aainstitutetext: Blackett Laboratory Imperial College London London, UK, SW7
2AZbbinstitutetext: Department of Applied Mathematics and Theoretical Physics
University of Cambridge Wilberforce Road, Cambridge, UK, CB3
OWAccinstitutetext: Department of Mathematics King’s College London The
Strand, London, UK, WC2R 2LS

# On disc-with-hole and disc-with-handle partition functions in bosonic string
theory

Brett Oertel a,c Omar Shahpo<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

Higher genus partition functions of string world sheets with boundaries are
relevant, e.g. for computation of quantum corrections to Wilson loop
expectation values. As a preparation for a possible study of strings in curved
space like AdS here we consider examples of genus one partition functions of
string world-sheets ending on a circle in the bosonic string theory in flat
space. We begin with the partition function for annular topology, writing it
as an integral over the modulus of the annulus. In the process, we compute the
determinant of the Laplacian on the annulus for Dirichlet-Dirichlet and
Dirichlet-Neumann boundary conditions. We then consider the case of the disc-
with-handle topology using the gluing method. We first write the partition
function using a Schottky parameterisation of the moduli space and then as an
integral over the period matrix.

Presented here is joint work done by the authors towards theses submitted in
partial fulfillment of the requirements for the MSci degree in Physics with
Theoretical Physics of Imperial College London.

##  1 Introduction

The correspondence between Wilson loops (WLs) in large $N$ $\mathcal{N}=4$ SYM
gauge theory and partition functions of $AdS_{5}\times S^{5}$ string world-
sheets ending on a fixed curve has been heavily studied since
PhysRevLett.80.4859 ; PhysRevD.60.125006 . A systematic discussion of leading
string inverse tension quantum corrections on disc topology appeared in
Drukker:2000ep  (following Kallosh:1998ji ; Forste:1999qn ). Some recent
papers discussing matching between string and strong-coupling gauge theory
results for circular WL include Medina-Rincon:2018wjs ; Giombi:2020mhz .

The next step is matching non-planar $1/N$ corrections to Wilson loops with
higher genus string partition functions. The former are known to all orders
for the circular WL case Drukker:2000rr  and a discussion of possible matching
was initiated in Giombi:2020mhz . Before trying to address the $AdS_{5}\times
S^{5}$ string case it is useful to review what can be said explicitly about
the corresponding corrections in the bosonic string theory in flat target
space. The $1/N$ correction to the circular WL expectation value111It is
absent in adjoint $SU(N)$ gauge theory but may be interpreted as due to a
fermion loop in fundamental representation in more general cases involving
orientifolds, see, e.g., Giombi:2020kvo . corresponds to the string partition
function on the annulus, i.e. to adding a hole (with Neumann b.c.) to the disc
(with Dirichlet b.c.) and integrating over the size of the hole. Similarly,
the $1/N^{2}$ correction to the WL corresponds to the string partition
function on a surface with topology of a disc with a handle.

We begin in Section 2 by considering the case of the partition function of the
world-sheet of annular topology with Dirichlet condition on one circle and
Neumann on the other. We first compute the determinant of the Laplacian on the
annulus for the case of Dirichlet-Dirichlet boundary conditions, which is a
simple modification of the determinant of the Laplacian on the torus, and then
for the case of the Dirichlet-Neumann conditions. The result in the latter
case agrees with that found in Graham:2018ibr  by a different method.

In Section 3 we tackle the case of the partition function on the disc with a
handle. This is done following the approach of BLAU1988285 , by a gluing a
disc with a handle to itself, obtaining a double torus, and then
parameterising the moduli of the disc-with-handle using the moduli of the
double torus. We compute the corresponding partition function using two
different parametrisations of the moduli space: first using the Schottky
parameters and then using the period matrix.

##  2 The Annulus

We work with the Polyakov action as given in Alvarez:1982zi :

$I^{\prime}[X,g]=\frac{1}{2}T\int_{\Sigma}d^{2}\xi\sqrt{g}g^{ab}\delta_{\mu\nu}\partial_{a}X^{\mu}\partial_{b}X^{\nu}+\phi_{0}\chi+\mu^{2}\int_{\Sigma}d^{2}\xi\sqrt{g}+I_{\mathrm{b}}.$
(1)

Here $\Sigma$ is a two-dimensional compact Riemannian manifold (in this
section with the topology of an annulus), the $X^{\mu}$ represent 26 real
scalar fields on $\Sigma$ which determine its embedding into
$\mathbb{R}^{26}$, $g$ is the Riemannian metric on $\Sigma$ and $\chi$ is the
Euler number of $\Sigma$. $T=\frac{1}{2\pi\alpha^{\prime}}$ is string tension.

We shall assume that the constants $\mu$ and the counter terms $I_{b}$ are
chosen so that the conformal anomaly cancels in 26 dimensions Alvarez:1982zi ,
and we drop these terms. The Euler number term provides the weighting of the
string partition function with string coupling that we shall suppress. Thus in
what follows we shall use

$I[X,g]=\frac{1}{2}T\int_{\Sigma}d^{2}\xi\sqrt{g}g^{ab}\delta_{\mu\nu}\partial_{a}X^{\mu}\partial_{b}X^{\nu}.$
(2)

The string world-sheet partition function is Alvarez:1982zi

$W=\frac{1}{V_{\mathrm{GC}\times\mathrm{Weyl}}}\int DgDX\ e^{-I[X,g]}$ (3)

The integral is over all embeddings $X^{\mu}$ of $\Sigma$ and all intrinsic
metrics $g$ on $\Sigma$ (both must satisfy boundary conditions to be
discussed). We divide by the volume of the classical gauge group
$V_{\mathrm{GC}\times\mathrm{Weyl}}$ where $\mathrm{GC}\times\mathrm{Weyl}$ is
the direct product of the general coordinate transformation and Weyl groups.

###  2.1 The parametrisation and boundary conditions

To parametrise the annulus we use a strip in the complex plane:

$\\{\xi^{i}\\},\hskip 14.22636pt\xi^{1},\xi^{2}\in\mathbb{R},\hskip
14.22636pt0\leq\xi^{2}\leq 1.$ (4)

This parametrises the annulus in the complex plane given by

$q\leq|z|\leq 1,\hskip 14.22636pt0<q<1\hskip 14.22636ptz\in\mathbb{C}$ (5)

using the map

$z=\exp(2\pi i(\xi^{1}+i\tau\xi^{2})),\hskip
14.22636pt\tau\equiv-\frac{1}{2\pi}\ln q.$ (6)

Under this map, which is periodic with period 1 in the $\xi^{1}$ direction,
the top boundary ($\xi^{2}=1$) of the strip is mapped to the inner boundary of
the annulus, while the bottom boundary ($\xi^{2}=0$) is mapped to the outer
boundary of the annulus. Since we are interested in the path integral over the
annulus as a one-loop correction to a phenomenological model of the Wilson
loop, we will use ‘modified Dirichlet’ Alvarez:1982zi  conditions for the
embedding functions $X^{\mu}$ on the outer boundary. After gauge fixing, these
become Dirichlet boundary conditions. On the other hand, the inner boundary is
not mapped to a fixed curve, and we simply choose Neumann boundary conditions
on it. Next, the $X^{\mu}$ and $g$ are obviously required to be periodic in
the $\xi^{1}$ direction:

$\displaystyle X^{\mu}(\xi^{1}+1,\xi^{2})=X^{\mu}(\xi^{1},\xi^{2}),$ (7)
$\displaystyle g_{ab}(\xi^{1}+1,\xi^{2})=g_{ab}(\xi^{1},\xi^{2}).$

On the upper boundary and lower boundaries $g_{ab}$ must satisfy
Alvarez:1982zi

$g_{ab}=\partial_{a}X^{\mu}\partial_{b}X_{\mu},$ (8)

The boundary conditions needed to compute the determinant of the Laplacian are
discussed below.

###  2.2 The path integral

To perform the $X^{\mu}$ integration one redefines the integration variable as
an expansion about the classical solution:

$X^{\mu}=X^{\mu}_{cl}+y^{\mu}.$ (9)

Here $X^{\mu}_{cl}$ is the solution to the classical equation of motion
obtained by varying the action. Note that this solution depends on $g$, thus
the answer will be a function of the metric and remains inside the integral
when we integrate over the metric. Then, the $X^{\mu}$ integration gives222We
set the string tension $T$ to 1 below as it can be scaled out of the partition
function.

$\displaystyle W_{g}$ $\displaystyle\equiv\int DX\
e^{-I[X,g]}=e^{-I[X_{cl},g]}\left(\mathrm{det}(\Delta_{g})\right)^{-13},$ (10)

where $\Delta_{g}$ is the ordinary Laplacian differential operator, i.e.

$\Delta_{g}\equiv-\frac{1}{\sqrt{g}}\partial_{a}\sqrt{g}g^{ab}\partial_{b}.$
(11)

The determinant of $\Delta_{g}$ is taken with Dirichlet boundary conditions.
Note that $X_{cl}$ depends on $g$ only through the moduli variables since $I$
is invariant under diffeomorphisms and conformal transformations. Our final
result for the partition function will be in the form of an integral over
these moduli. To perform the integration over $g$ we first note that any
metric on the annulus can be taken via a combined general coordinate and Weyl
transformation (respecting the above boundary conditions) to a metric of the
form (see, e.g., Rodrigues:1986us )

$g_{ab}=(d\xi^{1})^{2}+(\tau d\xi^{2})^{2}.$ (12)

Here $\tau=-\frac{1}{2\pi}\ln q$ is the Teichmuller parameter. The integral
over metrics then reduces (in $D=26$) to the integral over $\tau$

$\displaystyle W=$
$\displaystyle\frac{1}{2}\int_{0}^{\infty}d\tau\frac{1}{\sqrt{\tau}}\left(\frac{\det^{2}<T,\psi>}{\det<\psi,\psi>}\right)^{\frac{1}{2}}\left(\mathrm{det}^{\prime}P^{\dagger}_{1}P_{1}\right)^{\frac{1}{2}}$
(13)
$\displaystyle\times\left(\mathrm{det}(\Delta_{g})\right)^{-13}e^{-I[X_{cl},g]}.$

where we have used the following definitions:

$\displaystyle(T_{ab})$ $\displaystyle\equiv(\frac{\partial
g_{ab}}{\partial\tau}-\frac{1}{2}g_{ab}g^{cd}\frac{\partial
g_{cd}}{\partial\tau})$ (14) $\displaystyle(P_{1}\delta v)_{ab}$
$\displaystyle\equiv\nabla_{a}\delta v_{b}+\nabla_{b}\delta
v_{a}-\nabla_{c}\delta v^{c}g_{ab}$ $\displaystyle(P_{1}^{\dagger}\alpha)_{n}$
$\displaystyle\equiv-2\nabla^{m}\alpha_{mn}\ .$

$P_{1}$ acts on two-dimensional vectors (diffeomorphism generators) $\delta v$
and $P_{1}^{\dagger}$ acts on rank-2 traceless tensors $\alpha$. The norm on
rank-2 traceless tensors is defined by:

$<A,B>_{g}=\int_{\Sigma}d\sigma^{1}d\sigma^{2}\sqrt{g(\sigma)}\
g^{mi}(\sigma)g^{nj}(\sigma)\ A_{mn}(\sigma)B_{ij}(\sigma)$ (15)

and $\psi$ is any element in $\mathrm{Ker}(P_{1}^{\dagger})$. The Riemann-Roch
theorem implies Alvarez:1982zi :

$\mathrm{dim}\ \mathrm{Ker}(P_{1})-\mathrm{dim}\
\mathrm{Ker}(P_{1}^{\dagger})=3\chi(\Sigma)=0.$ (16)

From this we see that the conformal Killing vectors (CKV), i.e. vectors in in
$\mathrm{Ker}(P_{1})$, on the annulus form a one-dimensional real vector space
(corresponding to rigid translation in the $\xi^{1}$ direction). Also note
that the mapping class group is isomorphic to $\mathbb{Z}_{2}$ and is given by
the coordinate transformation Rodrigues:1986us

$\xi^{1}\rightarrow-\xi^{1},\hskip 14.22636pt\xi^{2}\rightarrow 1-\xi^{2}.$
(17)

The volume of the mapping class group is therefore $2$, and this is where the
factor of $\frac{1}{2}$ in front of the path integral comes from.

