
$R^{\tiny\\#⃝}$ to denote the sets of all group and core invertible elements
in $R$, respectively. ${\mathbb{C}}^{*}$ stands for the set of all nonzero
complex number and $A^{*}$ denotes the conjugate transpose $\overline{A}^{T}$
of the complex matrix $A$.

##  2\. Key lemmas

This section is to investigate elementary properties of the core inverse in a
*-ring which will be used in the sequel. We begin with

######  Lemma 2.1.

Let $p\in R$ be an idempotent, $a\in R^{\\#}$ and $pap^{\pi}=0$. If
$ap^{\pi}\in R^{\\#}$, then $(ap^{\pi})(ap^{\pi})^{\\#}=(aa^{\\#})p^{\pi}.$

###### Proof.

Since $pap^{\pi}=0$, we see that

$\begin{array}[]{lll}ap^{\pi}-(a^{\\#}p^{\pi})(ap^{\pi})^{2}&=&ap^{\pi}-(a^{\\#}a^{2})p^{\pi}\\\
&=&(a-a^{\\#}a^{2})p^{\pi}=0.\end{array}$

Hence, $ap^{\pi}=(a^{\\#}p^{\pi})(ap^{\pi})^{2}$. It follows that

$\begin{array}[]{lll}(ap^{\pi})(ap^{\pi})^{\\#}&=&(a^{\\#}p^{\pi})(ap^{\pi})^{2}(ap^{\pi})^{\\#}\\\
&=&(a^{\\#}p^{\pi})(ap^{\pi})=(aa^{\\#})p^{\pi},\end{array}$

as required.∎

######  Lemma 2.2.

Let $p\in R$ be a projection, $x\in R$ and $x=\left(\begin{array}[]{cc}a&0\\\
b&d\end{array}\right)_{p}$. Assume that $a\in R^{\\#}$. Then $x\in R^{\\#}$ if
and only if $d\in R^{\\#}$ and $d^{\pi}ba^{\pi}=0$. In this case,

$x^{\\#}=\left(\begin{array}[]{cc}a^{\\#}&0\\\
z&d^{\\#}\end{array}\right)_{p},$

where

$z=(d^{\\#})^{2}ba^{\pi}+d^{\pi}b(a^{\\#})^{2}-d^{\\#}ba^{\\#}.$

###### Proof.

This is the dual of [10, Theorem 2.1]. We easily obtained it by applying [10,
Theorem 2.1] to the opposite ring $M_{2}(R^{op})$.∎

######  Lemma 2.3.

Let $p\in R$ be a projection, $a\in R$ and $pap^{\pi}=0$. If $pap,ap^{\pi}\in
R^{\tiny\\#⃝}$ and $(ap^{\pi})^{\pi}p^{\pi}ap=0,$ then $a\in R^{\tiny\\#⃝}$
and $pa^{\tiny\\#⃝}p^{\pi}=0$.

###### Proof.

Since $pap^{\pi}=0$, we have $a=\left(\begin{array}[]{cc}pap&0\\\
p^{\pi}ap&ap^{\pi}\end{array}\right)_{p}.$ By virtue of Theorem 1.1,
$pap,ap^{\pi}\in R^{\\#}$. As $(ap^{\pi})^{\pi}p^{\pi}apap=0,$ it follows by
Lemma 2.2 that $a\in R^{\\#}$. Moreover, we have

$a^{\\#}=\left(\begin{array}[]{cc}(pap)^{\\#}&0\\\
z&(ap^{\pi})^{\\#}\end{array}\right)_{p},$

where

$z=[(ap^{\pi})^{\\#}]^{2}p^{\pi}ap(pap)^{\pi}+(ap^{\pi})^{\pi}p^{\pi}ap[(pap)^{\\#}]^{2}-(ap^{\pi})^{\\#}p^{\pi}ap(pap)^{\\#}.$

Since $p\in R$ is a projection, we have

$(pap)^{\tiny\\#⃝}=[p\cdot(pap)\cdot
p]^{\tiny\\#⃝}=p(pap)^{\tiny\\#⃝}p\subseteq pRp.$

Similarly,

$(ap^{\pi})^{\tiny\\#⃝}=(p^{\pi}ap^{\pi})^{\tiny\\#⃝}\subseteq
p^{\pi}Rp^{\pi}.$

Set

$x=\left(\begin{array}[]{cc}(pap)^{\tiny\\#⃝}&0\\\
-(ap^{\pi})^{\tiny\\#⃝}(p^{\pi}ap)(pap)^{\tiny\\#⃝}&(ap^{\pi})^{\tiny\\#⃝}\end{array}\right)_{p}.$

