A cyclic subgroup of a $p$-group

Question

Let $G\ne\{e\}$ be a $p$-group. Then there exists $x\in Z(G)$ such that $x\ne e$. Let $k>0$ such that $x^{p^k}=e$: i.e. let $p^k$ be the order of the element $x$. I would like to show that the order of the element $x^{p^{k-1}}$ is $p$. Obviously $$\left(x^{p^{k-1}}\right)^p=x^{p^{k-1}p}=x^{p^{k-1+1}}=x^{p^k}=e.$$

Let $n>0$ such that $\left(x^{p^{k-1}}\right)^n=e$. I am having difficulty showing that $p\leq n$. Any suggestions?

Answer 1

I will use this proposition from Dummit and Foote page 57:

Proposition 5. Let $G$ be a group, let $x\in G$ and let $a\in\mathbb{Z}-\{0\}$.

(2) If $|x|=n<\infty$, then $|x^{a}|=\frac{n}{(n,a)}$.

Since $|x|=p^{k}$, we have $|x^{p^{k-1}}|=\frac{p^{k}}{(p^{k},p^{k-1})}=\frac{p^{k}}{p^{k-1}}=p$.

If you do not want to use the proposition, then let $n$ be as you defined. Then $(x^{p^{k-1}})^{n}=x^{p^{k-1}n}=e$. Since $|x|=p^{k}$, we have $p^{k}\leq p^{k-1}n$. So $p\leq n$.

Answer 2

You want the minimum $m\in \mathbb{N}$ such that $p^{k-1} \cdot m = 0 \mod p$, because $p$ is a prime number and the unique prime decomposition $m$ must be $p$.