Let $G$ be a p-group. Proof that $Z(G)$ is cyclic if and only if $G$ has a unique normal subgroup $H$ of order $p$.
I am supposed to prove it without using Sylow theorems. I already prove the first implication, I am struggling with the reverse. Here is what I have so far:
I already know that $Z(G)$ is a subgroup of $G$ then $|Z(G)|=p^k$ for some $k\in \mathbb{Z}$. Also, p divides |Z(G)|, by Cauchy's theorem there exists $a\in Z(G)$ such that $|a|=p$. $Z(G)$ is abelian so every subgroup is normal, then $\langle a \rangle $ is a normal subgroup of $G$ of order p. Thus, $\langle a \rangle = H $ and $H\subset Z(G) $ .
From this point, I don't know what else to do. I was trying to prove that $Z(G)=H$, is this true? What else can I try?
Thank you for your help.
Let $H$ be a normal group of order $p$, $G/H$ is a $p$-group, whose center is not trivial and $p:G\rightarrow G/H$ the quotient map.
If $|G/H|>p$ and the its center $H_1$ is distinct of $G/H$, $p^{-1}(H_1)$ is a normal subgroup distinct of $H$. If $G/H$ is commutative, it contains a normal subgroup $H_1$ distinct of $G/H$, take $p^{-1}(H_1)$.
Group with order $p^2$ must be abelian . How to prove that?
If $|G/H|=p$, $G$ is Abelian it is isomorphic to $\mathbb{Z}/p\times\mathbb{Z}/p$ or to $\mathbb{Z}/(p^2)$
There exists only two groups of order $p^2$ up to isomorphism.
https://en.wikipedia.org/wiki/P-group#Non-trivial_center