Then we have

$\begin{array}[]{rll}ax&=&\left(\begin{array}[]{cc}pap&0\\\
p^{\pi}ap&ap^{\pi}\end{array}\right)_{p}\left(\begin{array}[]{cc}(pap)^{\tiny\\#⃝}&0\\\
-(ap^{\pi})^{\tiny\\#⃝}(p^{\pi}ap)(pap)^{\tiny\\#⃝}&(ap^{\pi})^{\tiny\\#⃝}\end{array}\right)_{p}\\\
&=&\left(\begin{array}[]{cc}(pap)(pap)^{\tiny\\#⃝}&0\\\
(ap^{\pi})^{\pi}p^{\pi}ap(pap)^{\tiny\\#⃝}&(ap^{\pi})(ap^{\pi})^{\tiny\\#⃝}\end{array}\right)_{p}\\\
&=&\left(\begin{array}[]{cc}(pap)(pap)^{\tiny\\#⃝}&0\\\
0&(ap^{\pi})(ap^{\pi})^{\tiny\\#⃝}\end{array}\right)_{p}.\end{array}$

Hence

$(ax)^{*}=\left(\begin{array}[]{cc}(pap)(pap)^{\tiny\\#⃝}&0\\\
0&(ap^{\pi})(ap^{\pi})^{\tiny\\#⃝}\end{array}\right)_{p}^{*}=ax.$

We further verify that

$\begin{array}[]{rll}axa&=&\left(\begin{array}[]{cc}(pap)(pap)^{\tiny\\#⃝}&0\\\
0&(ap^{\pi})(ap^{\pi})^{\tiny\\#⃝}\end{array}\right)_{p}\left(\begin{array}[]{cc}pap&0\\\
p^{\pi}ap&ap^{\pi}\end{array}\right)_{p}\\\ &=&a,\end{array}$

and so $a\in R^{(1,3)}$. According to Theorem 1.1, $a$ has core inverse.

Moreover, we have

$\begin{array}[]{rl}&a^{\tiny\\#⃝}=a^{\\#}ax=\left(\begin{array}[]{cc}(pap)^{\\#}&0\\\
z&(ap^{\pi})^{\\#}\end{array}\right)_{p}\left(\begin{array}[]{cc}(pap)(pap)^{\tiny\\#⃝}&0\\\
0&(ap^{\pi})(ap^{\pi})^{\tiny\\#⃝}\end{array}\right)_{p}\\\
=&\left(\begin{array}[]{cc}(pap)^{\\#}&0\\\
z(pap)(pap)^{\tiny\\#⃝}&(ap^{\pi})^{\\#}\end{array}\right)_{p}.\end{array}$

Therefore $pa^{\tiny\\#⃝}p^{\pi}=0$, as asserted.∎

######  Lemma 2.4.

Let $a\in R^{EP},b\in R^{\tiny\\#⃝}$. If $a^{\pi}b\in R^{D}$ and
$aba^{\pi}=0$, then the following are equivalent:

  1. (1)

$a+b\in R^{\tiny\\#⃝}$ and $a(a+b)^{\tiny\\#⃝}a^{\pi}=0$.

  2. (2)

$a(1+a^{\\#}b)\in R^{\tiny\\#⃝}$ and $b^{\pi}a^{\pi}b=0$.

###### Proof.

Let $p=aa^{\\#}$. Since $a\in R^{EP}$, we see that $p=aa^{\tiny\\#⃝}$ is a
projection. Clearly, $pbp^{\pi}=a^{\\#}(aba^{\pi})=0$. Then we have

$a=\left(\begin{array}[]{cc}a_{1}&0\\\
0&0\end{array}\right)_{p},b=\left(\begin{array}[]{cc}b_{1}&0\\\
b_{3}&b_{4}\end{array}\right)_{p}.$

Thus,

$a+b=\left(\begin{array}[]{cc}a_{1}+b_{1}&0\\\
b_{3}&b_{4}\end{array}\right)_{p}.$

Here,

$\begin{array}[]{rll}a_{1}+b_{1}&=&aa^{\\#}(a+b)aa^{\\#}=aa^{\\#}(a+b),\\\
b_{4}&=&a^{\pi}(a+b)a^{\pi}=ba^{\pi}.\end{array}$

In view of Lemma 2.1, we directly compute that

$\begin{array}[]{rll}b_{4}^{\\#}&=&(ba^{\pi})^{\\#}=b^{\\#}a^{\pi},\\\
b_{4}^{\pi}&=&a^{\pi}-b^{\\#}a^{\pi}ba^{\pi}=a^{\pi}-b^{\\#}ba^{\pi}=b^{\pi}a^{\pi}.\end{array}$

$(1)\Rightarrow(2)$ Set $x=a+b$ and $w=a_{1}+b_{1}$. Since
$aa^{\\#}(a+b)^{\tiny\\#⃝}a^{\pi}=0$, we may write
$x^{\tiny\\#⃝}=\left(\begin{array}[]{cc}x_{1}&0\\\
x_{3}&x_{4}\end{array}\right)_{p}.$ In light of Theorem 1.1, we have

$(xx^{\tiny\\#⃝})^{*}=xx^{\tiny\\#⃝},x(x^{\tiny\\#⃝})^{2}=x^{\tiny\\#⃝},x^{\tiny\\#⃝}x^{2}=x~{}\mbox{and}~{}xx^{\tiny\\#⃝}x=x.$

Hence, $(wx_{1})^{*}=wx_{1},wx_{1}^{2}=x_{1},x_{1}w^{2}=w.$ Therefore
$w=aa^{\\#}(a+b)=a(1+a^{\\#}b)\in R^{\tiny\\#⃝}$.

