Cauchy's theorem for groups use unclear

Question

Im reading about the Burnside's $p^aq^b$ theorem proof, and for the case when $b=0$ it uses that, since the order of the group is a power of a prime, by an elementary result of group theory we have that the center of the group is non trivial. Then we can choose $g\in Z(G)$, and we can take this element with order $p$ by the Cauchy's Theorem. The Cauchy theorem asserts that the exist some element for every divisor of the order of the group, but it doesn't say anything about the subgroups of the group. I don't see why the center must contain such an element. Thanks in andvance!

Answer 1

You know that your group $G$ has order $p^a$. You also know that the subgroup $Z(G) \subseteq G$ is nontrivial. Thus it has order $p^c$ for some $1 < c\leq a$. Hence by Cauchy it contains an element $g\in Z(G)$ of order $p$ (and in particular the subgroup $\Bbb Z/p \cong \langle p \rangle \subseteq Z(G)$).