###  2.3 The Determinant and the Weil-Petersson Measure

We can calculate the Weil-Petersson measure directly. We have

$(T_{ab})\equiv(\frac{\partial
g_{ab}}{\partial\tau}-\frac{1}{2}g_{ab}g^{cd}\frac{\partial
g_{cd}}{\partial\tau})=\frac{1}{\tau}\begin{pmatrix}-1&0\\\
0&\tau^{2}\end{pmatrix}.$ (18)

The boundary conditions and the symmetric and traceless requirement on
matrices in $\mathrm{Ker}(P_{1}^{\dagger}$) mean we must choose

$\psi=\begin{pmatrix}1&0\\\ 0&-\tau^{2}\end{pmatrix}.$ (19)

A direct calculation then gives

$\left(\frac{\det^{2}<T,\psi>}{\det<\psi,\psi>}\right)^{\frac{1}{2}}=\sqrt{\frac{2}{\tau}}.$
(20)

All that is left is to calculate the determinants of the Laplacians. Firstly,

$\displaystyle(P_{1}^{\dagger}P_{1}\delta v)_{a}$
$\displaystyle=-2\begin{pmatrix}\nabla_{g}^{2}&0\\\
0&\nabla_{g}^{2}\end{pmatrix}\begin{pmatrix}\delta v_{1}\\\ \delta
v_{2}\end{pmatrix}.$ (21)

We have chosen boundary conditions such that changes in the metric do not
change the normal direction at the boundary. This imposes Neumann boundary
conditions on $\delta v_{1}$ and Dirichlet on $\delta v_{2}$ Rodrigues:1986us
. Hence,

$\mathrm{det}^{\prime}(P^{\dagger}_{1}P_{1})=\mathrm{det}^{\prime}_{N}(-2\nabla_{g}^{2})\
\mathrm{det}^{\prime}_{D}(-2\nabla_{g}^{2}).$ (22)

Furthermore, the factor $(\mathrm{det}^{\prime}(\Delta_{g}))^{-13}$ which
arises due to the $X$-integration must be taken with Dirichlet boundary
conditions at the lower edge ($\xi^{2}=0$) and Neumann at the upper edge
($\xi^{2}=1$). Now we calculate each determinant individually. First, the
operator in the determinant

$\mathrm{det}^{\prime}_{N}(-2\nabla_{g}^{2})=\mathrm{det}^{\prime}_{N}(-2(\partial_{1}^{2}+\frac{1}{\tau^{2}}\partial_{2}^{2}))$
(23)

has eigenvectors

$\cos(2\pi m\xi^{1})\cos(\pi n\xi^{2}),\hskip 14.22636ptm,n\in\mathbb{Z}\hskip
14.22636ptm,n\geq 0,\hskip 14.22636pt(m,n)\neq(0,0)$ (24)

and the corresponding eigenvalues

$\displaystyle\lambda_{mn}$ $\displaystyle=\frac{8\pi^{2}}{\tau^{\prime
2}}|\tau^{\prime\prime}m+n|^{2},\hskip 14.22636pt\tau^{\prime}\equiv
2\tau,\tau^{\prime\prime}\equiv i\tau^{\prime}.$ (25)

These are identical to the eigenvalues encountered in the one-loop closed
string (torus) calculation in the case that the torus Teichmuller parameter is
purely imaginary. Such a determinant is calculated, for example, in
Nakahara:2003nw  using the Eisenstein series. This method can be immediately
adapted to obtain:

$\mathrm{det}^{\prime}_{N}(-2\nabla_{g}^{2})=\frac{1}{\sqrt{2}}2\tau|\eta(2i\tau)|^{2}=\sqrt{2}\tau|\eta(2i\tau)|^{2}.$
(26)

Next, the operator in the determinant

$\mathrm{det}^{\prime}_{D}(-2\nabla_{g}^{2})=\mathrm{det}^{\prime}_{D}(-2(\partial_{1}^{2}+\frac{1}{\tau^{2}}\partial_{2}^{2}))$
(27)

has eigenvectors

$\cos(2\pi m\xi^{1})\sin(\pi n\xi^{2}),\hskip
14.22636ptm,n\in\mathbb{Z},\hskip 14.22636ptn>0,\hskip 14.22636ptm\geq 0,$
(28)

and identical eigenvalues

$\displaystyle\lambda_{mn}$ $\displaystyle=\frac{8\pi^{2}}{\tau^{\prime
2}}|\tau^{\prime\prime}m+n|^{2},\hskip 14.22636pt\tau^{\prime\prime}\equiv
i2\tau.$ (29)

The only difference is that we allow $n=0$ so long as $m\neq 0$ in the Neumann
case. Hence we obtain

$\mathrm{det}^{\prime}_{N}(-2\nabla_{g}^{2})=\mathrm{det}^{\prime}_{D}(-2\nabla_{g}^{2})\prod_{m\neq
0}(2\pi^{2}m^{2}).$ (30)

We may regularize this factor as follows:

$\prod_{m\geq 0}(2\pi^{2}m^{2})=\exp(-\zeta^{\prime}_{\Delta}(0)),$ (31)

where

$\zeta_{\Delta}(s)\equiv\sum_{m\geq 1}(2\pi
m)^{-2s}=\frac{1}{(2\pi)^{-2s}}\zeta_{R}(2s).$ (32)

Then

$-\zeta^{\prime}_{\Delta}(0)=(2\zeta^{\prime}_{R}(0)-2\log(2\pi)\zeta_{R}(0))=0,$
(33)

and hence the factor regularizes to $1$.

Lastly, we need to find the determinant of the Laplacian on the annulus
$\mathrm{det}(\Delta_{g})$ where the inner boundary has Neumann boundary
conditions, and the outer boundary has Dirichlet boundary conditions. We have

$\mathrm{det}(\Delta_{g})=\mathrm{det}(-(\partial_{1}^{2}+\frac{1}{\tau^{2}}\partial_{2}^{2})).$
(34)

This has eigenvectors

$\cos(2\pi m\xi^{1})\sin(\pi(n+\frac{1}{2})\xi^{2}),\hskip
14.22636ptm,n\in\mathbb{Z},\hskip 14.22636ptm,n\geq 0$ (35)

with eigenvalues

$\displaystyle\lambda_{mn}$ $\displaystyle=\Big{(}(2\pi
m)^{2}+\frac{1}{\tau^{2}}(\pi(n+\frac{1}{2})\Big{)}^{2}.$ (36)

To calculate this determinant we will need to use a different method, relying
on contour integrals. Our method is similar to a spectral method used in
Kirsten:1999qjn . Consider the initial value problem

$\displaystyle\phi_{m,\mu}^{\prime\prime}(\xi^{2})=-\tau^{2}(\mu^{2}-4\pi^{2}m^{2})\phi_{m,\mu}(\xi^{2}),\hskip
14.22636pt\phi_{m,\mu}(0)=0,\hskip 14.22636pt\phi_{m,\mu}^{\prime}(0)=1.$ (37)

This has the unique solution

$\phi_{m,\mu}(\xi^{2})=\frac{1}{\tau\sqrt{\mu^{2}-4\pi^{2}m^{2}}}\sin((\tau\sqrt{\mu^{2}-4\pi^{2}m^{2}})\xi^{2}).$
(38)

In particular, for a given $m$ we will view $\phi^{\prime}_{m,\mu}(1)$ as a
meromorphic function of $\mu$:

$\phi^{\prime}_{m,\mu}(1)=\cos(\tau\sqrt{\mu^{2}-4\pi^{2}m^{2}}).$ (39)

This function has zeros exactly when

$\mu^{2}=(4\pi^{2}m^{2}+\frac{1}{\tau^{2}}(n+\frac{1}{2})^{2})\equiv\lambda_{mn},$
(40)

i.e. exactly when $\mu^{2}$ is an eigenvalue of $\Delta_{g}$. Hence, we will
consider the function

$\frac{d}{d\mu}\log(\phi^{\prime}_{m,\mu}(1))=\frac{\frac{d}{d\mu}\phi^{\prime}_{m,\mu}(1)}{\phi^{\prime}_{m,\mu}(1)}$
(41)

which clearly has a simple pole at each $\mu^{2}=\lambda_{mn}$ if we consider
the expansion of $\phi^{\prime}_{m,\mu}(1)$. Hence, we may write the zeta
function as

$\zeta_{\Delta}(s)=\frac{1}{2\pi
i}\int_{\Lambda}d\mu\mu^{-2s}\frac{d}{d\mu}\log(\phi^{\prime}_{0,\mu}(1))+\sum_{m=1}\frac{1}{2\pi
i}\int_{\Lambda}d\mu\mu^{-2s}\frac{d}{d\mu}\log(\phi^{\prime}_{m,\mu}(1))$
(42)

where $\Lambda$ is a contour enclosing all the positive simple poles, which
all lie on the real axis. Now we choose the contour $\Lambda$ to be
parametrised as

$\Lambda(k)=\epsilon+ik,\hskip 14.22636pt\epsilon,k\in\mathbb{R}$ (43)

where $\epsilon$ is a small positive number and $k$ runs from $-\infty$ to
$\infty$. Substituting this into the integral and taking the limit of
$\epsilon$ going to zero we obtain

$\displaystyle\zeta_{\Delta}(s)=$ $\displaystyle\frac{\sin(\pi
s)}{\pi}\int_{0}^{\infty}dkk^{-2s}\frac{d}{dk}\log(\phi^{\prime}_{0,ik}(1))$
(44) $\displaystyle+\sum_{m=1}m^{-2s}\frac{\sin(\pi
s)}{\pi}\int_{0}^{\infty}dkk^{-2s}\frac{d}{dk}\log(\phi^{\prime}_{m,imk}(1)),$

where

$\displaystyle\phi^{\prime}_{0,ik}(1)=\cos(\tau ik)=\cosh(\tau k),$ (45)
$\displaystyle\phi^{\prime}_{m,imk}(1)=\cos(\tau\sqrt{-m^{2}k^{2}-4\pi^{2}m^{2}})=\cosh(\tau\sqrt{m^{2}k^{2}+4\pi^{2}m^{2}}).$

Now we will regularise these integrals separately.

