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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!

Guillermo García Sáez
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  • The center is p-group hence every element has order a power of p. Hence some element has order p. (If $g$ has order $p^n$ then $g^{p^{n-1}}$ has order $p$) – lulu Jul 06 '23 at 16:38
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    Actually, Cauchy is overkill here since $Z(G)$ is an abelian finite $p$-group. – Anne Bauval Jul 06 '23 at 16:41
  • "The Cauchy theorem asserts that the exist some element for every divisor of the order of the group," Well, $p$ is a divisor of the order of $Z(G),$ isn't it? – Anne Bauval Jul 06 '23 at 16:43
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    That is not what Cauchy's Theorem says. If it said that "there exists some element for every divisor of the order of the group", then it would say that every group is cyclic. What it says is that for every prime divisor of the order of the group, there is at least one element of that prime order. – Arturo Magidin Jul 07 '23 at 02:47

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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 $\Bbb Z/p \cong \langle p \rangle \subseteq Z(G)$).

Jonas Linssen
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