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 $p$.
 A: 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.
A: Hint: Can you find an element of order $p$?
A: 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 $p$. 
A: 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 $p^k.$
A: 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 $G$.
A: I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is the same problem as Problem 5 on p.75 in Herstein's book.

Using Lemma 2.9.1 prove that a group of order $p^2$, where $p$ is a prime number, must have a normal subgroup of order $p$.

Lemma 2.9.1 on p.73 in Herstein's book:

If $G$ is a finite group, and $H\neq G$ is a subgroup of $G$ such that $o(G)\not\mid i(H)!$ then $H$ must contain a nontrivial normal subgroup of $G$. In particular, $G$ cannot be simple.

Note that $i(H):=(G:H)$.
My solution is here:

If $G$ is cyclic, $G=\{e,g,\dots,g^{p^2-1}\}$ for some $g\in G$.
Then, $o(g^p)=p$.
So, $G$ has a subgroup $H$ whose order is $p$.
Since $G$ is cyclic, $G$ is abelian.
So, $H$ is a normal subgroup of $G$ whose order is $p$.
If $G$ is not cyclic, $o(g)=p$ for all $g\in G-\{e\}$.
Let $H:=\{e,g,\dots,g^{p-1}\}$.
Then, $H\neq G$ and $o(G)=p^2\not\mid p!=i(H)!$.
So, by Lemma 2.9.1, $H$ must contain a nontrivial normal subgroup $H^{'}$ of $G$.
Since the set of all the subgroups of $H$ is $\{\{e\},H\}$, $H^{'}$ must be equal to $H$.
So, $o(H^{'})=p$.
So, $G$ has a normal subgroup of order $p$.

