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\begin{document}

\title{Orders For Which There Exist Exactly Two Groups}

\author{Clint Givens\\ 112123752\\ Adviser: Dr. Krishnan Shankar\\ May 12, 2006}

%\address{University of Oklahoma, Norman, OK 73069. \newline{\tt cgivens@ou.edu}}

\maketitle \normalsize \thispagestyle{empty}

\subsection*{\textsf{Introduction}}\hfill

A long-standing problem in group theory is to determine the number
of non-isomorphic groups of a given order. The inverse
problem--determining the orders for which there are a given number
of groups--has received considerably less attention. In this note,
we will give a characterization of those positive integers $n$ for
which there exist exactly $2$ distinct groups of order $n$ (up to
isomorphism). We call such numbers \emph{$2$-group numbers}.

Our work extends ideas from \cite{PS}. Following the authors, we say
a number $n=p_1^{a_1}\cdots p_r^{a_r}$ has \emph{nilpotent
factorization} if $p_i^k \not\equiv 1 \pmod{p_j}$ for all integers
$i$, $j$, and $k$ with $1\leq k\leq a_i$. In other words, no prime
power in the factorization is congruent to $1$ modulo any other
prime in the factorization. \cite{PS} show in their paper that a
number $n$ has nilpotent factorization if and only if every group of
order $n$ is nilpotent. Such $n$ they call, naturally enough,
\emph{nilpotent numbers}. \emph{Cyclic numbers} are those $n$ for
which every group of order $n$ is cyclic, and similarly for
\emph{abelian numbers}. The authors show that the cyclic numbers are
precisely the square-free nilpotent numbers, and the abelian numbers
are precisely the cube-free nilpotent numbers.

We will see that every $2$-group number is either a square-free
number that fails nilpotent factorization for exactly one pair of
primes, or an abelian number with exactly one square in its prime
factorization.

\subsection*{\textsf{Results}}\hfill
\begin{prop}
\label{p2abelian} Every group of order $p^2$ is abelian.
\end{prop}

\begin{proof}
Suppose $G$ is a non-abelian group of order $p^2$. Let $Z$ denote
the center of $G$. It is well known that $Z$ cannot be trivial in a
$p$-group. Since $G$ is non-abelian, $|Z|\neq p^2$, so by Lagrange's
Theorem we must have $|Z| = p$. But then $|G/Z| = p$. Let $h\in G$,
$h\not\in Z$. Then $G/Z = \{Z, hZ, h^2Z,\ldots,h^{p-1}Z\}$. These
cosets commute with one another and with $Z$ (since it is the
center), hence every element of $G$ commutes with every other, and
this contradicts our assumption that $G$ was non-abelian. (This
argument actually shows that $G/Z$ is never a non-trivial cyclic
group.)
\end{proof}

Now since every group of order $p^2$ is abelian, it follows by the
Classification Theorem for Finite Abelian Groups that $p^2$ is a
$2$-group number for every prime $p$, the two groups being
$\Z_{p^2}$ and $\Z_p\times \Z_p$.

\begin{prop}
If $n$ is not cube-free, then $n$ is at least a $5$-group number.
\end{prop}
\begin{proof}
Let $p$ be a prime. We will construct $5$ distinct groups of order
$p^3$. The Classification Theorem for Finite Abelian Groups tells us
there are three abelian groups of order $p^3$:
\begin{align*}
&\Z_p\times\Z_p\times\Z_p\\
&\Z_{p^2}\times\Z_p\\
&\Z_{p^3}
\end{align*}
Note that if $p=2$ we also have two nonabelian groups of order
$p^3$: the dihedral group $D_8$ and the quaternion group $Q_8$. We
claim that the same holds for $p$ an odd prime. Recall that the
center $Z$ of a $p$-group is non-trivial. So if $G$ is a nonabelian
group of order $p^3$, then either $|Z| = p$ or $|Z| = p^2$. In fact,
the latter case does not happen, for if so, we would have $G/Z$
cyclic of order $p$, a contradiction (see the proof of Proposition
\ref{p2abelian}). So $|Z| = p$. Now, we want to form two semidirect
products.

