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By a $p$-group, I mean a group $G$ of order $p^n$, for some prime $p$ and some $n>0$.

So far, I have managed to prove that $G$ must have a nontrivial centre $Z(G)$ and that $Z(G)$ must therefore be a p-group itself (as the order of $Z(G)$ must divide that of $G$). From here, how do I then prove that $G/Z(G)$ must always be abelian, and hence prove that $G$ is solvable as all p-groups G will have the normal series $G>Z(G)>(e)$?

Thank you

Stefan4024
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Pianoman1234
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1 Answers1

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We know that the center is an abelian subgroup in each subgroup. Then consider $G/Z(G)$, if it's abelian we're done. Otherwise take $Z(G/Z(G))$, which is a normal subgroup in $G/Z(G)$ and by the canonical homomorphism $\gamma: G \to G/Z(G)$ it's pre-image is normal in $G$ and let it denote by $Z_1(G)$. Note that $Z_1(G)/Z(G)$ is abelian, as a subgroup of $G/Z(G)$. Now consider $G/Z_1(G)$ and so on. Eventually you will find reach a point where $G/Z_n(G)$ is abelian.

We have that $Z_{i}(G) \not = Z_{i+1}(G)$ as $G/Z_i(G)$ is a p-group and it's center is nontrivial.

Maybe I should have said use the hint multiple times.

Anyway I think a better way to prove this is to make use of the Sylow's Theorem and use that $G$ has a normal subgroup $H_1$ of order $p^{n-1}$, which has a normal subgroup $H_2$ of order $p^{n-2}$ and so on.

Stefan4024
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  • Why does that mean G/Z(G) must be solvable then? I understand though that if Z(G) and G/Z(G) are solvable then G must therefore be solvable as well. – Pianoman1234 Oct 09 '16 at 09:01
  • @DanielePilkington-Scimone I will edit the proof to make things more clear. – Stefan4024 Oct 09 '16 at 09:15
  • thank you ffffttftffrdtrs – Pianoman1234 Oct 09 '16 at 09:18
  • Thank you for your clear explanation. Can I just ask at the end of the first paragraph, how do you know that you will reach a point where G/Zn(G) is abelian? – Pianoman1234 Oct 09 '16 at 09:33
  • @DanielePilkington-Scimone Note that $Z_n(G)$ is a p-group too and it's of an order greater than the one of $Z_{n-1}(G)$, which I explained in the second paragraph. So eventually you will reach $Z_k(G) = G$. Actually if $G/Z_{k-1}(G)$ is abelian, then $Z_k(G) = G$ – Stefan4024 Oct 09 '16 at 09:37
  • @DanielePilkington-Scimone Put as I've said I would rather use the second proof I written, using the Sylow's Theorem. – Stefan4024 Oct 09 '16 at 09:37
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    I understand that it would be easier to use the Sylow Theorems, it is just that in the book I am reading Sylow Theorems are mentioned in the next chapter after this exercise. – Pianoman1234 Oct 09 '16 at 09:43
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    Just as reference, I found a rather good proof here under Theorem 1: http://plaza.ufl.edu/thanos/Text%20Files/solvable.pdf – Pianoman1234 Oct 09 '16 at 10:04