Since $xx^{\tiny\\#⃝}x=x$, we have

$\begin{array}[]{rll}\left(\begin{array}[]{cc}w&0\\\
b_{3}&b_{4}\end{array}\right)_{p}&=&\left(\begin{array}[]{cc}w&0\\\
b_{3}&b_{4}\end{array}\right)_{p}\left(\begin{array}[]{cc}x_{1}&0\\\
x_{3}&x_{4}\end{array}\right)_{p}\left(\begin{array}[]{cc}w&0\\\
b_{3}&b_{4}\end{array}\right)_{p}\\\ &=&\left(\begin{array}[]{cc}xx_{1}w&0\\\
(b_{3}x_{1}+b_{4}x_{3})x+b_{4}x_{4}b_{3}&b_{4}x_{4}b_{4}\end{array}\right)_{p}.\end{array}$

Furthermore, we have

$\begin{array}[]{rll}\left(\begin{array}[]{cc}xx_{1}&0\\\
b_{3}x_{1}+b_{4}x_{3}&b_{4}x_{4}\end{array}\right)_{p}&=&\left(\begin{array}[]{cc}w&0\\\
b_{3}&b_{4}\end{array}\right)_{p}\left(\begin{array}[]{cc}x_{1}&0\\\
x_{3}&x_{4}\end{array}\right)_{p}\\\ &=&xx^{\tiny\\#⃝}=(xx^{\tiny\\#⃝})^{*}\\\
&=&\left(\begin{array}[]{cc}(xx_{1})^{*}&(b_{3}x_{1}+b_{4}x_{3})^{*}\\\
0&(b_{4}x_{4})^{*}\end{array}\right)_{p}.\end{array}$

Thus

$b_{3}=(b_{3}x_{1}+b_{4}x_{3})w+b_{4}x_{4}b_{3},b_{3}x_{1}+b_{4}x_{3}=0.$

We infer that $b_{3}=b_{4}x_{4}b_{3}$, and so
$(p^{\pi}-bb^{\\#}p^{\pi})baa^{\\#}=0$. This implies that
$b^{\pi}a^{\pi}baa^{\\#}=0$. On the other hand,
$b^{\pi}a^{\pi}ba^{\pi}=b^{\pi}ba^{\pi}-b^{\pi}a^{\\#}(aba^{\pi})=0$.
Therefore $b^{\pi}a^{\pi}b=b^{\pi}a^{\pi}b(aa^{\\#}+a^{\pi})=0$.

$(2)\Rightarrow(1)$ We verify that

$\begin{array}[]{lll}p^{\pi}[1-b_{4}^{\\#}b_{4}]b_{3}&=&a^{\pi}[1-b_{4}^{\\#}b_{4}]a^{\pi}baa^{\\#}\\\
&=&a^{\pi}[1-(ba^{\pi})(ba^{\pi})^{\\#}]a^{\pi}baa^{\\#}\\\
&=&a^{\pi}[1-bb^{\\#}a^{\pi}]a^{\pi}baa^{\\#}\\\
&=&a^{\pi}b^{\pi}a^{\pi}baa^{\\#}=0.\end{array}$

Hence,

$\begin{array}[]{rll}p(a+b)p^{\pi}&=&a^{\\#}(aba^{\pi})=0,\\\
p(a+b)p&=&a+aa^{\\#}baa^{\\#}=a(1+a^{\\#}b)\in R^{\tiny\\#⃝},\\\
(a+b)p^{\pi}&=&ba^{\pi}\in R^{\tiny\\#⃝}.\end{array}$

Since every core invertible element has group inverse, by Lemma 2.2, $a+b\in
R^{\\#}$. Moreover, we derive

$\begin{array}[]{rll}((a+b)p^{\pi})^{\pi}p^{\pi}(a+b)p&=&(1-(a+b)p^{\pi}((a+b)p^{\pi})^{\\#})p^{\pi}(a+b)p\\\
&=&(1-p^{\pi}(a+b)(a+b)^{\\#})p^{\pi}(a+b)p\\\
&=&p^{\pi}(1-(a+b)(a+b)^{\\#})p^{\pi}(a+b)p\\\
&=&p^{\pi}(1-b_{4}^{\\#}b_{4})p^{\pi}(a+b)p=0.\end{array}$

According to Lemma 2.3, $a+b\in R^{\tiny\\#⃝}$ and
$a(a+b)^{\tiny\\#⃝}a^{\pi}=a^{\\#}p(a+b)^{\tiny\\#⃝}p^{\pi}=0$.∎

##  3\. The main result

We have at our disposal all the information necessary to prove our main
result.