For the first one, consider $\aut(\Z_{p^2})$. We know that it is
isomorphic to $\Z_{p^2}^{\times}$, and a theorem of Gauss
(\cite[Theorem 42]{S}) tells us that $\Z_m^{\times}$ is cyclic if
and only if $m=2$, $4$, $p^k$, or $2p^k$ for some odd prime $p$ and
$k\geq 1$. Since this is the case here, we have
$\aut(\Z_{p^2})\simeq \Z_{p^2-p}$, and, because $\aut(\Z_{p^2})$ is
cyclic, it will have exactly one subgroup $H$ of order $p$. Hence we
may map $\Z_p$ nontrivially into $H$ by sending a fixed generator
$g$ of $\Z_p$ into any generator of $H$. This gives a semidirect
product $\Z_{p^2}\rtimes \Z_p$ of order $p^3$.

For the second semidirect product, look at $\aut(\Z_p\times \Z_p)$.
We may also view $\Z_p\times \Z_p$ as $GL_2(\F_p)$ the
$2$-dimensional vector space over $\F_p$, the finite field with $p$
elements. It has order $(p^2-1)(p^2-p) = p(p-1)^2(p+1)$. By Cauchy's
Theorem, then, it has a subgroup of order $p$. Mapping a generator
of $\Z_p$ into this subgroup gives another semidirect product,
$(\Z_p\times\Z_p)\rtimes\Z_p$ of order $p^3$.

Now in the general case where $n = p^3 m$, we have at least five
distinct groups:
\begin{align*}
&\Z_p\times\Z_p\times\Z_p\times\Z_m\\
&\Z_{p^2}\times\Z_p\times\Z_m\\
&\Z_{p^3}\times\Z_m\\
&(\Z_{p^2}\rtimes\Z_p)\times\Z_m\\
&[(\Z_p\times\Z_p)\rtimes\Z_p]\times\Z_m.
\end{align*}
\end{proof}

We would now like to limit the possibilities for $2$-group numbers
further, namely to $n$ which are \emph{almost square-free}, by which
we mean numbers which are either square-free or whose prime
factorization contains just one square.

\begin{prop}
If the prime factorization of $n$ contains two distinct squares,
then $n$ is at least a $4$-group number.
\end{prop}
\begin{proof}
Let $n = p^2 q^2 m$, where $p$ and $q$ are primes and $m\geq 1$ is
an integer. If $p|m$ or $q|m$, then $n$ is not cube-free, and hence
at least a $5$-group number. So assume $(p,m) = (q,m) = 1$. Then we
may write four distinct abelian groups:
\begin{align*}
&\Z_p\times\Z_p\times\Z_q\times\Z_q\times\Z_m\\
&\Z_{p^2}\times\Z_q\times\Z_q\times\Z_m\\
&\Z_p\times\Z_p\times\Z_{q^2}\times\Z_m\\
&\Z_{p^2}\times\Z_{q^2}\times\Z_m.
\end{align*}
Again, glancing at orders of elements convinces one that these
groups are distinct.
\end{proof}

\begin{rem}
It is possible for an $n$ with two distinct squares in its
factorization to be exactly a $4$-group number. This will happen if
and only if $n$ has nilpotent factorization: if it does, then it is
an abelian number, and the above groups are the only groups of order
$n$; and if it does not, then there are the above groups plus at
least one non-nilpotent group. An example of an $n$ with exactly $4$
groups is $n = 5^2 7^2 = 1125$. (This is by no means, however, the
smallest $4$-group number.)
\end{rem}

We are almost ready for the characterization of $2$-group numbers;
we just need one more well-known result as a lemma. The proof is
elementary, but relatively involved, so we will omit it here. For
details, see \cite[Example 10.46]{V}.

\begin{lem}
Let $G$ be a finite group of order $|G|=n$, and let $p$ be the
smallest prime dividing $n$. If $G$ contains a subgroup $H$ of index
$p$, then $H\trianglelefteq G$.
\end{lem}

Now, at last, we come to the classification of $2$-group numbers.

\begin{thm}
$n$ is a $2$-group number if and only if one of the following holds:\\
(1) $n$ is square-free, and fails to have nilpotent factorization
for exactly one pair of primes; or\\
(2) $n=p^2 q_1\cdots q_r$ where $p, q_1,\ldots,q_r$ are distinct
primes (we allow $r=0$, in which case $n = p^2$) and $n$ is an
abelian number (which means $n$ has nilpotent factorization).
\end{thm}
\begin{proof}
We have already seen that $n$ must be almost square-free. Suppose
first that $n$ is, in fact, square-free. If $n$ fails to have
nilpotent factorization in more than one way, then, following the
method of Pakianathan and Shankar, we may construct two
nonisomorphic, non-nilpotent groups of order $n$. Together with
$\Z_n$, this gives three groups already. So we need to show that if
$n$ fails to have nilpotent factorization in exactly one way, then
there are only two groups of order $n$: namely, $\Z_n$ and
\[
(\Z_{p_1}\rtimes \Z_{p_2})\times \Z_m,
\]
where $m= n/p_1 p_2$. (This being the non-nilpotent group given by
the method in \cite{PS}.) The difficulty will be showing that there
is precisely one non-abelian group of order $n$. This we will do
using induction on $r$, the number of prime factors of $n$.