####  2.3.1 The m=0 term

First, note that

$\frac{d}{dk}\log(\phi^{\prime}_{0,ik}(1))=\frac{d}{dk}\log(\cosh(\tau
k))=\tau\tanh(\tau k).$ (46)

This has the following asymptotic expansion:

$\tau\tanh(\tau k)=\tau+\mathcal{O}(\exp(-2\tau k)).$ (47)

Hence, we will write

$\frac{\sin(\pi s)}{\pi}\int_{0}^{\infty}dkk^{-2s}(\tau\tanh(\tau
k)-\tau+\tau)=Z_{0}(s)+A_{0}(s),$ (48)

where

$\displaystyle Z_{0}(s)\equiv\frac{\sin(\pi
s)}{\pi}\left(\int_{0}^{1}dkk^{-2s}(\tau\tanh(\tau
k))+\int_{1}^{\infty}dkk^{-2s}(\tau\tanh(\tau k)-\tau)\right),$ (49)
$\displaystyle A_{0}(s)\equiv\frac{\sin(\pi
s)}{\pi}\int_{1}^{\infty}dkk^{-2s}\tau.$

$Z_{0}(s)$ is not divergent at $s=0$ and we find

$\displaystyle Z^{\prime}_{0}(0)$
$\displaystyle=\int_{0}^{1}dk\frac{d}{dk}(\log(\cosh(\tau
k)))+\int_{1}^{\infty}dk\frac{d}{dk}(\log(\cosh(\tau k))-\tau k)$ (50)
$\displaystyle=-\log(2)+\tau.$

On the other hand, we may regularise $A_{0}(s)$ as

$\displaystyle A_{0}(s)=\frac{\tau\sin(\pi s)}{\pi(2s-1)}\implies{}$
$\displaystyle A_{0}^{\prime}(0)=-\tau.$ (51)

In total, for the $m=0$ term we obtain

$\left.\frac{\sin(\pi
s)}{\pi}\int_{0}^{\infty}dkk^{-2s}\frac{d}{dk}\log(\phi^{\prime}_{0,ik}(1))\right|_{s=0}=-\log(2).$
(52)

####  2.3.2 The m > 0 terms

Again, we have

$\displaystyle\frac{d}{dk}\log(\phi^{\prime}_{m,ik}(1)))=\frac{d}{dk}\log(\cosh(\tau
m\sqrt{k^{2}+4\pi^{2}})),$ (53)

and

$\displaystyle\sum_{m=1}m^{-2s}\frac{\sin(\pi
s)}{\pi}\int_{0}^{\infty}dkk^{-2s}\frac{d}{dk}\log(\phi^{\prime}_{m,imk}(1))=Z_{m}(s)+A_{m}(s).$
(54)

We have defined

$\displaystyle Z_{m}(s)\equiv$
$\displaystyle\sum_{m=1}\left(m^{-2s}\frac{\sin(\pi
s)}{\pi}\int_{0}^{1}dkk^{-2s}\frac{d}{dk}(\log(\cosh(\tau
m\sqrt{k^{2}+4\pi^{2}}))\right.$ (55) $\displaystyle+m^{-2s}\frac{\sin(\pi
s)}{\pi}\int_{1}^{\infty}dkk^{-2s}\frac{d}{dk}(\log(\cosh(\tau
m\sqrt{k^{2}+4\pi^{2}})-m\tau k)$ $\displaystyle-\left.m^{-2s}\frac{\sin(\pi
s)}{\pi}\int_{1}^{\infty}(1-2\pi)\tau m\exp(-k+1)\right),$ $\displaystyle
A_{m}(s)\equiv$ $\displaystyle\frac{\tau\sin(\pi
s)}{\pi}(\frac{1}{2s-1}+(1-2\pi))\zeta_{R}(2s-1).$

The terms that we have added and subtracted are specifically chosen so that
$Z_{m}(s)$ is well defined in a neighbourhood of $s=0$. Hence we simply
differentiate and set $s=0$, obtaining

$\displaystyle Z_{m}^{\prime}(0)=-\sum_{m=1}\log(1+\exp(-4\pi\tau m)).$ (56)

Lastly, the divergent piece $A_{m}(s)$ can now be regularized in the obvious
manner, by analytically continuing the Riemann zeta function:

$A^{\prime}_{m}(0)=\frac{\pi\tau}{6}.$ (57)

Hence, we may combine all results to obtain

$\displaystyle\zeta_{\Delta}^{\prime}(0)=-\log(2)+\frac{\pi\tau}{6}-\sum_{m=1}\log(1+\exp(-4\pi\tau
m)),$ (58)
$\displaystyle\textrm{det}(\Delta_{g})=2\exp(\frac{-\pi\tau}{6})\prod_{m=1}(1+\exp(-4\pi\tau
m)).$

The final expression for the partition function then reads

$W=\int_{0}^{\infty}d\tau\
|\eta(2i\tau)|^{2}\left(\frac{1}{2}\exp(\frac{\pi\tau}{6})\prod_{m=1}(1+\exp(-4\pi\tau
m))^{-1}\right)^{13}e^{-\pi(1-q^{2})}.$ (59)

Note that since we parametrised the annulus by a rectangular strip, we can
check this result by looking for calculations of the determinant of the
Laplacian on the rectangle with appropriate boundary conditions. Indeed, this
is done using a different method in Brink:1973rq , and we find agreement
between their result and ours. Note also that here we have used the fact that
the classical action $I[X_{cl},g]$ is simply the minimal area of the annulus
as a function of the modulus parameter:

$I[X_{cl},g]=\pi(1-q^{2}).$ (60)

The integral over $\tau$ is formal as it is divergent for large $\tau$
corresponding to small $q=e^{-2\pi\tau}$, i.e. the integral is divergent when
the inner radius of the annulus goes to zero.

##  3 The disc-with-handle

Let us now compute the partition function for the case where $\Sigma$ is
topologically a disc-with-handle, with boundary again given by a fixed circle
in spacetime. Here again the partition function is given by

$W=\frac{1}{V_{\mathrm{GC}\times\mathrm{Weyl}}}\int
DgDX\exp\left(-\frac{1}{2}\int_{\Sigma}d^{2}\xi\sqrt{g}g^{ab}\delta_{\mu\nu}\partial_{a}X^{\mu}\partial_{b}X^{\nu}\right)\
.$ (61)

There are no CKV on the disc-with-handle so that we get in general
Alvarez:1982zi :

$W=\int_{\mathscr{F}}d\mu_{\mathrm{WP}}\sqrt{\det(P_{1}^{\dagger}P_{1})}\left(\det(\Delta)_{g}\right)^{-13}\exp(-I(X_{cl},g)).$
(62)

Here $d\mu_{\mathrm{WP}}$ is the Weil-Petersson measure on the moduli space of
$\Sigma$ and the integration is over a fundamental region $\mathscr{F}$.
$P_{1}^{\dagger}P_{1}$ and $\Delta$ are the vector and scalar Laplacians on
$\Sigma$ respectively, $g$ is a choice of gauge-fixed metric, and
$I_{\mathrm{cl}}(X,g)$ is the classical action. The boundary conditions are
the same as the outer boundary in the annulus case: $\det(\Delta)_{g}$ is
calculated using Dirichlet boundary conditions and
$\det(P_{1}^{\dagger}P_{1})$ using mixed boundary conditions.

###  3.1 Schottky doubles

To evaluate $W$ we need to parametrise the moduli space of $\Sigma$ and
evaluate determinants on $\Sigma$. In fact, it will be easier to work with the
Schottky double of $\Sigma$, which will be a surface without boundary denoted
$\Sigma^{2}$; we will closely follow the approach outlined in BLAU1988285 .
$\Sigma^{2}$ is topologically constructed by attaching two copies of $\Sigma$,
denoted $\Sigma$ and $\Sigma^{*}$, with opposite orientation along the
boundaries. Since $\Sigma$ is a disc-with-handle, $\Sigma^{2}$ is therefore a
double-torus. The key point is that the moduli space and determinants on
$\Sigma$ can then be related to those on $\Sigma^{2}$ in a manner we will now
describe. First, recall that the moduli and Teichmuller spaces on $\Sigma$ can
be defined as BLAU1988285

$\mathscr{M}(\Sigma)\equiv\frac{C(\Sigma)}{\mathrm{Diff}(\Sigma)},\hskip
14.22636pt\mathscr{T}(\Sigma)\equiv\frac{C(\Sigma)}{\mathrm{Diff_{0}}(\Sigma)},$
(63)

where $\mathrm{Diff(\Sigma)}$ is the group of orientation preserving
diffeomorphisms on $\Sigma$, $\mathrm{Diff}_{0}(\Sigma)$ is the subgroup
connected to the identity, and $C(\Sigma)$ is the set of almost complex
structures on $\Sigma$ (which is in bijection with the set of classes of
metrics where metrics in a given class differ only by a Weyl transformation).
Now, there exists a natural choice of involution $I$ on the constructed
topological space $\Sigma^{2}$ such that

$I^{2}=1,\hskip 14.22636ptI(\Sigma)=\Sigma^{*},\hskip
14.22636ptI(\Sigma^{*})=\Sigma,\hskip
14.22636ptI|_{\partial\Sigma}=I|_{\partial\Sigma^{*}}=1.$ (64)

Furthermore, if the surface $\Sigma$ comes with a choice of almost complex
structure $J$, then we may define an almost complex structure $\tilde{J}$ on
$\Sigma^{2}$ as BLAU1988285

$\tilde{J}_{p}=J_{p}\ \textrm{if}\ p\in\Sigma,\hskip
14.22636pt\tilde{J}_{p}=-(I_{*}\circ J_{p}\circ I_{*})_{p}\ \textrm{if}\
p\in\Sigma^{*}.$ (65)

Then, $I$ is in addition anti-conformal. On the other hand, an arbitrary
almost complex structure $J$ on $\Sigma^{2}$ may or may not make $\Sigma^{2}$
the double of a surface; it will be considered the double of the surface
$\Sigma$ if $J$ is such that the corresponding involution $I$ is anti-
conformal. In particular, in BLAU1988285  the following important facts are
noted: any metric on $\Sigma$ is equivalent in $\mathscr{T}(\Sigma)$ to a
metric which extends smoothly to a metric on $\Sigma^{2}$ which is invariant
under $I$, and any metric on $\Sigma^{2}$ invariant under $I$ (up to a
diffeomorphism and Weyl transformation) is equivalent in
$\mathscr{T}(\Sigma^{2})$ to a metric exactly invariant under $I$. Hence,
$\mathscr{T}(\Sigma)$ is a natural subset of $\mathscr{T}(\Sigma^{2})$. In
fact, we will therefore be able to write the partition function over $\Sigma$
in terms of objects defined on $\Sigma^{2}$. In particular, the determinants
of interest on $\Sigma$ can be expressed in terms of determinants on
$\Sigma^{2}$ using the following formulas BLAU1988285 :

$\displaystyle\det(\Delta)_{g,\Sigma}=\left(\frac{\det^{\prime}(\Delta)_{g,\Sigma^{2}}}{\int_{\Sigma^{2}}\sqrt{g}}\right)^{\frac{1}{2}}(R_{\Sigma^{2},I}(J))^{\frac{1}{2}},$
(66)
$\displaystyle\det(P_{1}^{\dagger}P_{1})_{g,\Sigma}=(\det(P_{1}^{\dagger}P_{1})_{g,\Sigma^{2}})^{\frac{1}{2}}.$

Here $R_{\Sigma^{2},I}(J)$ is BLAU1988285

$R_{\Sigma^{2},I}(J)=\det[(1+\Gamma)\textrm{Im}\ \tau+(1-\Gamma)(\textrm{Im}\
\tau)^{-1}]$ (67)

where $\tau$ and $\Gamma$ are defined as follows. First, we define a canonical
homology basis of $A$ cycles and $B$ cycles on our Schottky double
$\Sigma^{2}$. In general, the involution $I$ will take $A$ cycles to $A$
cycles and $B$ cycles to $B$ cycles. Then we define $\Gamma$ as the matrix
corresponding to how the involution pushes forward A cycles to other A cycles:

$I_{*}A_{i}=\Gamma_{ij}A_{j}.$ (68)

In addition, if we let $\\{\omega_{i}\\}$ be a basis of holomorphic
differentials on the Schottky double, normalised to the A-cycles in the sense
that

$\int_{A_{i}}\omega_{j}=\delta_{ij},$ (69)

then we define the ‘period matrix’ $\tau$ as

$\int_{B_{i}}\omega_{j}=\tau_{ij}.$ (70)

Now we turn to the Weil-Petersson measure $d\mu_{\mathrm{WP}}$ on $\Sigma$.
Let $\\{\mu_{i}\\}$ and $\\{S_{i}\\}$ be bases for the spaces of Beltrami and
quadratic differentials on $\Sigma^{2}$ which are even under the involution
$I$. Each Beltrami differential $\mu_{i}$ corresponds to a tangent vector in
the Teichmuller space which we write as $\frac{d}{dm_{i}}$. Then the Weil-
Petersson measure on $\Sigma$ can be written as BLAU1988285

