I'm being asked to show: Show a finite $p$-group $G$ has a supersolvable series, i.e. a normal series

$$G=G_0\ge G_1\ge\cdots\ge G_m=1$$

such that each factor group is cyclic and each $G_i$ is normal in $G$.

Here are some facts about $p$-groups I know which may or may not help:

(1) $G$ has order $p^n$ for some $n$

(2) $G$ has a normal subgroup of order $p^k$ for each $0\le k\le n$

(3) Given any proper subgroup of order $p^k$ it is contained within some subgroup of order $p^{k+1}$.

I'm not seeing how to put these facts together to show $G$ has a supersolvable series, or if I need anything else. I'd also prefer a push in the right direction, rather than just being given the answer.


A finite $p$-group has a non-trivial center, and any finite abelian group whose order is divisible by $p$ has an element of order $p$. Let $H$ be the cyclic group generated by $p$. Then $G/H$ is a finite $p$ group, so argue inductively.

  • $\begingroup$ This worked. Thanks! $\endgroup$ – Alex Mathers Mar 7 '16 at 1:44
  • $\begingroup$ You're quite welcome. $\endgroup$ – Tim Raczkowski Mar 7 '16 at 1:46

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