The first case is with $r=2$. Let $G$ be a non-abelian group of
order $n = pq$ for some primes $p, q$ with $p\equiv 1\pmod{q}$. By
the lemma, a Sylow $p$-subgroup is normal, hence unique. Call it
$N$, and let $H$ be any Sylow $q$-subgroup. Since $p$ and $q$ are
prime, it follows from Lagrange's Theorem that $N\cap H = \{1\}$.
Also, $NH$ is a subgroup of order
\[
|NH| = \frac{|N||H|}{|N\cap H|} = pq = n,
\]
so in fact $NH = G$. We can then write $G$ as an (internal)
semidirect product $G=N\rtimes H$. Moreover, we claim any two such
semidirect products are isomorphic. If this is the case, we will
have shown there is exactly one non-abelian group of order $n=pq$.
In fact, it will be helpful later on to understand $\aut(G)$, so we
will investigate its structure now, and on the way see that the
claim is true. Let $n_q$ denote the number of Sylow $q$-subgroups of
$G$. Then by the Sylow theorems, we know that $n_q\equiv 1\pmod{q}$
and $n_q|p$. However, $n_q$ cannot equal $1$, since this would mean
$H$ was normal, and then we would have $G=N\times H$ a direct
product of cyclic groups, contradicting $G$ non-abelian. Thus there
are exactly $p$ Sylow $q$-subgroups.

Fix $n$ and $h$, generators for $N\simeq\Z_p$ and $H\simeq\Z_q$
respectively. Recalling that every element of a semidirect product
$N\rtimes H$ can be written uniquely as the product of an element of
$N$ with one of $H$, we see that every element of $G$ can be
uniquely written in the form $n^a h^b$ for some $a,b$ satisfying
$0\leq a\leq p-1$ and $0\leq b\leq q-1$. Let $\phi\in\aut(G)$. Since
$\phi$ fixes orders of elements, it must map $n\mapsto n^s$ and
$h\mapsto\tilde{h}$ where $1\leq s\leq p-1$ and $\tilde{h}$ is a
generator for some Sylow $q$-subgroup. In this case for a typical
element $n^a h^b\in G$ we have
\begin{align*}
\phi(n^a h^b) &= \phi(n)^a\phi(h)^b\\
&=(n^s)^a \tilde{h}^b\\
&=(n^a)^s \tilde{h}^b
\end{align*}
Not every such map will give an automorphism, however; to see which
ones do, we note that $\phi$ is a homomorphism, so we must have
\[
\phi(n^a h^b n^{a'} h^{b'})=\phi(n^a h^b)\phi(n^{a'}h^{b'})
\]
for all $a,a',b,b'\in\Z$. Using $\phi(n^a h^b)=(n^a)^s \tilde{h}^b$
and the fact that $h^b n^{a'} h^{-b}\in N$ (since $N$ is normal), we
obtain for the LHS:
\begin{align*}
\phi(n^a h^b n^{a'} h^{b'}) &= \phi(n^a h^b n^{a'} h^{-b} h^b h^{b'})\\
&= \phi(n^a h^b n^{a'} h^{-b} h^{b+b'})\\
&= (n^a h^b n^{a'} h^{-b})^s \tilde{h}^{b+b'}\\
&= n^{as} (h^b n^{a'} h^{-b})^s \tilde{h}^{b+b'}\\
&= n^{as} h^b n^{a's} h^{-b} \tilde{h}^{b+b'}.
\end{align*}