$d\mu_{\mathrm{WP},\Sigma}=\frac{\det<S_{i},\mu_{j}>}{(\det<S_{i},S_{j}>)^{\frac{1}{2}}}\prod_{i=1}^{3}dm_{i}.$
(71)

This measure is on a subset of the Teichmuller space of $\Sigma^{2}$ which is
isomorphic to the Teichmuller space of $\Sigma$ (in particular, it is on the
subset of $\mathscr{T}(\Sigma^{2})$ which is invariant under $I$). Now we
compile results, obtaining

$\displaystyle
W_{\Sigma}=\int_{\mathscr{F}}d\mu_{\mathrm{WP},\Sigma}(R_{\Sigma^{2},I}(J))^{\frac{-13}{2}}\
\left[(\det(P_{1}^{\dagger}P_{1})_{g,\Sigma^{2}})^{\frac{1}{2}}\left(\frac{\det^{\prime}(\Delta)_{g,\Sigma^{2}}}{\int_{\Sigma^{2}}\sqrt{g}}\right)^{-13}\right]^{\frac{1}{2}}\
e^{-I(X_{cl},g)}.$ (72)

At this stage we review our strategy to obtain an explicit representation of
the partition function on $\Sigma$. We wish to write the above integrand in
terms of moduli parameters on $\Sigma^{2}$ and restrict the integration to the
subset of $\mathscr{T}(\Sigma^{2})$ which is isomorphic to
$\mathscr{T}(\Sigma)$ (in fact we restrict further to the moduli spaces). It
is in fact easy to carry this out if one notes that the partition function on
$\Sigma^{2}$ is given by Alvarez:1982zi

$W_{\Sigma^{2}}=\int_{\mathscr{F}_{\Sigma^{2}}}d\mu_{\mathrm{WP},\Sigma^{2}}(\det(P_{1}^{\dagger}P_{1})_{g,\Sigma^{2}})^{\frac{1}{2}}\left(\frac{2\pi}{\int_{\Sigma^{2}}\sqrt{g}}\mathrm{det}^{\prime}(\Delta)_{g,\Sigma^{2}}\right)^{-13}\
.$ (73)

This is the 2-loop closed bosonic string partition function, and its value is
well-known from the literature. We see that a detailed analysis of the above
integrand in terms of a good set of moduli parameters, combined with knowledge
of how the region of integration should be restricted, will allow us to
immediately obtain $W_{\Sigma}$.

###  3.2 Schottky Parametrisation

The double torus can be represented as the complex plane (union infinity) with
two pairs of non-intersecting circles cut out such that in each pair the two
circles are isometric to each other; here, we give a summary of this
representation as described, e.g., in TseytlinRG . The handles on the double
torus are formed by identification of the boundaries of each pair of circles.
To make this identification, we associate with each pair of circles
$(I_{1},I^{\prime}_{1})$ and $(I_{2},I^{\prime}_{2})$ a Mobius transformation
$T_{i}:\mathbb{C}\rightarrow\mathbb{C}$ where

$\displaystyle T_{i}(z)\equiv\frac{A_{i}z+B_{i}}{C_{i}z+D_{i}},\hskip
14.22636ptA_{i}D_{i}-B_{i}C_{i}=1,\hskip 14.22636pti\in\\{1,2\\},$ (74)
$\displaystyle T_{i}(I_{i})=I^{\prime}_{i},$

i.e. $T_{i}$ maps the circle $I_{i}$ to the isometric circle $I^{\prime}_{i}$.
This leads to the fact that the isometric circles are given by

$I_{i}=\\{|C_{i}z+D|=1\\},\hskip 14.22636ptI^{\prime}_{i}=\\{|C_{i}z-A|=1\\}.$
(75)

The Schottky group $G$ is the group generated by these Mobius transformations.
Then, the fundamental region of the double torus is the region of the complex
plane exterior to the four circles $I_{1}$, $I^{\prime}_{1}$, $I_{2}$ and
$I^{\prime}_{2}$.

Furthermore, it can be shown that any double torus can be represented by the
fundamental region of such a Schottky group. Since each of the Mobius
transformations $T_{i}$ can be parametrised by three complex numbers
$(\xi_{i},\eta_{i},k_{i})$ where $k_{i}$ is the multiplier of $T_{i}$ and
$\xi_{i}$ and $\eta_{i}$ are the repulsive and attractive fixed points of
$T_{i}$, we have 6 complex parameters. However, if two sets of parameters are
related by an overall $\mathrm{SL}(2,\mathbb{C})$ transformation, then the
corresponding Riemann surfaces relate to the same point in moduli space.

This reduces the number of complex parameters to 3 as expected. In fact, we
will write the integrand in a $\mathrm{SL}(2,\mathbb{C})$ invariant form, and
divide out by the volume of $\mathrm{SL}(2,\mathbb{C})$. Now we write down the
partition function for the 2-loop closed bosonic string DiVecchia:1987uf

$\displaystyle W_{\Sigma^{2}}=$
$\displaystyle\frac{1}{\Omega}\int_{\mathrm{mod}(\Sigma^{2})}\prod_{i=1}^{2}\left(\frac{d^{2}k_{i}d^{2}\xi_{i}d^{2}\eta_{i}}{|k_{i}|^{4}|\eta_{i}-\xi_{i}|^{4}}|1-k_{i}|^{4}\right)(\det\
\mathrm{Im}\ \tau)^{-13}$ (76) $\displaystyle\ \ \ \ \
\times\prod^{\prime}_{\alpha}\prod_{n=1}^{\infty}|1-k_{\alpha}^{n}|^{-48}\prod^{\prime}_{\alpha}|1-k_{\alpha}|^{-4},$

where $\prod_{\alpha}^{\prime}$ is over all primitive elements of the Schottky
group and $\Omega$ is the volume of the $\mathrm{SL}(2,\mathbb{C})$ group. We
now assume that the Weil-Petersson measure in this expression is

$d\mu_{\mathrm{WP},\Sigma^{2}}=\prod_{i=1}^{2}\left(\frac{d^{2}k_{i}d^{2}\xi_{i}d^{2}\eta_{i}}{|k_{i}|^{4}|\eta_{i}-\xi_{i}|^{4}}|1-k_{i}|^{4}\right).$
(77)

####  3.2.1 Schottky Parametrisation of $\Sigma^{2}$

We will express the disc-with-handle as the upper half complex plane with two
isometric holes cut out. In this model, the boundary of the disc is given by
the real axis, and the handle results from identification of the boundaries of
the two circles. The opposite orientation copy is the lower complex plane with
two isometric holes cut out.

To be precise, the involution corresponds to complex conjugation. We form the
Schottky double $\Sigma^{2}$ by attaching these two surfaces along the
boundary, i.e. along the real axis, obtaining the complex plane with two pairs
of isometric circles cut out, i.e. a double torus. In particular, we see that
each point in the moduli space of $\Sigma$ corresponds to a point in moduli
space of $\Sigma^{2}$ such that

$I_{1}=(I_{2})^{*},\hskip 14.22636ptT_{1}(z)=(T_{2}(z^{*}))^{*},\hskip
14.22636ptk_{1}\backslash\xi_{1}\backslash\eta_{1}=k_{2}^{*}\backslash\xi_{2}^{*}\backslash\eta_{2}^{*}.$
(78)

Figure 1: $\Sigma$ and its opposite orientation copy are related by the
involution $I$ (shown in green), which is simply complex conjugation.

Thus, the Weil-Petersson measure for $\Sigma$ is given by

$d\mu_{\mathrm{WP},\Sigma}=\frac{d^{2}k_{1}d^{2}\xi_{1}d^{2}\eta_{1}}{|k_{1}|^{4}|\eta_{1}-\xi_{1}|^{4}}|1-k_{1}|^{4},$
(79)

whilst the integrand derived from the result for $\Sigma^{2}$ is simply
restricted to the appropriate subset of moduli space. Thus, we obtain the
result

$\displaystyle W_{\Sigma}=$
$\displaystyle\frac{1}{\sqrt{\Omega}}\int_{\mathrm{mod}(\Sigma)}\frac{d^{2}k_{1}d^{2}\xi_{1}d^{2}\eta_{1}}{|k_{1}|^{4}|\eta_{1}-\xi_{1}|^{4}}|1-k_{1}|^{4}(\frac{1}{2\pi}R_{\Sigma^{2},I}(J))^{-\frac{13}{2}}$
(80) $\displaystyle\bigg{(}(\det\ \mathrm{Im}\ \tau)^{-13}\
\prod^{\prime}_{\alpha}\prod_{n=1}^{\infty}|1-k_{\alpha}^{n}|^{-48}\prod^{\prime}_{\alpha}|1-k_{\alpha}|^{-4}\bigg{)}^{\frac{1}{2}}\
e^{-I_{\mathrm{cl}}(X,g)}\ ,$

where the term is brackets is calculated for the case where the relations in
78 hold.

###  3.3 Period Matrix Parametrisation

Alternatively, the method in BLAU1988285  used to calculate the partition
function of the three-holed sphere can be immediately adapted to the disc-
with-handle. Here one uses the period matrix to parametrise the moduli space.
Up to a constant independent of the metric one has BLAU1988285

$\displaystyle
d\mu_{\mathrm{WP},\Sigma}\left({(\det(P_{1}^{\dagger}P_{1})_{g,\Sigma^{2}})^{\frac{1}{2}}}\left(\frac{\det^{\prime}(\Delta)_{g,\Sigma^{2}}}{\det\
\textrm{Im}\ \tau\int_{\Sigma^{2}}\sqrt{g}}\right)^{-13}\right)^{\frac{1}{2}}$
$\displaystyle=$ $\displaystyle\underbrace{\prod_{1\geq i\geq j\geq
2}\\{\frac{1}{2}(1+I^{*})d\ \mathrm{Im}\ \tau+\frac{1}{2}(1+I^{*})d\
\mathrm{Re}\ \tau\\}_{ij}}_{\text{Weil-Petersson Measure}}\prod_{\alpha\
\textrm{even}}|\theta^{2}\\{\alpha\\}(0,\tau)|^{-1}$ (81)

where the second product is of the ten even genus two theta functions. Note
that the Weil-Petersson measure for $\Sigma$ is given in terms of the
imaginary and real parts of the period matrix $\tau$ of the doubled surface
$\Sigma^{2}$. The integration, however, is over the subset of the Teichmuller
space of $\Sigma^{2}$ which corresponds to the Teichmuller space of $\Sigma$.
To calculate the Weil-Petersson measure, we use the fact that

$I^{*}\tau=-\Gamma^{T}\bar{\tau}\Gamma,\hskip
14.22636pt\Gamma=\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}.$ (82)

Note that the Teichmuller space of the double torus is the Siegel upper half-
space with the subspace of diagonal period matrices removed Rodrigues:1986us .
Denoting $\tau$ as

$\tau=\begin{pmatrix}a+ib&c+id\\\ c+id&e+if\end{pmatrix},$ (83)

we obtain

$\prod_{1\geq i\geq j\geq 2}\\{\frac{1}{2}(1+I^{*})d\ \mathrm{Im}\
\tau+\frac{1}{2}(1+I^{*})d\ \mathrm{Re}\ \tau\\}_{ij}=\frac{1}{2}dx\wedge
dy\wedge dz,$ (84)

where we have made a change of variables

$x=b+f,\ \ \ \ y=d,\ \ \ \ z=a-e.$ (85)

We find now that

$\displaystyle R_{\Sigma^{2},I}(J)$
$\displaystyle=\frac{2}{bf-d^{2}}((b+d)^{2}+(d+f)^{2})$ (86)
$\displaystyle=\frac{4x^{2}+16y(x+y)}{x^{2}-(b-f)^{2}-4y^{2}}.$

Compiling everything,

$W_{\Sigma}\propto\frac{1}{\Omega^{\prime}}\int
dxdydz\left(x^{2}+4y(x+y)\right)^{-\frac{13}{2}}\prod_{\alpha\
\textrm{even}}|\theta^{2}\\{\alpha\\}(0,\tau)|^{-1}\
e^{-I_{\mathrm{cl}}(X,g)}.$ (87)

Note that we are integrating over all period matrices of the form

$\tau=\begin{pmatrix}\frac{1}{2}(z+ix)&iy\\\
iy&\frac{1}{2}(z+ix)\end{pmatrix}$ (88)

subject to the condition that $\tau$ is in the Siegel upper half space with
the subspace of diagonal period matrices removed Rodrigues:1986us . We
integrate over Teichmuller space and then ‘divide out’ by the volume
$\Omega^{\prime}$ of the mapping class group of the disc-with-handle, which is
the 3-stranded braid group. As before, the classical action is a function of
the moduli, however, we do not evaluate it in this thesis.