######  Theorem 3.1.

Let $a,b\in R^{EP}$ and $a^{\pi}b,b^{\pi}a\in R^{D}$. If
$aba^{\pi}=bab^{\pi}=0$, then the following are equivalent:

  1. (1)

$a+b\in R^{\tiny\\#⃝},a(a+b)^{\tiny\\#⃝}a^{\pi}=b(a+b)^{\tiny\\#⃝}b^{\pi}=0$.

  2. (2)

$aa^{\\#}b+bb^{\\#}a\in R^{\tiny\\#⃝}$ and
$a^{\pi}b^{\pi}a=b^{\pi}a^{\pi}b=0$.

###### Proof.

$(1)\Rightarrow(2)$ In view of Lemma 2.4, $a(1+a^{\\#}b)\in R^{\tiny\\#⃝}$ and
$b^{\pi}a^{\pi}b=0$. Analogously, $a^{\pi}b^{\pi}a=0$. As in the proof in
Lemma 2.4, we see that
$(a(1+a^{\\#}b))^{\tiny\\#⃝}=aa^{\\#}(a+b)^{\tiny\\#⃝}aa^{\\#}$. Let
$q=aa^{\\#}bb^{\\#}$. Since $aba^{\pi}=0$, we have
$b=\left(\begin{array}[]{cc}aa^{\\#}b&0\\\
a^{\pi}baa^{\\#}&ba^{\pi}\end{array}\right)_{aa^{\\#}}$. In view of [13, Lemma
2.1], $ba^{\pi}\in R^{\\#}$. Then $aa^{\\#}b\in R^{\\#}$ and
$(aa^{\\#}b)^{\\#}=aa^{\\#}b^{\\#}$ by [10, Theorem 2.3]. Hence
$q^{2}=q=(aa^{\\#}b)(aa^{\\#}b)^{\\#}$. Thus

$\begin{array}[]{rll}q(a(1+a^{\\#}b))(1-q)&=&aa^{\\#}bb^{\\#}ab^{\pi}+aa^{\\#}bb^{\\#}aa^{\\#}b(1-aa^{\\#}bb^{\\#})\\\
&=&aa^{\\#}bb^{\\#}aa^{\\#}b-aa^{\\#}bb^{\\#}aa^{\\#}b[1-a^{\pi}]bb^{\\#})\\\
&=&0,\\\
q(a(1+a^{\\#}b))^{\tiny\\#⃝}(1-q)&=&qaa^{\\#}(a+b)^{\tiny\\#⃝}aa^{\\#}(1-q)\\\
&=&qaa^{\\#}(a+b)^{\tiny\\#⃝}aa^{\\#}(1-aa^{\\#}bb^{\\#})\\\
&=&qaa^{\\#}(a+b)^{\tiny\\#⃝}aa^{\\#}b^{\pi}\\\
&=&qaa^{\\#}(a+b)^{\tiny\\#⃝}b^{\pi}-qaa^{\\#}(a+b)^{\tiny\\#⃝}a^{\pi}b^{\pi}\\\
&=&aa^{\\#}bb^{\\#}aa^{\\#}(a+b)^{\tiny\\#⃝}b^{\pi}\\\ &=&0.\end{array}$

We may write

$a+aa^{\\#}b=\left(\begin{array}[]{cc}c_{1}&0\\\
c_{2}&c_{3}\end{array}\right)_{q},(a+aa^{\\#}b)^{{}^{\tiny\\#⃝}}=\left(\begin{array}[]{cc}x_{1}&0\\\
x_{2}&x_{3}\end{array}\right)_{q}.$

As in the proof of Lemma 2.4, we prove that $c_{1}^{\tiny\\#⃝}=x_{1}$.
Moreover, we have

$\begin{array}[]{rll}c_{1}&=&aa^{\\#}bb^{\\#}a[1+a^{\\#}b]aa^{\\#}bb^{\\#}\\\
&=&aa^{\\#}bb^{\\#}abb^{\\#}+aa^{\\#}bb^{\\#}aa^{\\#}baa^{\\#}bb^{\\#}\\\
&=&aa^{\\#}bb^{\\#}a+aa^{\\#}bb^{\\#}aa^{\\#}b\\\
&=&aa^{\\#}(1-b^{\pi})a+aa^{\\#}bb^{\\#}(1-a^{\pi})b\\\
&=&aa^{\\#}a-aa^{\\#}b^{\pi}a+aa^{\\#}bb^{\\#}b-aa^{\\#}bb^{\\#}a^{\pi}b\\\
&=&a-b^{\pi}a+aa^{\\#}b\\\ &=&aa^{\\#}b+bb^{\\#}a.\end{array}$

Therefore $aa^{\\#}b+bb^{\\#}a\in R^{\tiny\\#⃝}$.