The RHS gives
\begin{align*}
\phi(n^a h^b)\phi(n^{a'}h^{b'}) &=
n^{as}\tilde{h}^{b}n^{a's}\tilde{h}^{b'}\\
&=n^{as}\tilde{h}^{b}n^{a's}\tilde{h}^{-b}\tilde{h}^{b}\tilde{h}^{b'}\\
&=n^{as}\tilde{h}^{b}n^{a's}\tilde{h}^{-b}\tilde{h}^{b+b'}.
\end{align*}
Canceling where possible, we see that the LHS=RHS if and only if
\begin{align*}
& \ \ \ \ h^b n^{a's} h^{-b} = \tilde{h}^{b}n^{a's}\tilde{h}^{-b}\\
\Longleftrightarrow &\ \ \ \  \tilde{h}^{-b} h^b n^{a's} =
n^{a's}\tilde{h}^{-b} h^b,
\end{align*}
which says precisely that $\tilde{h}^{-b} h^b$ and $n^{a's}$
commute. But this means that $\tilde{h}^{-b} h^b$ is in the
centralizer subgroup $Z(n^{a's})$, and since $G$ is non-abelian, we
must have simply $Z(n^{a's}) = N$. This shows that $\phi$ is a
homomorphism if and only if $\tilde{h}^{-b} h^b\in N$, for all
$b\in\Z$. We now claim this inclusion holds for all $b$ if and only
if it holds for $b=1$. Another way of saying this is that, if
$\tilde{h}^{-1} = n^{\ell} h^{-1}$ for some $\ell$, then
$\tilde{h}^{-b} h^b\in N$ for all $b$. Simply induct on $b$: if
$\tilde{h}^{-b} = n^k h^{-b}$ for some $k\in N$, then multiply on
the left by $\tilde{h}^{-1} = n^{\ell} h^{-1}$ to obtain
\begin{align*}
\tilde{h}^{-1}\tilde{h}^{-b} &= n^{\ell} h^{-1} n^k h^{-b}\\
&= n^{\ell} h^{-1} n^k h h^{-1} h^{-b},
\end{align*}
which shows that $\tilde{h}^{-(b+1)} h^{b+1}\in N$.

So now we know that $\phi$ is a homomorphism if and only if there is
some $\ell\in\Z$ such that $\tilde{h}^{-1}=n^{\ell}h^{-1}$. But
since each element of $G$ has a unique representation as $n^a h^b$,
then for a given $\ell$ ($0\leq\ell\leq p-1$), this equation
uniquely determines an $\tilde{h}$ for which the map $\phi$ is a
homomorphism sending $n\mapsto n^s$ and $h\mapsto
\tilde{h}=(n^{\ell}h^{-1})^{-1}$. So finally we can count the number
of possible automorphisms: we can pick any $s$ with $0\leq s\leq
p-1$, and any $\ell$ with $0\leq\ell\leq p-1$, for a total of
$p(p-1)$ choices. So $|\aut(G)| = p(p-1)$. Note also that we can
send $h$ to a member of any Sylow $q$-subgroup we like. This shows
that each of the internal semidirect products mentioned above is
isomorphic, because the construction depends only on the Sylow
$q$-subgroup $H$ chosen, and this completes the base case of the
induction (at last!).

Having shown that $n=pq$ is a $2$-group number if and only if it
fails nilpotent factorization, now assume $n=p_1\cdots p_r$ is the
product of three or more distinct primes and fails nilpotent
factorization for exactly one pair. Relabeling if necessary, we may
assume the failing pair is $p_1$ and $p_2$, with $p_2\equiv
1\pmod{p_1}$. To complete the induction, we need to show there is
exactly one non-abelian group of order $n$. The argument here turns
out to require rather more sophisticated tools, so we will simply
cite the results we need rather than giving a full development.
First, it is well known that every group of square-free order is
solvable (see, for instance, \cite[10.1.10]{R}--the proof relies on
a tool called the transfer). We will also use a theorem of P. Hall
(\cite[9.1.7]{R}) which says that if $G$ is a finite solvable group,
$|G| = n$, and $k$ is any divisor of $n$ such that $\gcd(k,n/k) =
1$, then $G$ contains a subgroup of order $k$, and any two such
subgroups are conjugate.