##  4 Conclusion

In this paper, we have reduced the infinite dimensional path integral for the
partition functions of the string world-sheets ending on a fixed circular
curve with annular topology and with the disc-with-handle topology to finite
dimensional integrals over moduli parameters. For the annular topology, the
partition function reduced to a one dimensional (divergent) integral over the
inner radius of the annulus. To do this we calculated the determinant of the
Laplacian operator on the annulus for the cases of Dirichlet-Dirichlet and
Dirichlet-Neumann boundary conditions, which are factors that appear in the
integrand.

For the case of the disc-with-handle, we pursued two different methods in
computing the partition function. In both cases, we utilised the Schottky
double of the surface, that is the surface obtained by gluing the disc-with-
handle to itself. This eliminates the problem of having to deal with a
boundary on the surface. We then used the fact that determinants on a Schottky
double are the square of Laplacian determinants on the single copy of the
surface, and that we can write the Weil-Petersson measure on the single-copy
surface via the Weil-Petersson measure on the doubled surface. We obtained the
final result twice, first using Schottky parameters of the moduli space, and
then using the period matrix.

###### Acknowledgements.

The authors would like to thank A. Tseytlin for taking them on for a summer
project, which later developed into a Master’s project. His continued guidance
and support have been invaluable. We are also grateful for illuminating
correspondence with R. Russo. The work of OS has been supported by the
Theoretical Physics Undergraduate Summer Research Project funding. The work of
BO has been supported by the Malcolm Weir foundation’s summer funding.

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# Groups whose non-normal subgroups are either nilpotent or minimal non-
nilpotent

Nasrin Dastborhan, Hamid Mousavi∗ Department of Pure Mathematics, Faculty of
Mathematical Sciences, University of Tabriz, Tabriz, Iran
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

Let $\mathfrak{Nil}$ be a class of nilpotent groups and $G$ be a
non-$\mathfrak{Nil}$ group. We call $G$ a meta-$\mathfrak{Nil}$-Hamiltonian
group if any of its non-$\mathfrak{Nil}$ subgroups is normal. Also, we call
$G$ is a para-$\mathfrak{Nil}$-Hamiltonian group if every non-normal subgroup
of $G$ is either an $\mathfrak{Nil}$-group or a minimal non-$\mathfrak{Nil}$
group. In this paper we investigate the class of finitely generated
meta-$\mathfrak{Nil}$-Hamiltonian and para-$\mathfrak{Nil}$-Hamiltonian
groups.

###### Key words and phrases:

Meta-$\mathfrak{Nil}$-Hamilponian, para-$\mathfrak{Nil}$-Hmiltonian

###### 2020 Mathematics Subject Classification:

20D05, 20D99

∗ Corresponding author

##  1\. Introduction

Let $\mathfrak{X}$ be a class of groups and $G$ be a non-$\mathfrak{X}$ group.
the group $G$ is said to be meta-$\mathfrak{X}$-Hamiltonian if any of its
non-$\mathfrak{X}$ subgroups is normal. Also, we say $G$ is
para-$\mathfrak{X}$-Hamiltonian if every non-normal subgroup of $G$ is either
an $\mathfrak{X}$-group or a minimal non-$\mathfrak{X}$-group. If
$\mathfrak{X}$ is a class of groups that is subgroup closed, the
non-$\mathfrak{X}$ condition is necessary.

If $\mathfrak{A}$ is the class of Abelian groups, the class of
meta-$\mathfrak{A}$-Hamiltonian is called metahamiltonian and the class of
para-$\mathfrak{A}$-Hamiltonian groups is called parahamiltonian.
Parahamiltonian groups were first introduced by Romalis and Sesekin and they
proved that if a parahamiltonian group is soluble, then its derived length is
at most three and its commutator subgroup is finite of prime-power order[13,
14, 15]. Then, these groups attracted the attention of many researchers, as
Giovanni and others.

A group $G$ is called locally graded if every non-trivial finitely generated
subgroup of $G$ contains a proper subgroup of finite index. The results of
Romalis and Sesekin [13, 14, 15] show that a locally graded metahamiltonian
group is necessarily soluble. Therefore, it has a finite commutator subgroup.
Also, an infinite locally graded group whose proper subgroups are
metahamiltonian is metahamiltonian [5, Theorem 3.1]. However, the
metahamiltonian groups are not necessarily soluble. For example, Tarski
groups, i.e. infinite simple groups whose proper non-trivial subgroups have
prime order.

Let $\mathfrak{X}$ be a class of groups, a group $G$ is said to be minimal
non-$\mathfrak{X}$ if it is not an $\mathfrak{X}$-group but all its proper
subgroups belong to $\mathfrak{X}$. Moreover, $G$ is called biminimal
non-$\mathfrak{X}$ if it is neither an $\mathfrak{X}$-group nor a minimal
non-$\mathfrak{X}$-group, but each proper subgroup of $G$ either belongs to
$\mathfrak{X}$ or is a minimal non-$\mathfrak{X}$-group. Atlihan and Giovanni
showed that every locally graded biminimal non-abelian group is finite and its
order is divisible by at most three prime numbers[1, Theorem 1]. They also
showed that $G/\Phi(G)\cong A_{5}$, for any finite insoluble biminimal non-
abelian group $G$. We will show that such a group is necessarily is isomorphic
to $A_{5}$(Corollary 3.8).

Parahamiltonian groups were first introduced by Atlihan and Giovanni[1]. They
proved that the commutator subgroup of a locally graded parahamiltonian group
is finite [1, Lemma 5]. They also showed that the parahamiltonian group which
is locally graded and insoluble, is finite and $\pi(G)=\\{2,3,5\\}$. We will
show that such a group is necessarily is isomorphic to $A_{5}$ (Corollary
3.9).

Let $\mathfrak{Nil}$ be the class of nilpotent groups. The most natural case
for metahamiltonian and parahamiltonian generalization are
meta-$\mathfrak{Nil}$-Hamiltonian and para-$\mathfrak{Nil}$-Hamiltonian
groups. In this paper we investigate the class of finitely generated
meta-$\mathfrak{Nil}$-Hamilton and para-$\mathfrak{Nil}$-Hamilton groups. It
will be proved that every finitely generated locally graded biminimal non-
nilpotent group, is finite and its order is divisible by at most three prime
numbers and every finite insoluble biminimal non-nilpotent group is isomorphic
to $A_{5}$. Also, any locally graded insoluble
para-$\mathfrak{Nil}$-Hamiltonian group is isomorphic to $A_{5}$ or
$\operatorname{SL}(2,5)$, if it has a finite minimal non-nilpotent subgroup.
In particular, this applies to any para-$\mathfrak{Nil}$-Hamiltonian group
that is insoluble and finite. In the case of locally graded
meta-$\mathfrak{Nil}$-Hamiltonian groups, we show that they are soluble. Some
of our techniques are taken from [1].

##  2\. Primary lemmas

In this article, we will use finite non-abelian minimal simple groups and the
structure of subgroups of $\operatorname{PSL}(2,p)$ and the Suzuki groups
$\operatorname{Sz}(q)$. So, in this section, we will discuss some important
structure theorems.

######  Theorem 2.1.

[17, Theorems 6.25 , 6.26] Let $q$ be a power of the prime $p$. Then, a
subgroup of $\operatorname{PSL}(2,q)$ is isomorphic to one of the following
groups.

  * (i)

The dihedral groups of order $2(q\pm 1)/d$ and their subgroups where
$d=(2,q-1)$.

  * (ii)

A group $H$ of order $q(q-1)/d$ and its subgroups. A Sylow $p$-subgroups $P$
of $H$ is elementary abelian, $P\unlhd H$, and the factor group $H/P$ is a
cyclic group of order $(q-1)/d$.

  * (iii)

$A_{4}$ whenever $p\neq 2$ or $n$ is even.

  * (iv)

$S_{4}$ whenever $q^{2}\equiv 1\pmod{16}$.

  * (v)

$A_{5}$ whenever $q(q^{2}-1)\equiv 0\pmod{5}$.

  * (vi)

$\operatorname{PSL}(2,r)$ where $r$ is a power of $p$ such that, $q=r^{m}$.

  * (vii)

$\operatorname{PGL}(2,r)$ where $q=r^{m}$ is odd and $m$ is even.

######  Theorem 2.2.

[7, Theorem XI.3.10] We put $q=2^{2m+1}$, $r=2^{m}$ and $G=Sz(q)$.

  * (a)

$G$ possesses cyclic Hall subgroups $U_{1}$ and $U_{2}$ of orders $q+2r+1$ and
$q-2r+1$.

  * (b)

If $1\neq u\in U_{i}$, $\mathcal{C}_{G}(u)=U_{i}$. Also
$|\mathcal{N}_{G}(U_{i}):U_{i}|=4$ and $\mathcal{N}_{G}(U_{i})=\langle
U_{i},t_{i}\rangle$ where $u^{t_{i}}=u^{q}$ for all $u\in U_{i}$. In
particular, $\mathcal{N}_{G}(U_{i})$ is a Frobenius group with Frobenius kerne
$U_{i}$.

######  Theorem 2.3.

[18, Corollary 1] Every finite minimal simple group is isomorphic to one of
the following minimal simple groups:

  * (a)

$\operatorname{PSL}(2,2^{p})$, $p$ any prime.

  * (b)

$\operatorname{PSL}(2,3^{p})$, $p$ any odd prime.

  * (c)

$\operatorname{PSL}(2,p)$, $p$ any prime exceeding $3$ such that
$p^{2}+1\equiv 0\pmod{5}$

  * (d)

$\operatorname{Sz}(2^{p})$, $p$ any odd prime.

  * (e)

$\operatorname{PSL}(3,3)$.

######  Theorem 2.4.

[12, 12.1.5] If $M$ is a maximal subgroup of a locally nilpotent group $G$,
then $M$ is normal in $G$. Equivalently $G^{\prime}\leqslant\Phi(G)$.

##  3\. The finite case

######  Lemma 3.1.

Let $G$ be a finite group such that
$G/\Phi(G)\cong\mathbb{Z}_{p}\rtimes\mathbb{Z}_{q}$. Then, for natural numbers
$m$ and $n$, $G\cong\mathbb{Z}_{p^{m}}\rtimes\mathbb{Z}_{q^{n}}$.

###### Proof.

Let $P\in\operatorname{Syl}_{p}(G)$ and $Q\in\operatorname{Syl}_{q}(G)$. Since
$P\Phi(G)\trianglelefteq G$, by Frattini argument, $P\trianglelefteq G$, so
$\Phi(P)\leqslant P\cap\Phi(G)$. If $\Phi(P)\neq P\cap\Phi(G)$, then there is
a maximal subgroup $M$ of $P$ such that $P=M(P\cap\Phi(G))$ and so
$G=PQ=M(P\cap\Phi(G))Q$, thus $G=MQ$, a contradiction. Therefore
$\Phi(P)=P\cap\Phi(G)$ and so
$P/\Phi(P)\cong\dfrac{P\Phi(G)}{\Phi(G)}\cong\mathbb{Z}_{p}$, hence $P$ is
cyclic.

