$|G| = pq$ with $p$, $q$ primes not necessarily distinct: case $p=q$

Question

I'd like to review the reasoning of this question for the case $p=q$. It says

If $|G|=pq$ where $p$ and $q$ are primes that are not necessarily distinct. Prove that $G$ is abelian or $Z(G)=1$.

This question is already answered here for the case $p \ne q$.
The reasoning is:
$Z(G)$ is a subgroup of $G$ and then $|Z(G)|$ is $1,p,q$ or $pq$. If $|Z(G)| = p$ then $G/Z(G)$ has order $q$ and therefore is cyclic, which means $G$ is abelian and that is a contradiction because in that case we would have $Z(G)=G$.
We conclude $G$ is abelian or $Z(G)=1$.

Now, if $p=q$ what part of the previous reasoning fails? I mean, we still could say "since $|G| = pp$ then $|Z(G)|$ is $1,p$ or $pp$. If $|Z(G)|=p$ then $G/Z(G)$ has prime order and hence is cyclic" and conclude again $G$ is abelian or $Z(G)=1$. However if $G$ has order $p^2$ ($p$ prime) then $Z(G) \ne 1$, according to a further result.
So as much as possible, without involving the class equation I'd like to know what is wrong with that proof for the case $p=q$ (actually this problem is in Dummit&Foote and it is supposed to be solved without class equation).

Nicky Hekster · Answer 1 · 2018-05-09T15:47:09.460

2

If you insist on proving $Z(G)$ is non-trivial in case $|G|=p^2$, without an appeal to the Class Equation, you can reason as follows.

Assume that $|G|=p^2$ and $G$ is not abelian. Then every non-trivial element of $G$ must have order $p$ (if not, then there is some element of order $p^2$ and hence $G$ would be cyclic, whence abelian). Now observe that it is impossible that for all $x \neq 1$, $\langle x \rangle$ is normal in $G$. This would imply $G$ to be abelian: let $x,y \in G-\{1\}$ and put $M=\langle x \rangle$, $N=\langle y \rangle$. Then $M,N \lhd G$, $M \cap N=1$, so $|MN|=\frac{|M|\cdot|N|}{|M \cap N|}=p^2$ and hence $G=MN$. Since $M \cap N=1$, it follows that all elements of $M$ and $N$ commute and hence $G$ is abelian.

Pick a non-identity element $x \in G$ with $H=\langle x \rangle$ not normal. Next, pick a $y \in G-N_G(H)$ with $H \neq H^y$. Then, since $H$ is cyclic of order $p$ and $y \notin N_G(H)$, we must have $H \cap H^y=1$. But then the set $|HH^y|=\frac{|H|\cdot|H^y|}{|H \cap H^y|}=p^2$, implying $G=HH^y$. So $y=hy^{-1}ky$, for some $h,k \in H$. This yields $y=kh \in H$, a contradiction.

So $G$ must be abelian and $Z(G)=G$.

edited May 09 '18 at 15:47

answered May 08 '18 at 11:13

Nicky Hekster

49,281

You are welcome, hope you learned something! This solution you would never see in text books, you know. – Nicky Hekster May 08 '18 at 22:15
Indeed. I am not familiar with the notation $H^y$. Is that the group generated by $y$? – user8989 May 08 '18 at 22:40
No that is $H^y=y^{-1}Hy={y^{-1}hy : h \in H }$, the conjugate subgroup of $H$. – Nicky Hekster May 08 '18 at 23:01
I'm sorry I am having trouble understanding why $H \cap H^y=1$. I am trying this: by contradiction assume there exists $a\in \mathbb{Z}$ such that $y^{-1}x^ay = x^b$ for some $b \ne p$. I guess it is related to $y \in N_G(H)$ in that case? thanks in advance – user8989 May 09 '18 at 00:44
You have a point +1 for you, excellent. I adapt the proof. – Nicky Hekster May 09 '18 at 06:47
thanks a lot for your help – user8989 May 11 '18 at 04:56

score 1 · Accepted Answer · answered May 08 '18 at 05:23

No part fails. To prove the result you want, there is no need to split into cases. (If fact, in the link you provided, none of the answers assume $p\neq q$ as far as I can see. In that sense, your question is an exact duplicate of that one.)

On the other hand, it turns out that when $p=q$, the case $Z(G)=1$ cannot happen, which is an interesting and important fact, but not what is asked here. (The usual proof of this uses the class equation.)

$|G| = pq$ with $p$, $q$ primes not necessarily distinct: case $p=q$

2 Answers2