The cyclic subgroups of $p^2$ order non-cyclic group are normal

Question

I’m having a hard time on proving that every cyclic subgroup of $p^2$ order group is a normal subgroup, where $p$ is a prime number. I’m not going to use the truth that $p^2$ order group are abelian, since this is what I want to show through the proof.

I tried to show that $gag^{-1}$ belongs to $\langle a\rangle$, where $g$ is in $G$ but not in $\langle a\rangle$, any hints on how to check this? For $p=2$, I can check this by contradiction, but for higher $p$, I have no idea, maybe I should use induction?

Answer 1

Let the group be $G$ of order $p^2$. You see any subgroup of $G$ can have order $1,p$ or $p^2$. If the order of the cyclic subgroup is $1$ or if it is $p^2$ then we are done .

Now we look at the cyclic subgroup $H=\langle a\rangle$ of order $p$ . We are to show that $H$ is a normal subgroup of $G$. If not then $\exists g\in G$ such that $gag^{-1}\not\in H$ Observe that order or $gag^{-1}$ is $p$, also observe that $ \langle gag^{-1} \rangle \cap H =\{e\}$.

Then the cosets of $ \langle gag^{-1} \rangle $ in $G$ are $ \langle gag^{-1} \rangle,a\langle gag^{-1} \rangle,a^2 \langle gag^{-1} \rangle,\cdots,$ and $ a^{p-1}\langle gag^{-1} \rangle $. Now $g^{-1}\in a^{i}\langle gag^{-1} \rangle$ for some $1\le i \le p-1$.

$\therefore g^{-1}= a^{i}(gag^{-1})^j$ for some $1\le j\le p-1 $. Then $$g^{-1}= a^iga^jg^{-1}\implies e= a^iga^j\implies g=a^{-i-j}\in H \implies gag^{-1}\in H $$.

So a contradiction. Hence we are done.

Answer 2

Let $H$ be a subgroup of $G$, $|H|$ divides $|G|$ implies that $|H|=1,p,p^2$.

Suppose $|H|=p$ and consider the action of $f:G\rightarrow \text{Sym}(G/H)$ on the quotient space $G/H$ such that $g(aH)=gaH$. Since its image is contained in the symmetric group of $G/H$ which has order $p!$, the kernel of $f$ is not trivial. Let $g$ in $\ker(f)$ different of the neutral, $gH=H$ implies that $g\in H$ and $g$ generates $H$ since the order of $H$ is prime.

Answer 3

It's pretty easy to see that a group of order $p^2$ is abelian, as @ArturoMagidin points out.

For the proof, using the class equation one can see that the center is nontrivial (this true of any $p$-group). Then $G/Z(G)$ is cyclic, and the result follows.

Alternatively, and this was also touched upon in the comments, if the index of a subgroup is the smallest prime dividing the order of the group, then the subgroup is normal, by a theorem.

Answer 4

Let $H$ be a subgroup of order $p$. By contradiction, suppose that $H$ is non-normal; then, there is $\tilde g\in G$ such that $gH\tilde g^{-1}\ne H$, and hence $H\cap \tilde gH\tilde g^{-1}$ is trivial (Lagrange). But then the $G$-action by left multiplication on the left quotient set $G/H$ (whose kernel is $\bigcap_{g\in G}gHg^{-1}$) is faithful, namely $G$ embeds into $S_{[G:H]}=S_p$: contradiction, because $p\nmid(p-1)!$, and hence $p^2\nmid p!$. Therefore, all the subgroups of order $p$ are normal.