Since $\frac{Q}{Q\cap\Phi(G)}\cong\mathbb{Z}_{q}$, $Q\cap\Phi(G)$ is maximal
in $Q$, so $\Phi(Q)\leqslant Q\cap\Phi(G)$. Now for any maximal subgroup $M$
of $Q$, $PM$ is maximal in $G$. As

$Q\cap\Phi(G)\leqslant Q\cap PM=(P\cap Q)M=M,$

then $Q\cap\Phi(G)\leqslant\Phi(Q)$. Therefore, $\Phi(Q)=Q\cap\Phi(G)$. So,
similar to the previous case, $Q/\Phi(Q)\cong\mathbb{Z}_{q}$ and so $Q$ is
cyclic. Hence, for some $n,m\in\mathbb{N}$,

$G=PQ\cong\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q^{m}}\;\&\;\Phi(G)=\mathbb{Z}_{p^{n-1}}\times\mathbb{Z}_{q^{m-1}}.$

∎

######  Lemma 3.2.

Let $G$ be a finite group and $\Phi(G)\leqslant H\leqslant G$. If $H$ and
$H/\Phi(G)$ are minimal non-nilpotent, then $\Phi(G)\leqslant\Phi(H)$.

###### Proof.

Let $M$ be a maximal subgroup of $H$. If $\Phi(G)\nleqslant M$, then
$H=M\Phi(G)$ and so $H/\Phi(G)\cong M/(M\cap\Phi(G))$ is nilpotent, which is a
contradiction. So $\Phi(G)\leqslant\Phi(H)$. ∎

######  Theorem 3.3.

Let $G$ be a finite group such that $G/\Phi(G)\cong A_{5}$. If $G$ is
para-$\mathfrak{Nil}$-Hamiltonian, then $G\cong A_{5}$ or
$G\cong\operatorname{SL}(2,5)$.

###### Proof.

Suppose that $H$ and $K$ are two subgroups of $G$ such that $H/\Phi(G)\cong
S_{3}$ and $K/\Phi(G)\cong D_{10}$. By Lemma 3.1, $H$ and $K$ are minimal non-
nilpotent. According to Lemma 3.2, $\Phi(G)\leqslant\Phi(H),\Phi(K)$. Since
$H$ and $K$ are noncyclic, therefore, $\Phi(H)=\Phi(K)=\Phi(G)$. Suppose that
$H=\langle x,t_{1}\rangle$ and $K=\langle y,t_{2}\rangle$ where $t_{1}$ and
$t_{2}$ are $2$-elements, $|x|=3^{m}$ and $|y|=5^{n}$ where $m,n\in N$.
According to Lemma 3.1, $\Phi(H)=\langle x^{3},t_{1}^{2}\rangle$,
$\Phi(K)=\langle y^{5},t_{2}^{2}\rangle$. Therefore, $x^{3}=y^{5}=1$ and
$\Phi(G)$ is a cyclic $2$-subgroup. Since, $G/\mathcal{C}_{G}(\Phi(G))$ is
abelian, thus $G=\mathcal{C}_{G}(\Phi(G))$ and so $Z(G)=\Phi(G)$. As $G$ is
perfect, the sequence

$M(G)\to M(G/Z(G))\to Z(G)\to 1,$

is exact by [8, Theorem 2.5.6], where $M(G)$ is the Schur multiplier of $G$.
As $M(A_{5})\cong\mathbb{Z}_{2}$, then $|Z(G)|=2$ and $|G|=120$. Among the
three insoluble groups of order 120, only $\operatorname{SL}(2,5)$ satisfies
all conditions. ∎

######  Lemma 3.4.

Let $G$ be a finite minimal simple group. If $G$ is biminimal non-nilpotent,
then $G\cong A_{5}$.

###### Proof.

Assume that $G\cong\operatorname{PSL}(2,2^{p})$. According to Theorem 2.1, $G$
has two dihedral subgroups of orders $2(2^{p}-1)$ and $2(2^{p}+1)$. These
subgroups are minimal non-nilpotent if both $2^{p}-1$ and $2^{p}+1$ are prime.
Since $2^{p}+1$ is prime whenever $p=2$, therefore
$G\cong\operatorname{PSL}(2,4)\cong A_{5}$ .

Suppose that $G\cong\operatorname{PSL}(2,3^{p})$. Similar to the previous
case, $G$ has two dihedral subgroups of orders $3^{p}-1$ and $3^{p}+1$. So
both $(3^{p}\pm 1)/2$ are either prime or a power of $2$ greater than $8$. Let
$(3^{p}+1)/2=2^{\ell}$ where $\ell\geq 3$. So $q=3^{p}=2^{\ell+1}-1$.
Therefore

$q^{2}=3^{2p}=2^{2\ell+2}-2^{\ell+2}+1\equiv 1\pmod{16}.$

Thus by Theorem 2.1(iv), $G$ has a subgroup isomorphic to $S_{4}$ that is
neither nilpotent nor minimal non-nilpotent. Similarly, if $3^{p}-1$ is a
power of $2$, again we will reach to contradiction caused by the existence of
a subgroup isomorphic to $S_{4}$. So both $(3^{p}\pm 1)/2$ are prime.
According to the relation

$(3^{p}+1)/2=(3^{p}-1)/2+1$

necessarily $(3^{p}-1)/2=2$, a contradiction.

Now suppose that $G\cong\operatorname{PSL}(2,p)$. Similar to the above
argument, $p\pm 1$ is not a power of $2$ greater than $8$, so $(p\pm 1)/2$ is
necessarily prime. Since $(p+1)/2=(p-1)/2+1$, $p-1=4$ and so $p=5$, which
contradicts $p\neq 5$ by Theorem 2.3(c).

Using the ${\rm GAP}$ software, it is determined that
$G\not\cong\operatorname{PSL}(3,3)$. Finally assume that
$G\cong\operatorname{Sz}(2^{p})$, where $p$ is an odd prime. By Theorem 2.2,
$G$ has two cyclic Hall subgroups $U_{1}$ and $U_{2}$ of odd orders
$n_{1}=2^{p}+2^{m+1}+1$ and $n_{2}=2^{p}-2^{m+1}+1$, respectively, where
$m=(p-1)/2$. Also, $\mathcal{N}_{G}(U_{i})$ for $i=1,2$ is a Frobenius group
with kernel $U_{i}$, which is non-nilpotent. (The center of a Frobenius group
is trivial). Since $|\mathcal{N}_{G}(U_{i}):U_{i}|=4$, then
$\mathcal{N}_{G}(U_{i})$ has a dihedral subgroup of order $2n_{i}$, which is
non-nilpotent. Therefore, $\mathcal{N}_{G}(U_{i})$ is not a minimal non-
nilpotent. So $G\not\cong\operatorname{Sz}(2^{p})$. ∎

######  Lemma 3.5.

Let $G$ be a finite minimal insoluble para-$\mathfrak{Nil}$-Hamiltonian group.
Then, either $G\cong A_{5}$ or $G\cong\operatorname{SL}(2,5)$.

###### Proof.

Since $G$ is minimal insoluble, $G/\Phi(G)$ is a minimal simple group.
Therefore by Lemma 3.4 $G/\Phi(G)\cong A_{5}$. Now the statement follows from
Theorem 3.3 ∎

######  Theorem 3.6.

Let $G$ be a finite insoluble para-$\mathfrak{Nil}$-Hamiltonian group. Then,
$G\cong A_{5}$ or $\operatorname{SL}(2,5)$. In particular, the unique prime
divisors of the order of G are $2$, $3$ and $5$.

###### Proof.

It is clear that every insoluble subgroup of $G$ is normal. Let $N$ be a
minimal insoluble normal subgroup of $G$. According to Lemma 3.5, $N$ is
isomorphic to $A_{5}$ or $\operatorname{SL}(2,5)$.

First, suppose that $N\cong A_{5}$. If $\mathcal{C}_{G}(N)\neq 1$, as
$N\cap\mathcal{C}_{G}(N)=1$, the group $G$ has a non-normal subgroup
isomorphic to $A_{4}\times\mathcal{C}_{G}(N)$ which is not minimal non-
nilpotent, a contradiction. Hence, $\mathcal{C}_{G}(N)=1$, and $N\leqslant
G\leqslant S_{5}$ by $N/C$-theorem. Therefore $G=N\cong A_{5}$.

Now assume that $N\cong\operatorname{SL}(2,5)$. If
$Z=Z(N)\neq\mathcal{C}_{G}(N)$, then $G/Z$ has a subgroup isomorphic to
$A_{5}\times\mathcal{C}_{G}(N)/Z$, which we reach to contradiction, similar to
the previous case. Therefore, $Z(G)=\mathcal{C}_{G}(N)=Z(N)$, for $|Z(N)|=2$.
Using $N/C$-theorem, $G/Z$ is isomorphic to a subgroup of
$\operatorname{Aut}(N)\cong S_{5}$. As $G/Z\not\cong S_{5}$ and has a subgroup
isomorphic to $A_{5}$, thus $G/Z\cong A_{5}$. Hence $|G|=|N|$ and so
$G\cong\operatorname{SL}(2,5)$. ∎

######  Corollary 3.7.

Let $G$ be a finite insoluble group. If $G$ is a biminimal non-nilpotent
group, then $G$ is isomorphic to $SL(2,5)$ or $A_{5}$.

Since
$Q_{8}\leqslant\operatorname{SL}(2,3)\ntrianglelefteq\operatorname{SL}(2,5)$
and $\operatorname{SL}(2,3)$ is not minimal non-abelian, therefore Lemma 1 and
Theorem 2 of [1] can be improved as follows.

######  Corollary 3.8.

Let $G$ be a finite insoluble biminimal non-abelian group. Then, $G\cong
A_{5}$.

######  Corollary 3.9.

Let $G$ be a locally graded insoluble parahamiltonian group. Then, $G\cong
A_{5}$.

###### Proof.

According to [1, Theorem 2], $G$ is a finite insoluble group. Now, the
statement holds by Theorem 3.6. ∎

A subgroup $H$ of a group $G$ is called permutable (or quasinormal) if for all
subgroups $K$ of $G$, $HK=KH$. Obviously any normal subgroup is permutable. By
[2, Lemma 2.3], if all non-nilpotent subgroups of finite group $G$ are
permutable, then $G$ is soluble. So we can conclude that finite
meta-$\mathfrak{Nil}$-Hamiltonian groups are soluble.

######  Theorem 3.10.

Let $G$ be a finite meta-$\mathfrak{Nil}$-Hamiltonian group. Then $G$ is
soluble.

Although the proof for the above theorem is not necessary, we can provide a
straightforward proof.

Since every finite minimal non-nilpotent group is soluble, it can be assumed
that $G$ has a proper non-nilpotent subgroup. Suppose that $X\leqslant G$ is
non-nilpotent. So $G/X$ is a finite Dedekind group and $\gamma_{3}(G)\leqslant
X$. Therefore, $\gamma_{3}(G)$ is contained in the intersection of all non-
nilpotent subgroups of $G$. Since $\gamma_{3}(G)$ is nilpotent or minimal non-
nilpotent, then it is soluble, which will cause $G$ to be soluble.

##  4\. Para-$\mathfrak{Nil}$-Hamiltonian and biminimal non-nilpotent groups

Let $G$ be a group and

$G=\gamma_{1}(G)\geq\gamma_{2}(G)\geq\cdots\geq\gamma_{n}(G)\geq\cdots,$

is its lower central series.The intersection $\bigcap_{i\geq 1}\gamma_{i}(G)$
is denoted by $\gamma_{\infty}(G)$.