$(2)\Rightarrow(1)$ Let $q=aa^{\\#}bb^{\\#}$. Then $q^{2}=q\in R$. Then

$\begin{array}[]{rll}q&=&aa^{\\#}bb^{\\#}aa^{\\#}bb^{\\#}\\\
&=&(1-a^{\pi})bb^{\\#}aa^{\\#}bb^{\\#}\\\
&=&bb^{\\#}aa^{\\#}bb^{\\#}-a^{\pi}bb^{\\#}aa^{\\#}\\\
&=&bb^{\\#}aa^{\\#}bb^{\\#}.\end{array}$

Since $a,b\in R^{EP}$, we have
$(aa^{\\#})^{*}=aa^{\\#},(bb^{\\#})^{*}=bb^{\\#}.$ Then $q^{*}=q$, i.e., $q\in
R$ is a projection. In view of Lemma 2.4, it will suffice to prove that
$a+aa^{\\#}b=a(1+a^{\\#}a)\in R^{\tiny\\#⃝}$. We check that

$\begin{array}[]{rll}qa(1-q)&=&aa^{\\#}bb^{\\#}a[1-aa^{\\#}bb^{\\#}]\\\
&=&aa^{\\#}bb^{\\#}a-aa^{\\#}bb^{\\#}a\\\
&=&aa^{\\#}(1-b^{\pi})a-aa^{\\#}bb^{\\#}a\\\
&=&a-(1-a^{\pi})b^{\pi}a-aa^{\\#}bb^{\\#}a\\\
&=&a-b^{\pi}a-aa^{\\#}(1-b^{\pi})a\\\ &=&-b^{\pi}a+(1-a^{\pi})b^{\pi}a\\\
&=&0,\\\ qaa^{\\#}b(1-q)&=&aa^{\\#}bb^{\\#}aa^{\\#}b[1-aa^{\\#}bb^{\\#}]\\\
&=&aa^{\\#}bb^{\\#}[aa^{\\#}b-aa^{\\#}baa^{\\#}bb^{\\#}]\\\
&=&aa^{\\#}bb^{\\#}[aa^{\\#}b-aa^{\\#}b(1-a^{\pi})bb^{\\#}]\\\
&=&aa^{\\#}bb^{\\#}[aa^{\\#}b-aa^{\\#}b^{2}b^{\\#}-aa^{\\#}ba^{\pi}bb^{\\#}]\\\
&=&0,\\\ \end{array}$
$\begin{array}[]{rll}(1-q)aa^{\\#}bq&=&[1-aa^{\\#}bb^{\\#}]aa^{\\#}baa^{\\#}bb^{\\#}\\\
&=&[1-aa^{\\#}bb^{\\#}]aa^{\\#}b\\\
&=&aa^{\\#}b-aa^{\\#}bb^{\\#}(1-a^{\pi})b\\\
&=&aa^{\\#}(1-b^{\pi})a^{\pi})b\\\ &=&0,\\\
(1-q)aa^{\\#}b(1-q)&=&[1-aa^{\\#}bb^{\\#}]aa^{\\#}b[1-aa^{\\#}bb^{\\#}]\\\
&=&[aa^{\\#}b-aa^{\\#}bb^{\\#}(1-a^{\pi})b][1-aa^{\\#}bb^{\\#}]\\\
&=&[aa^{\\#}bb^{\\#}a^{\pi})b][1-aa^{\\#}bb^{\\#}]\\\
&=&[aa^{\\#}(1-b^{\pi})a^{\pi})b][1-aa^{\\#}bb^{\\#}]\\\ &=&0.\end{array}$

Then

$a=\left(\begin{array}[]{cc}a_{1}&0\\\
a_{3}&a_{4}\end{array}\right)_{q},aa^{\\#}b=\left(\begin{array}[]{cc}b_{1}&0\\\
0&0\end{array}\right)_{q}.$

Here $a_{1}=qaq,a_{3}=q^{\pi}aq,a_{4}=q^{\pi}aq^{\pi}$. Then

$a+aa^{\\#}b=\left(\begin{array}[]{cc}a_{1}+b_{1}&0\\\
a_{3}&a_{4}\end{array}\right)_{q},$

where

$\begin{array}[]{rll}a_{4}&=&(1-aa^{\\#}bb^{\\#})a(1-aa^{\\#}bb^{\\#})\\\
&=&(1-aa^{\\#}bb^{\\#})ab^{\pi}\\\ &=&ab^{\pi})-aa^{\\#}bb^{\\#}ab^{\pi}\\\
&=&ab^{\pi}\in R^{\tiny\\#⃝}.\end{array}$

Set $z=a+aa^{\\#}b$. Then
$zq^{\pi}=[a+aa^{\\#}b][1-aa^{\\#}bb^{\\#}]=ab^{\pi}+aa^{\\#}b-aa^{\\#}baa^{\\#}bb^{\\#}=ab^{\pi}+aa^{\\#}b-aa^{\\#}b(1-a^{\pi})bb^{\\#}=ab^{\pi}\in
R^{\tiny\\#⃝}$. We verify that

