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 $p$.
Best Answer
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.