Minimal non-nilpotent groups and especially groups which are not nilpotent of
class $k$ but whose proper subgroups are nilpotent of class $k$, were studied
by Newman and Wiegold [11]. The following two important theorems are about
finitely generated non-nilpotent groups.

######  Theorem 4.1.

[11, Lemma 3.2] Let G be a finitely generated non-nilpotent group all of whose
proper subgroups are locally nilpotent. If $G/\gamma_{\infty}(G)$ is non-
trivial, then $G$ is finite.

######  Theorem 4.2.

[11, Theorem 3.3] The Frattini factor group of an infinite finitely generated
minimal non-nilpotent group is non-abelian and simple.

If the finitely generated minimal non-nilpotent group $G$ has a subgroup of
finite index, then it has a normal subgroup $N$ of finite index. Since $G/N$
is finite and soluble, therefore $\gamma_{2}(G)$ will be a proper subgroup of
$G$ and then $G$ will be finite by Theorem 4.1. Then, the infinite finitely
generated minimal non-nilpotent group has no subgroups of finite index,
therefore, it is not locally graded. Thus, we will have the following result.

######  Corollary 4.3.

If a finitely generated minimal non-nilpotent group has a subgroup of finite
index, then it is finite. In particular, every locally graded finitely
generated minimal non-nilpotent group is finite.

Smith [16, Theorem 1] showed that if every proper subgroup of a torsion free
locally nilpotent group is nilpotent, then the group must be finite. Also he
proved that, if $G$ is a soluble minimal non-nilpotent group that has no
maximal subgroup, then $G$ has no proper subgroup of finite index [16, Theorem
3.1-v].

######  Lemma 4.4.

Let $G$ be a finitely generated locally graded group. If $G$ is infinite, then
the finite residual $J$ of $G$ has infinite index.

###### Proof.

Assume that $J$ has finite index in $G$. Then $J$ is finitely generated and
hence it contains a proper subgroup $K$ such that $|G:K|$ is finite.
Therefore, $|G:{\rm Core}_{G}(K)|$ is also finite and so $J\leqslant{\rm
Core}_{G}(K)\leqslant K$ , which is a contradiction. ∎

The following lemma is a generalization of [1, Lemma 4] by combining the parts
of the proofs of [1, Lemma 4, Theorem 2].

######  Lemma 4.5.

Let $\mathfrak{X}$ be a class of groups that is subgroup closed and let $G$ be
a para-$\mathfrak{X}$-Hamiltonian group. If $X\leqslant G$ which neither is
normal nor $\mathfrak{X}$-group then $X$ is maximal in its normal closure
$X^{G}$. Furthermore, if $X$ is finite and $G$ is locally graded, then

  * (i)

$X^{G}$ is finite;

  * (ii)

either $X\trianglelefteq X^{G}$ or $G$ is finite.

###### Proof.

Let $X<Y\leqslant G$. If $Y\ntrianglelefteq G$, then either $Y\in\mathfrak{X}$
or $Y$ is minimal non-$\mathfrak{X}$ which implies that $X\in\mathfrak{X}$, a
contradiction. Therefore, $Y\trianglelefteq G$ and hence $X^{G}\leqslant Y$.
Then $X$ is maximal in $X^{G}$.

(i) Now suppose that $X$ is finite. Then $X^{G}$ is finitely generated, for
$X$ is maximal in $X^{G}$. Let $Y$ be any normal subgroup of finite index of
$X^{G}$. If $Y\leqslant X$ then $X^{G}$ is finite. Otherwise $X^{G}=YX$ and
hence $|X^{G}:Y|\leq|X|$. Therefore, the finite residual of $X^{G}$ has finite
index, then $X^{G}$ is finite by Lemma 4.4.

(ii) Let $X$ be a proper subgroup of $\mathcal{N}_{G}(X)$. Then
$\mathcal{N}_{G}(X)\trianglelefteq G$. Therefore
$X^{G}\leqslant\mathcal{N}_{G}(X)$ and so $X\trianglelefteq X^{G}$. Suppose
that $X=\mathcal{N}_{G}(X)$, as $X^{G}$ is finite and $X^{G}\trianglelefteq G$
then $X$ has finitely many conjugates in $G$. Hence $|G:\mathcal{N}_{G}(X)|$
is finite, and so $G$ is finite. ∎

If a group $G$ satisfies one of the following conditions, then its commutator
subgroup $G^{\prime}$ is finite:

  * (i)

$|H^{G}:H|$ is finite, for every subgroup $H$ of $G$ [4, Theorem 5.70].

  * (ii)

$G$ is a locally graded and metahamiltonian [5, Theorem 2.1].

  * (iii)

$G$ is an insoluble locally graded parahamiltonian [1, Theorem 2.1].

In case (iii), it follows from Corollary 3.9 that $G$ is isomorphic to
$A_{5}$. In the following theorem, we extend the finiteness of $G^{\prime}$
with an additional condition for para-$\mathfrak{Nil}$-Hamiltonian groups.

######  Theorem 4.6.

Let $G$ be a para-$\mathfrak{Nil}$-Hamiltonian group. If $G$ has a finitely
generated minimal non-nilpotent subgroup $X$ which its normal closure $X^{G}$
is locally graded, then $G^{\prime}$ is finite. Furthermore, if
$X\trianglelefteq G$ or $X\ntrianglelefteq G$ and $G$ is infinite, then $G$ is
soluble.

###### Proof.

According to Corollary 4.3, $X$ is finite and hence is soluble. If
$X\trianglelefteq G$, then every subgroup of $G/X$ is normal, hence $G$ is
soluble. Since $(G/X)^{\prime}\cong G^{\prime}/(G^{\prime}\cap X)$ is finite,
then $G^{\prime}$ is finite. Now suppose that $X\ntrianglelefteq G$. As $X$ is
maximal in $X^{G}$, according to Lemma 4.5, $X^{G}$ is finite. Since every
subgroup of $G/X^{G}$ is normal, $(G/X^{G})^{\prime}$ is finite and hence
$G^{\prime}$ is finite. Suppose that $G$ is infinite. According to Lemma 4.5,
$X\trianglelefteq X^{G}$. Therefore, $X^{G}$ is soluble and hence $G$ is
solvable. ∎

######  Corollary 4.7.

Let $G$ be an insoluble para-$\mathfrak{Nil}$-Hamiltonian group. If $G$ has a
finitely generated minimal non-nilpotent subgroup whose normal closure is
locally graded, then $G\cong A_{5}$ or $G\cong\operatorname{SL}(2,5)$.

###### Proof.

Let $X$ be a finitely generated minimal non-nilpotent subgroup of $G$.
According to Theorem 4.6, $X\ntrianglelefteq G$ and $G$ is finite. Hence the
statement follows by Theorems 3.6. ∎

Every locally graded biminimal non-abelian group is finite and its order is
divisible by at most three prime numbers [1, Theorem 1]. The similar result
holds for finitely generated locally graded biminimal non-nilpotent groups.

######  Theorem 4.8.

Let $G$ be a finitely generated locally graded biminimal non-nilpotent group.
Then $G$ is finite, and its order is divisible by at most three prime numbers.

###### Proof.

Since $G$ is locally graded, it has a proper normal subgroup $N$ of finite
index. It is obvious that $N$ is finitely generated. If $N$ is non-nilpotent,
it is finite by Corollary 4.3, therefore the statement holds. So assume that
$N$ is nilpotent. Then $N$ is supersoluble by [12, 5.4.6(ii)]. Since $G$ is a
biminimal non-nilpotent group, it has a proper subgroup $M$ that is minimal
non-nilpotent. Also, $M$ is a maximal subgroup of $G$. We show that $M$ has
finite index, thus it is finitely generated, and according to Corollary 4.3,
$G$ will be finite.

It can be assumed that $N\nleqslant M$, then $G=NM$ and hence $N\cap
M\trianglelefteq G$. Let $H$ be the maximal subgroup of $N$ containing $N\cap
M$. Since $N$ is supersoluble, $H$ has finite index in $N$ [12, 5.4.7], also
$N\cap M\leqslant H_{G}$. where $H_{G}={\rm Core}_{G}(H)$. If $N\cap M\neq
H_{G}$ then $G=H_{G}M$. Now for each $x\in N\backslash H$, there are $h\in
H_{G}$ and $m\in M$, where $x=hm$. Then $h^{-1}x=m\in N\cap M\leqslant H$,
which is a contradiction. Therefore, $N\cap M=H_{G}$ and $M$ has finite index.

The statement holds for insoluble group $G$ by Corollary 3.7. So suppose that
$G$ is a soluble group and $H$ is a non-nilpotent subgroup of $G$. Therefore,
$H$ is minimal non-nilpotent and is maximal in $G$. As $|H|$ has two prime
factors and $|G:H|$ is a power of a prime number, so the order of $G$ has at
most three prime factors. ∎

######  Corollary 4.9.

Let $G$ be a locally graded biminimal non-nilpotent group. If $G$ has a
finitely generated non-nilpotent subgroup, then $G$ is finite.

###### Proof.

Let $X$ be a finitely generated non-nilpotent subgroup of $G$. According to
Theorem 4.8, it can be assumed that $X$ is a proper subgroup of $G$.
Therefore, $X$ is minimal non-nilpotent and so it is a maximal subgroup of
$G$. By Corollary 4.3, $X$ is finite, therefore $G$ is finitely generated. Now
$G$ is finite by Theorem 4.8. ∎

Suppose that $G$ is a locally graded infinite group whose subgroups are either
nilpotent or minimal non-nilpotent. According to Corollary 4.9, if $G$ is not
nilpotent, then it is neither finite nor does it have any non-nilpotent
subgroups that are finitely generated. Therefore, it is locally nilpotent.

######  Corollary 4.10.

Let $G$ be an infinite locally graded group whose subgroups are either
nilpotent or minimal non-nilpotent. Then $G$ is locally nilpotent.

##  5\. Meta-$\mathfrak{Nil}$-Hamiltonian groups

######  Lemma 5.1.

Let $G$ be a biminimal non-nilpotent locally nilpotent group. Then $G$ is
meta-$\mathfrak{Nil}$-Hamiltonian.

###### Proof.

If $X$ is a proper non-nilpotent subgroup of $G$, then $X$ is a maximal
subgroup of $G$. Hence $\Phi(G)\leqslant X$. On the other hand, since $G$ is a
locally nilpotent, $G^{\prime}\leqslant X$ and consequently $X$ is normal. ∎

######  Lemma 5.2.

Let $G$ be a finitely generated meta-$\mathfrak{Nil}$-Hamiltonian group. If
every subgroup of $G$ is locally nilpotent, then either $G$ is finite or
$G/\Phi(G)$ is the infinite simple non-abelian group.

###### Proof.

If $G$ is an infinite, minimal non-nilpotent group, then it is simple non-
abelian according to Theorem 4.2. Otherwise $G$ has a non-nilpotent subgroup
$X$. Since $G/X$ is a Dedekind group, then $\gamma_{3}(G)\leqslant X$.
Therefore, $G/\gamma_{\infty}(G)$ is non-trivial. Now the statement follows
from Theorem 4.1. ∎

Therefore, every finitely generated locally nilpotent group whose non-
nilpotent subgroups are normal, is either finite or infinite insoluble.

The following lemma holds with a completely similar proof from [6, Lemma 3.1].

######  Lemma 5.3.

Let $\mathfrak{X}$ be a class of groups that is subgroup closed. If every
finitely generated subgroup of $G$ is meta-$\mathfrak{X}$-Hamiltonian, then
$G$ is meta-$\mathfrak{X}$-Hamiltonian.

