3

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.

Choxom
  • 43
  • I forgot the hypothesize that H is normal. ( fixed up) – Choxom Mar 25 '20 at 20:22
  • If $G/Z(G)$ is cyclic, then $G$ is abelian. Also, every nontrivial finite $p$-group has a nontrivial center. Hence, if $p^m$ is the order of the center of a nonabelian $p$-group of order $p^n$, then we must have $2 \le m \le n-2$. – Geoffrey Trang Mar 25 '20 at 20:33
  • 1
    $Z(G)$ can have only one subgroup of order $p$, and for an abelian $p$-group, this mean that $Z(G)$ is cyclic. – Arturo Magidin Mar 26 '20 at 02:12
  • @ArturoMagidin Thank you, this is much simpler. The proof is here :[https://math.stackexchange.com/questions/948722/p-prime-group-of-order-pn-is-cyclic-iff-it-is-an-abelian-group-having-a-un] – Choxom Mar 26 '20 at 06:04

1 Answers1

2

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

Tsemo Aristide
  • 87,475
  • 1
    To conclude is it correct saying : Then G is isomorphic to $\mathbb{Z}/(p^2)$ because of $ \mathbb{Z}/p \times \mathbb{Z}/p $ having at least 2 different normal subgroups of order p ( ${ (0_p ,a) : a \in \mathbb{Z}/p }$ and ${ ( a, 0_p) : a \in \mathbb{Z}/p }$) ? – Choxom Mar 25 '20 at 22:22