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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 $G^{(n)} := <a>$ (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!

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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? – user641 Oct 18 '12 at 20:20
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 ...). – Nicky Hekster Oct 18 '12 at 21:10
@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... – nate Oct 18 '12 at 22:05
What do you need a subgroup of order $\,p\,$ for in this problem? Of course this follows at once from Cauchy's theorem... – DonAntonio Oct 18 '12 at 22:10
@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? – nate Oct 18 '12 at 22:17

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