###### Proof.

Suppose that $X$ is a non-$\mathfrak{X}$ subgroup of $G$ and $\mathfrak{N}$ is
the set of finitely generated non-$\mathfrak{X}$ subgroups of $X$. Since any
finitely generated subgroup of $X$ is meta-$\mathfrak{X}$-Hamiltonian, so is
non-$\mathfrak{X}$ subgroup of $X$, thus $\mathfrak{N}$ is non-empty. Now for
any $Y\in\mathfrak{N}$, $\langle Y,g\rangle$ is finitely generated for all
$g\in G$, so $Y^{g}=Y$, then $Y\trianglelefteq G$. Also for any $g\in X$,
$\langle Y,g\rangle\in\mathfrak{N}$, so $X=\langle
Y\,|\,Y\in\mathfrak{N}\rangle\trianglelefteq G$. ∎

######  Lemma 5.4.

Let $\mathfrak{X}$ be a class of groups that is subgroup closed and $G$ be a
meta-$\mathfrak{X}$-Hamiltonian group. Suppose that $J$ is finite residual of
$G$, $\mathfrak{N}$ is the class of non-$\mathfrak{X}$ subgroups of $G$ and
$I=\bigcap_{X\in\mathfrak{N}}X$.

  * (i)

$\gamma_{3}(G)\leqslant I$ and $I$ is either an $\mathfrak{X}$-group or a
minimal non-$\mathfrak{X}$ group.

  * (ii)

If $G$ is not an $\mathfrak{X}$-by-finite, then $\gamma_{3}(G)\leqslant
I\leqslant J$.

###### Proof.

(i) Assume that $X\in\mathfrak{N}$. Since $G/X$ is a Dedekind group,
$\gamma_{3}(G)\leqslant X$. Therefore, $\gamma_{3}(G)\leqslant I$. Now suppose
that $X\leqslant I$ is a non-$\mathfrak{X}$-group, then $I=X$ and so $I$ is a
minimal non-$\mathfrak{X}$-group.

(ii) Suppose that $G$ is not an $\mathfrak{X}$-by-finite group. Let
$\mathfrak{F}$ be the class of subgroups of finite index of $G$. Therefore,
none of the members of $\mathfrak{F}$ is an $\mathfrak{X}$-group, so
$\mathfrak{F}\leqslant\mathfrak{N}$. Thus,

$\gamma_{3}(G)\leqslant I\leqslant\bigcap_{N\in\mathfrak{F}}N=J.$

∎

Since meta-$\mathfrak{Nil}$-Hamiltonian groups are non-nilpotent, according to
(ii), they are not residually finite.

######  Theorem 5.5.

Let $G$ be a finitely generated meta-$\mathfrak{Nil}$-Hamiltonian group. If
$G$ has a torsion-free nilpotent subgroup $H$ of finite index, then $G\cong
H_{G}\times K$, where $K$ is finite non-nilpotent soluble and $H_{G}$ is a
free abelian group of finite rank. In particular, $G$ is soluble and
$G^{\prime}$ is finite.

###### Proof.

Suppose that $H$ is a torsion-free nilpotent subgroup of finite index in $G$.
Hence $|G:H_{G}|$ is also finite. In particular, $N=H_{G}$ is finitely
generated torsion-free nilpotent. Suppose that $p$ and $q$ are primes such
that $q>p>|G/N|$. Since $|N:N^{pq}|$ is finite, $|G/N^{pq}|$ is also finite.
Since $|\bar{G}:\bar{N}|=|G/N|<p$, where $\bar{G}=G/N^{pq}$, then
$|\bar{G}:\bar{N}|$ and $|\bar{N}|$ are relatively prime,
$\bar{G}\cong\bar{N}\rtimes\bar{K}$ where $\bar{K}\cong G/N$, by Schur-
Zassenhaus Theorem.

Since Sylow subgroups of $\bar{N}$ are characteristic, ${\bar{N}}^{p}\bar{K}$
and ${\bar{N}}^{q}\bar{K}$ are proper subgroups of $\bar{G}$. If, for example,
${\bar{N}}^{p}\bar{K}$ is nilpotent, then $K\cap N^{p},[K,N^{p}]\leqslant
N^{pq}$. Since $N$ is a finitely generated torsion-free nilpotent group,
according to [12, 5.2.21 or 5.4.17],
$\bigcap_{q>p}N^{pq}\leqslant\bigcap_{q>p}N^{q}=1$. Therefore $N^{p}\cap
K=[K,N^{p}]=1$ and hence $G\cong N^{p}\times K$ is nilpotent, because
$\bar{K}\cong K$. This is a contradiction. Therefore, ${\bar{N}}^{p}\bar{K}$
and ${\bar{N}}^{q}\bar{K}$ both are non-nilpotent and so are normal in
$\bar{G}$. Hence $\bar{K}\trianglelefteq\bar{G}$, thus $\bar{K}$ acts
trivially on $\bar{N}$. Then $K\cap N,[K,N]\leqslant N^{pq}\leqslant N^{q}$.
So again, we will have $G\cong N\times K$. Since $G/K\cong N$ is torsion-free
Dedekind group, it is abelian. Therefore, $N$ is a free abelian group of
finite rank. By Theorem 3.10, $K$ is soluble. ∎

The group $\mathbb{Z}\times S_{3}$ is a simplest example of Theorem 5.5.

######  Corollary 5.6.

Let $G$ be a finitely generated meta-$\mathfrak{Nil}$-Hamiltonian group and
$H\leqslant G$ is nilpotent. If $|G:H|$ is finite, then $G$ is soluble.
Furthermore, if $G/T$ is not nilpotent, where $T$ is the torsion subgroup of
$H_{G}$, then $G^{\prime}$ is finite.

###### Proof.

According to Theorem 3.10, we can assume that $G$ is infinite. As $|G:H|$ is
finite, $H_{G}$ is a finitely generated nilpotent group, thus $T$ is finite,
for $H_{G}$ is polycyclic. Hence $H_{G}/T\neq 1$ is a torsion free nilpotent
subgroup of $G/T$. If $G/T$ is nilpotent, so $G$ is soluble. Otherwise,
according to Theorem 5.5, $(G/T)^{\prime}$ is a finite soluble group.
Therefore $G^{\prime}$ is a finite soluble group and so $G$ is soluble. ∎

If a meta-$\mathfrak{Nil}$-Hamiltonian group $G$ has a finite non-nilpotent
subgroup $X$, since $G/X$ is a Dedekind group, then $G^{\prime}$ is finite.

######  Corollary 5.7.

Let $G$ be a locally graded meta-$\mathfrak{Nil}$-Hamiltonian group. If $G$
has a finitely generated minimal non-nilpotent subgroup, then $G^{\prime}$ is
finite.

If $G$ is insoluble such that all whose non-nilpotent subgroups are
permutable, then $G^{\prime\prime}$ is perfect by [2, Lemma 2.1]. In the
following propositions with meta-$\mathfrak{Nil}$-Hamiltonian assumption, we
will have a stronger result for $G^{\prime\prime}$.

######  Proposition 5.8.

Let $G$ be an insoluble meta-$\mathfrak{Nil}$-Hamiltonian group. Then the
following statements are equivalent.

  * (i)

$G$ is perfect.

  * (ii)

$G$ is minimal non-nilpotent.

  * (iii)

$G$ is infinite finitely generated and $G/\Phi(G)$ is non-abelian simple.

###### Proof.

(i)$\rightarrow$(ii) Assume that $G$ is perfect and $X\leqslant G$ is non-
nilpotent. Since $G/X$ is Dedekind group we have the contradiction
$G^{\prime}\neq G$, if $X\neq G$. Therefore, $G$ is minimal non-nilpotent.

(ii)$\rightarrow$(iii) Suppose that $G$ is minimal non-nilpotent. If $G$ is
not finitely generated, $G^{\prime}$ is abelian, which is a contradiction.
Therefore, $G$ is infinite finitely generated and so, $G/\Phi(G)$ is an
infinite non-abelian simple group by Theorem 4.2.

(iii)$\rightarrow$(i) straightforward. ∎

######  Proposition 5.9.

Let $G$ be an insoluble meta-$\mathfrak{Nil}$-Hamiltonian group. If $G$ is not
perfect, then $G^{\prime\prime}=\gamma_{3}(G)$ is a perfect minimal non-
nilpotent finitely generated group and its Frattini factor group is infinite
non-abelian simple.

###### Proof.

According to Proposition 5.8, $G$ has a proper non-nilpotent subgroup.
Therefore, $\gamma_{3}(G)$ is included in all non-nilpotent subgroups of $G$.
Since $G^{\prime\prime}\leqslant\gamma_{3}(G)$ is non-nilpotent,
$G^{\prime\prime}=\gamma_{3}(G)$ is minimal non-nilpotent by Lemma 5.4. By
reusing of Proposition 5.8, we obtain the result. ∎

Atlihan and et al. proved that if all non-nilpotent subgroups of locally
graded group $G$ are permutable, then $G$ is soluble (see [2, Theorems A, B]
and [3, Theorem 1.1]). Therefore a locally graded
meta-$\mathfrak{Nil}$-Hamiltonian group is soluble. In the following lemma we
have a simple proof for solubility of locally graded
meta-$\mathfrak{Nil}$-Hamiltonian groups.

######  Lemma 5.10.

Let $G$ be a locally graded meta-$\mathfrak{Nil}$-Hamiltonian group. Then $G$
is soluble.

###### Proof.

Suppose that $G$ is insoluble, so it does not have a nilpotent subgroup of
finite index (by using Theorem 3.10). According to [10, Corollary 2.3], $G$
has a soluble subgroup $M$ of finite index. Therefore, $M$ is non-nilpotent
and so is normal, which contradicts the insolubility of $G$. ∎

######  Theorem 5.11.

Let $G$ be a finitely generated locally graded
meta-$\mathfrak{Nil}$-Hamiltonian group. If $G^{\prime}$ is not torsion, it
has a nilpotent maximal subgroup of infinite order and index. In particular,
the Frattini subgroup and finite normal subgroups of $G$ are nilpotent.
According to the symbols of Lemma 5.4,

  * (i)

If $G/X$ is non-abelian for some non-nilpotent subgroup $X$, then $J$ is
nilpotent. Therefore, if $J$ is non-nilpotent, $G^{\prime}\leqslant I$;

  * (ii)

If $I$ is non-nilpotent, then $G^{\prime}$ is either nilpotent or minimal non-
nilpotent, $G^{\prime\prime}\leqslant\gamma_{\infty}(G)$ is abelian, and for
some prime $p$, $G^{\prime}/\gamma_{\infty}(G)$ is a finite cyclic $p$-group.

###### Proof.

Since $G$ is non-nilpotent finitely generated soluble by Lemma 5.10, it has a
non-normal nilpotent maximal subgroup $N$ (by [9, 4.4.5(ii)]). By Corollary
5.6, $N$ is not of finite index, so every finite normal subgroup of $G$ is
contained in $N$ and so is nilpotent. If $N$ is finite, since for any subgroup
$X$ of finite index, $X$ is non-nilpotent, so $G=XN$. Then $|G:X|\leq|N|$,
thus $G$ has a finite number of subgroups of finite index. Therefore, $J$ has
finite index, which contradicts Lemma 4.4.

(i) In this case, the $G/J$ is non-abelian. If $J$ is non-nilpotent then $G/J$
is a finitely generated Dedekind torsion group, so it is finite which
contradicts Lemma 4.4. Therefore, $J$ is nilpotent.

(ii) According to Lemma 5.4, $I$ is minimal non-nilpotent, so
$G^{\prime}\leqslant I$ by (i) and $I$ is not finitely generated by Corollary
4.3. Therefore, $I^{\prime}$ is abelian and for some prime $p$, $I/I^{\prime}$
is a finite cyclic $p$-group (see [11, 4.7, 4.8]), thus
$I^{\prime}=\gamma_{3}(I)\leqslant\gamma_{3}(G)$ and so
$G^{\prime\prime}\leqslant I^{\prime}\leqslant\gamma_{\infty}(G)$. Therefore
$G^{\prime\prime}$ is abelian and $G^{\prime}/\gamma_{\infty}(G)$ is a finite
cyclic group. ∎

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