$\begin{array}[]{lll}qzq&=&qaq+qaa^{\\#}bq\\\
&=&aa^{\\#}bb^{\\#}a+aa^{\\#}bb^{\\#}aa^{\\#}b\\\
&=&aa^{\\#}(1-b^{\pi})a+aa^{\\#}bb^{\\#}(1-a^{\pi})b\\\
&=&a-(1-a^{\pi})b^{\pi}a+aa^{\\#}b-aa^{\\#}(1-b^{\pi})a^{\pi}b\\\
&=&a-b^{\pi}a+aa^{\\#}b+aa^{\\#}b^{\pi}a^{\pi}b\\\ &=&aa^{\\#}b+bb^{\\#}a\in
R^{\tiny\\#⃝}.\end{array}$

Moreover, we have

$\begin{array}[]{rll}qzq^{\pi}&=&qa(1-q)+qaa^{\\#}b(1-q)=0,\\\
(zq^{\pi})^{\pi}q^{\pi}zq&=&q^{\pi}a_{4}^{\pi}a_{3}\\\
&=&q^{\pi}(ab^{\pi})^{\pi}q^{\pi}aq\\\
&=&q^{\pi}[1-aa^{\\#}b^{\pi}][1-aa^{\\#}bb^{\\#}]abb^{\\#}\\\
&=&q^{\pi}[1-aa^{\\#}b^{\pi}][a-aa^{\\#}bb^{\\#}a]\\\
&=&q^{\pi}[a-[1-a^{\pi}]b^{\pi}a][1-a^{\\#}bb^{\\#}a]\\\
&=&[1-a^{\\#}bb^{\\#}]bb^{\\#}[1-aa^{\\#}bb^{\\#}]a\\\
&=&[1-a^{\\#}bb^{\\#}][bb^{\\#}-bb^{\\#}aa^{\\#}bb^{\\#}]a\\\
&=&[1-a^{\\#}bb^{\\#}][1-b^{\pi}]a^{\pi}bb^{\\#}a\\\
&=&[1-a^{\\#}bb^{\\#}]a^{\pi}[1-b^{\pi}]a\\\ &=&0.\end{array}$

In light of Lemma 2.4, $a+aa^{\\#}b=z\in R^{\tiny\\#⃝}$. Additionally, we have

$\begin{array}[]{lll}(a+aa^{\\#}b)^{\tiny\\#⃝}&=&\left(\begin{array}[]{cc}(a_{1}+b_{1})^{\tiny\\#⃝}&0\\\
v&a_{4}^{\tiny\\#⃝}\end{array}\right)_{q},\end{array}$

where $v=-a_{4}^{\tiny\\#⃝}a_{3}(a_{1}+b_{1})^{\tiny\\#⃝}.$

In light of Lemma 2.4, $a+b\in R^{\tiny\\#⃝}$. By the argument above, we have

$\begin{array}[]{c}aa^{\\#}bb^{\\#}(a+aa^{\\#}b)^{\tiny\\#⃝}(1-aa^{\\#}bb^{\\#})=0,\\\
(a+aa^{\\#}b)^{\tiny\\#⃝}=aa^{\\#}(a+b)^{\tiny\\#⃝}aa^{\\#}.\end{array}$

Therefore we get

$\begin{array}[]{rll}ba^{\pi}(a+b)^{\tiny\\#⃝}b^{\pi}&=&baa^{\\#}(a+b)^{\tiny\\#⃝}b^{\pi}\\\
&=&baa^{\\#}(a+b)^{\tiny\\#⃝}(a^{\pi}+aa^{\\#})b^{\pi}\\\
&=&b[(bb^{\\#}aa^{\\#}][aa^{\\#}(a+b)^{\tiny\\#⃝}aa^{\\#}]aa^{\\#}b^{\pi}\\\
&=&b[aa^{\\#}-b^{\pi}aa^{\\#}][aa^{\\#}(a+b)^{\tiny\\#⃝}aa^{\\#}]aa^{\\#}b^{\pi}\\\
&=&baa^{\\#}bb^{\\#}aa^{\\#}][aa^{\\#}(a+b)^{\tiny\\#⃝}aa^{\\#}]aa^{\\#}b^{\pi}\\\
&=&b[aa^{\\#}bb^{\\#}](a+b)^{\tiny\\#⃝}[aa^{\\#}(1-aa^{\\#}bb^{\\#})]\\\
&=&b[aa^{\\#}bb^{\\#}](a+b)^{\tiny\\#⃝}[1-aa^{\\#}bb^{\\#}]\\\
&=&0,\end{array}$

as asserted.∎

######  Corollary 3.2.

Let $a,b\in R^{EP}$ and $a^{\pi}b,b^{\pi}a\in R^{D}$. If
$aa^{\\#}b=bb^{\\#}a\in R^{\tiny\\#⃝}$ and $\frac{1}{2}\in R$, then $a+b\in
R^{\tiny\\#⃝}$.

