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If a central series is considered as $$G = G_0 \supset G_1 \supset \cdots \supset G_m = \{1\}$$ such that $$G_{i+1} \triangleleft G_i$$ and $$G_i/G_{i+1} \subset Z(G/G_{i+1})$$ then,

Show that finite p-groups admit central series.


What I have so far is this.

If the order of group G is 1 - nothing to prove

If the order of group G is p, then certainly $\{1\} \triangleleft G$ and $G/\{1\} \subset Z(G/\{1\})$

If the order of group G is p^n, then it has a nontrivial center, $Z(G_n) \neq \{1\}$. Choose an element of the center, $a \in Z(G_n)$ s.t. $ord(a) = p$, then let $G^{(n)} := $ (because $p \mid p^n$ and using Cauchy's thm)

Then $G^{(2)} \triangleleft G_2$ and $ord(G_2/G^{(2)})=p$ (using the class equation)


This is where I seem to run out of steam. Does anyone have any suggestions?

Thanks a lot!

  • 2
    Note that a finite group fails to have a central series, only when $G/Z_n(G)$ is eventually centerless (here $Z_n$ is the $n$th center). Now what do you know about the centers of $p$-groups?2012-10-18
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    Use induction on $|G|$ and the fact that since $G$ is a $p$-group $Z(G)$ isn't trivial (hence by induction $G/Z(G)$ has a central series ...).2012-10-18
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    @SteveD I know that a finite p-group has a nontrivial center; I'm also thinking of having to show that there is a subgroup of order $p$ for the p-group...2012-10-18
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    What do you need a subgroup of order $\,p\,$ for in this problem? Of course this follows at once from Cauchy's theorem...2012-10-18
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    @DonAntonio well I was thinking I needed to construct a chain of subgroups but are you suggesting I consider the chain as something else, like factor groups or centers?2012-10-18
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    No. I'm stronlgy advicing you to do what Nicky hinted at you. Inductively, $\,G/Z(G)/,$ has a central series which is *exactly* what you need!2012-10-18

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