Now consider two cases: first, suppose $p_1$ is the smallest prime
dividing $n$. Then, by Hall's theorem, we know there is a subgroup
$H$ of index $p_1$. Since $H$ is square-free with nilpotent
factorization, it is cyclic,
$H\simeq\Z_{p_2}\times\cdots\times\Z_{p_r}$. In particular, $H$ is
normal so we may form an (internal) semidirect product
$\Z_{p_1}\rtimes H$. Now consider the automorphism group of $H$.
Since any automorphism preserves orders of elements, given any
direct product whose factors have relatively prime orders, the group
of automorphisms is simply the direct product of the automorphism
groups of the factors. In this case,
\[
\aut(H) \simeq \aut{(\Z_{p_2})}\times\cdots\times\aut{(\Z_{p_r})}.
\]
Since the $p_i$'s are prime, their automorphism groups are cyclic of
order $p_i - 1$.  By assumption, $p_1|p_i-1$ if and only if $i=2$.
So the only way we may form a non-trivial semidirect product
$\Z_{p_1}\rtimes H$ is by mapping $\Z_{p_1}$ into the unique
$p_1$-subgroup of the cyclic group $\aut(\Z_{p_2})$, which gives
exactly one non-abelian group of order $n$.

In the second case, $p_1$ is not the smallest prime dividing $n$.
Necessarily $p_1 < p_2$, as $p_2\equiv 1\pmod{p_1}$; hence the
smallest prime is neither of these. Say it's $p_3$. Again, by Hall's
theorem and the lemma, we know there is a normal subgroup $H$ of
index $p_3$, so we can form an (internal) semidirect product
$\Z_{p_3}\rtimes H$. Now by induction there are exactly $2$ groups
of order $n/p_3$, the cyclic group and one group $G$ given by a
semidirect product, so $H$ must be one of these. If $H$ is cyclic,
then the semidirect product $\Z_{p_3}\rtimes H$ must be trivial,
since $\aut(H)$ does not have a subgroup of order $p_3$. This
trivial semidirect product gives one group of order $n$, namely
$\Z_n$. The other possibility is that
\[
H = (\Z_{p_1}\rtimes
\Z_{p_2})\times\Z_{p_4}\times\cdots\times\Z_{p_r}.
\]
But in this case, since the automorphism group of a direct product
is the direct product of the automorphism groups (as long as the
groups' orders are relatively prime!), we have
\[
\aut(H) = \aut(\Z_{p_1}\rtimes
\Z_{p_2})\times\aut(\Z_{p_4})\times\cdots\times\aut(\Z_{p_r}).
\]
Here is where it's useful to know that $|\aut(H)| = p(p-1)$, as this
means $|\aut(H)|$ is not divisible by $p_3$. In particular, the
semidirect product $\Z_{p_3}\rtimes H$ must again be trivial, so
that actually
\[
\Z_{p_3}\rtimes
H=(\Z_{p_1}\rtimes\Z_{p_2})\times\Z_{p_3}\times\cdots\times\Z_{p_r}.
\]
This is precisely the semidirect product from \cite{PS}, and since
these are the only two groups possible, $n$ is indeed a $2$-group
number.

In terms of our goal of classifying the $2$-group numbers, this
leaves only the relatively trivial case where $n$ is not
square-free. Indeed, in this case we know from previous work that
$n=p^2 q_1\cdots q_r$ with $p, q_1,\ldots,q_r$ distinct primes. Put
$m = q_1\cdots q_r$. The Classification Theorem for finite abelian
groups gives us two abelian groups of order $n$:
\begin{align*}
& \Z_p\times \Z_p\times \Z_m\ \ \ \textrm{and}\\
& \Z_{p^2}\times Z_m\
\end{align*}
If $n$ is an abelian number, then these are precisely all the groups
of order $n$, so $n$ is indeed a $2$-group number. Conversely, if
$n$ is not an abelian number, then there is some non-abelian group
of order $n$ as well, so $n$ is not a $2$-group number. This
completes the proof.
\end{proof}

\begin{ack}
The author would like to thank Dr.\ Krishnan Shankar of the
University of Oklahoma for many invaluable discussions.
\end{ack}

%\bibliographystyle{alpha}{ABcd}
\begin{thebibliography}{PS}

\bibitem[PS]{PS} %[PS00]
J.\ Pakianathan and K.\ Shankar, \textit{Nilpotent Numbers}, Amer.
Math. Monthly, August--September 2000, 631--634.

\bibitem[R]{R}
Derek J.S.\ Robinson, \textit{A Course in the Theory of Groups, 2nd
ed.}, New York, Springer-Verlag, 1996.

\bibitem[S]{S}
Daniel Shanks, \textit{Solved and Unsolved Problems in Number
Theory}, Washington D.C., Spartan, 1962.

\bibitem[V]{V}
E.B.\ Vinberg, \textit{A Course in Algebra}, Providence, RI, Amer.
Math. Soc., 2001.

\end{thebibliography}
\end{document}