###### Proof.

Since $aa^{\\#}b=bb^{\\#}a\in R^{\tiny\\#⃝}$ and $\frac{1}{2}\in R$, we have
$aa^{\\#}b+bb^{\\#}a=2aa^{\\#}b\in R^{\tiny\\#⃝}$. Also we have
$aba^{\pi}=aa^{\\#}aba^{\pi}=abb^{\\#}aa^{\pi}=0$. Similarly,
$bab^{\pi}=bb^{\\#}bab^{\pi}=baa^{\\#}bb^{\pi}=0$. Moreover,
$a^{\pi}b^{\pi}a=a^{\pi}bb^{\\#}a=a^{\pi}aa^{\\#}b=0.$ Likewise,
$b^{\pi}a^{\pi}b=0.$ This completes the proof by Theorem 3.1.∎

######  Corollary 3.3.

Let $a,b\in R^{EP}$. If $ab=ba$ and $a^{*}b=ba^{*}$, then the following are
equivalent:

  1. (1)

$a+b\in R^{\tiny\\#⃝}.$

  2. (2)

$aa^{\\#}b+bb^{\\#}a\in R^{\tiny\\#⃝}$.

###### Proof.

Since $ab=ba$, by [4, Theorem 2.2], $a^{\\#}b=ba^{\\#}$. Hence,
$a^{\pi}b=ba^{\pi}$, and so $a^{\pi}b\in R^{D}$. Likewise, $b^{\pi}a\in
R^{D}$. Moreover, we have $aba^{\pi}=bab^{\pi}=0$.

$(1)\Rightarrow(2)$ Since $ab=ba$ and $a^{*}b=ba^{*}$, we have
$a^{\pi}(a+b)=(a+b)a^{\pi}$. Since $a\in R^{EP}$, we have
$(a^{\pi})^{*}=a^{\pi}$ by [14, Lemma 2.1]. Hence
$a^{\pi}(a+b)^{*}=(a+b)^{*}a^{\pi}$. In light of [3, Theorem 3.2], we have
$a^{\pi}(a+b)^{\tiny\\#⃝}=(a+b)^{\tiny\\#⃝}a^{\pi}$. Accordingly,
$a(a+b)^{\tiny\\#⃝}a^{\pi}=0$. Likewise, $b(a+b)^{\tiny\\#⃝}b^{\pi}=0$.
Therefore $aa^{\\#}b+bb^{\\#}a\in R^{\tiny\\#⃝}$ by Theorem 3.1.

$(2)\Rightarrow(1)$ Clearly, $b^{\pi}a^{\pi}b=b^{\pi}(ba^{\pi})=0$. Likewise,
$b^{\pi}a^{\pi}b=0$. This completes the proof by Theorem 3.1.∎

##  4\. applications

Let $M=\left(\begin{array}[]{cc}A&B\\\
C&D\end{array}\right)\in{\mathbb{C}}^{(m+n)\times(m+n)},$ where
$A\in{\mathbb{C}}^{m\times m},B\in{\mathbb{C}}^{m\times
n},C\in{\mathbb{C}}^{n\times m},D\in{\mathbb{C}}^{n\times n}.$ We now applying
the foregoing to present the conditions under which the block complex $M$ has
core inverse.

######  Lemma 4.1.

Let $A$ and $D$ have core inverses. If $D^{\pi}C=0$, then
$\left(\begin{array}[]{cc}A&0\\\ C&D\end{array}\right)$ has core inverse.

###### Proof.

This is obvious by [13, Theorem 2.9].∎

######  Lemma 4.2.

Let $A,B\in{\mathbb{C}}^{n\times n}$ be EP and $AB=\lambda BA$ for some
$\lambda\in{\mathbb{C}}^{*}$. If $AA^{\\#}B+BB^{\\#}A$ has core inverse, then
$A+B$ has core inverse.

###### Proof.

Since every square complex matrix has Drazin inverse,
$A^{\pi}B,B^{\pi}A\in({\mathbb{C}}^{n\times n})^{D}$. Since $AB=\lambda BA$
for some $\lambda\in{\mathbb{C}}^{*}$, we verify that $ABA^{\pi}=(\lambda
BA)A^{\pi}=0$. Likewise, $BAB^{\pi}=0$. Since $AB=\lambda BA$, by virtue of
[4, Theorem 2.2], $A^{D}B=\lambda^{-1}BA^{D}$ and $AB^{D}=\lambda^{-1}B^{D}A$.
Then $B^{\pi}A=A-B(B^{D}A)=A-\lambda(BA)B^{D}=A-ABB^{D}=AB^{\pi}$, and so
$A^{\pi}B^{\pi}A=A^{\pi}AB^{\pi}=0$. Likewise, $B^{\pi}A^{\pi}B=0$. Therefore
$A+B\in{\mathbb{C}}^{n\times n}$ has core inverse by Theorem 3.1.∎

We are ready to prove:

######  Theorem 4.3.

Let $A,D$ and $BC$ be EP. If $(BC)^{\pi}A=0,C(BC)^{\pi}=0,(CB)^{\pi}D$
$=0,B(CB)^{\pi}=0,AB=\lambda BD$ and $DC=\lambda CA$ for some
$\lambda\in{\mathbb{C}}^{*}$, then $M$ has core inverse.

