# Every group of order $p^2$ has a normal subgroup of order $p$.

Prove that if $p$ is a prime and $G$ is a group of order $p^\alpha$ for some $\alpha \in \mathbb{Z}^+$, then every subgroup of index $p$ is normal in $G$. Deduce that every group of order $p^2$ has a normal subgroup of order $p$.

If $G$ is a finite group of order $n$ and $p$ is the smallest prime dividing $|G|$, then any subgroup of index $p$ is normal. Since $p$ is a prime, $p$ is the smallest prime dividing $|G|$, hence every subgroup of index $p$ is normal in $G$.

For second one, I can show until that any subgroup $H$ with order $p$ is normal in $G$. But I was not able to come up with one, except for using Sylow's theorem. But this excercise appears before that. Maybe I'll end up proving Sylow's theorem to show the existence of order $p$ subgroup?

My very basic solution $\ \ \$ $^\dagger$Compare to Mariano's fancy congugacy

Given $p$ a prime and $G$ is a group of order $p^2$, then every subgroup of index $p$ is normal in $G$. This is equivalent to say that any subgroup of order $p^2/p = p$ is normal in $G$.

So now we set out to show that $G$ must have a subgroup with order $p$.

Suffice to show that $G$ must have a cyclic subgroup with order $p$. The order of a cyclic group is equal to the order of its generator. Since by Lagrange's Theorem, the order of the subgroup divides the group, we know that the order of cyclic subgroup is either 1, $p$, or $p^2$. Hence, the order of the generator needs to be 1, $p$, or $p^2$ respectively.

When the order of generator is 1, iff the generator is identity. We skip this case and show a cyclic subgroup of order $p$ exist when the order of generator is $p$ or $p^2$. If the order of the generator is $p$, we are done.

Finally, we check that the order of the generator is $p^2$. However, we know that all order $p^2$ cyclic group is isomorphic to $\mathbb{Z}/p^2\mathbb{Z}$. And in this group, $p$ has order $p$. Hence in any group isomorphic to $\mathbb{Z}/p^2\mathbb{Z}$, it has an element with order $p$. Hence every group of order $p^2$ has a normal subgroup of order $p$.

Let $G$ be a group of order $p^2$. Let $c_1$, $\dots$, $c_r$ be the conjugacy classes of elements of $G$. The order of each class divides the order of $G$ and $$\sum_{i=1}^r|c_i|=p^2.$$ There is at least one conjugacy class with exactly one element: that of the identity element of $G$. Use this to show that the center $Z$ of $G$ is non-trivial. Now either $Z=G$, and therefore $G$ is abelian, or $Z$ is of order $p$ and $G/Z$ cyclic. In this last case, $G$ is also abelian.

Therefore our group $G$ is abelian, and things got easier.

• Oh! Awesome Mariano! – Tumbleweed Sep 17 '13 at 3:45
• But I think what I eventually did is different. My method looks more rudimentary, hope it is alright...? – Tumbleweed Sep 17 '13 at 3:46
• Also... Can this be solved directly via Cauchy's Theorem..? – Tumbleweed Sep 17 '13 at 3:54
• Which one of Cauchy's theorems? – Mariano Suárez-Álvarez Sep 17 '13 at 3:57
• The one says $p \;|\; |G|$ then $G$ has an element of order $p$. – Tumbleweed Sep 17 '13 at 4:06

Hint: Can you find an element of order $p$?

• Hi Serkan, yes, that is exactly the thing left to do. But I tried all kinds, but still was not able to come up with one so far.. – Tumbleweed Sep 17 '13 at 2:27
• Oh but I have been trying to find a group with order $p$.. – Tumbleweed Sep 17 '13 at 2:28
• I see every element can have order $1,p,p^2$. 1 is contradiction, $p$ we are done, but it is not obvious to me that $p^2$ does not work. – Tumbleweed Sep 17 '13 at 2:45
• No I was not able to come up with one, except for using Sylow's theorem. But this excercise appears before that. Maybe I'll end up proving Sylow's theorem to show the existence of order $p$ subgroup? – Tumbleweed Sep 17 '13 at 2:57
• @Tumbleweed To expand on the hint: if you have an element of order $p^2$, and your group is of order $p^2$, then your group is a very specific group. Can you see which one? – Nick Peterson Sep 17 '13 at 3:01

Here's my solution:

Consider the center $Z(G)$. The following are its possible order: $1,p,p^{2}$

Since $|G|=p^{2}$ then G is a $p-group$ and $p-groups$ have non-trivial center.

This eliminates the possibility that $|Z(G)|=1$.

If $|Z(G)|=p$ then $|G/Z(G)|=p\implies G/Z(G)$ is cyclic $\implies G$ is abelian which is a contradiction since $|Z(G)|=p$.

Therefore $|Z(G)|=p^{2}$ and $G$ is abelian.

Since $G$ is abelian we know that the converse of Lagrange's Theorem holds and we also know that all subgroups of an abelian group are normal.

Thus $G$ has a normal subgroup of order $p$.

Actually, there is something stronger.

If $|G|=p^n$ where $p$ is a prime then for every $k$ s.t. $0\leq k\leq n$ there is a normal subgroup with order $p^k.$

Indeed, answer is a corollary of the fact that every group action induces a homomorphism. Let $\left| G \right| = {p^k}$ and $\left[ {G:H} \right] = p$. Let $X$ be all left cosets of $H$ in $G$. Then, $\left| X \right| = p$. The action of $G$ on $X$ by left translation induces a homomorphism $\sigma :G \to {S_p}$ defined by $g \to {\varphi _g}$ , where ${\varphi _g}\left( {aH} \right) = \left( {ga} \right)H$. Observe that $Ker\sigma \le H$. Since $\left| G \right| = {p^k}$, $\frac{{\left| G \right|}}{{\left| {Ker\sigma } \right|}} = {p^i}$ for some $i \in \left\{ {0,1, \ldots ,k} \right\}$. Since $\frac{{\left| G \right|}}{{\left| {Ker\sigma } \right|}}$ divides $\left| {{S_p}} \right| = p!$ and ${p^i}$ doesn't divide $p!$ for all integer $i \ge 2$, we get that $\frac{{\left| G \right|}}{{\left| {Ker\sigma } \right|}} = 1$ or $p$. If $\frac{{\left| G \right|}}{{\left| {Ker\sigma } \right|}} = 1$, then $G = Ker\sigma$. But this contradicts with $Ker\sigma \le H < G$. Then, $\frac{{\left| G \right|}}{{\left| {Ker\sigma } \right|}} = p$. Hence, clearly, $Ker\sigma = H$.

If $\left| G \right| = {p^2}$, every subgroup of order $p$ has index $p$ and so is normal in $G$.