###### Proof.

Write $M=P+Q$, where

$P=\left(\begin{array}[]{cc}A&0\\\
0&D\end{array}\right),Q=\left(\begin{array}[]{cc}0&B\\\
C&0\end{array}\right).$

Since $A$ and $D$ are EP, so is $P$.

Clearly, we have

$(Q^{2})^{D}=\left(\begin{array}[]{cc}BC&0\\\
0&CB\end{array}\right)^{D}=\left(\begin{array}[]{cc}(BC)^{D}&0\\\
0&(CB)^{D}\end{array}\right).$

Hence,

$Q^{D}=Q(Q^{2})^{D}=\left(\begin{array}[]{cc}0&B(CB)^{D}\\\
C(BC)^{D}&0\end{array}\right).$

Thus,

$Q^{\pi}=\left(\begin{array}[]{cc}(BC)^{\pi}&0\\\
0&(CB)^{\pi}\end{array}\right).$

Hence, $QQ^{D}=Q^{D}Q,Q^{D}=Q^{D}QQ^{D}.$ Moreover, we have

$\begin{array}[]{lll}QQ^{\pi}&=&\left(\begin{array}[]{cc}0&B\\\
C&0\end{array}\right)\left(\begin{array}[]{cc}(BC)^{\pi}&0\\\
0&(CB)^{\pi}\end{array}\right)\\\ &=&\left(\begin{array}[]{cc}0&B(CB)^{\pi}\\\
C(BC)^{\pi}&0\end{array}\right)=0;\end{array}$

whence $Q=Q^{2}Q^{D}$. We infer that $Q$ has group inverse and

$QQ^{\\#}=\left(\begin{array}[]{cc}BC(BC)^{D}&0\\\
0&CB(CB)^{D}\end{array}\right).$

Since $BC$ and $CB$ are EP, we see that $((BC)(BC)^{\\#})^{*}=(BC)(BC)^{\\#}$
and $((CB)(CB)^{\\#})^{*}=(CB)(CB)^{\\#}$. This implies that
$(QQ^{\\#})^{*}=QQ^{\\#}$, and so $Q$ has core inverse by Theorem 1.1.

It is easy to verify that

$PQ=\left(\begin{array}[]{cc}0&AB\\\
DC&0\end{array}\right)=\lambda\left(\begin{array}[]{cc}0&BD\\\
CA&0\end{array}\right)=\lambda QP.$

Moreover, we compute

$\begin{array}[]{rll}PP^{\\#}Q&=&\left(\begin{array}[]{cc}AA^{\\#}&0\\\
0&DD^{\\#}\end{array}\right)\left(\begin{array}[]{cc}0&B\\\
C&0\end{array}\right)\\\ &=&\left(\begin{array}[]{cc}0&0\\\
DD^{\\#}C&0\end{array}\right),\\\
QQ^{\\#}P&=&\left(\begin{array}[]{cc}BC(BC)^{\\#}&0\\\
0&CB(CB)^{\\#}\end{array}\right)\left(\begin{array}[]{cc}A&0\\\
0&D\end{array}\right)\\\ &=&\left(\begin{array}[]{cc}BC(BC)^{\\#}A&0\\\
0&CB(CB)^{\\#}D\end{array}\right).\end{array}$

Thus,

$\begin{array}[]{rll}PP^{\\#}Q+QQ^{\\#}P&=&\left(\begin{array}[]{cc}BC(BC)^{\\#}A&0\\\
DD^{\\#}C&CB(CB)^{\\#}D\end{array}\right)\\\
&=&\left(\begin{array}[]{cc}(I_{m}-(BC)^{\pi})A&0\\\
DD^{\\#}C&(I_{n}-(CB)^{\pi})D\end{array}\right)\\\
&=&\left(\begin{array}[]{cc}A&0\\\ DD^{\\#}C&D\end{array}\right).\end{array}$

Clearly, $D^{\pi}(DD^{\\#}C)=0$. Since $A$ and $D$ have core inverses, by
virtue of Lemma 4.1, $\left(\begin{array}[]{cc}A&0\\\
DD^{\\#}C&D\end{array}\right)$ has core inverse. Therefore
$PP^{\\#}Q+QQ^{\\#}P$ has core inverse. By using Lemma 4.2, $M=P+Q$ has core
inverse, hence the result. ∎

######  Corollary 4.4.

Let $A,D$ be EP and $BC$ be invertible. If $AB=\lambda BD$ and $DC=\lambda CA$
for some $\lambda\in{\mathbb{C}}^{*}$, then $M$ has core inverse.

###### Proof.

Since $BC$ be invertible, then so is $CB$. Hence, $(BC)^{\pi}=0$ and
$(CB)^{\pi}=0$. This completes the proof by Theorem 4.3.∎

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