The question: Let $G$ be group and $H$ a normal subgroup of $G$. Let $p$ be a prime number. If $G/H$ and $H$ are $p$-groups, then $G$ is a $p$-group.

Attempt: We can define the cyclic subgroup $<Ha>$ of $G/H$, for some $a \in G$. By Lagranges Theorem, $ord(Ha)=p$ is a factor of $ord(G/H)$. However, $Ha$ has prime order. So, $(Ha)^{p} = Ha^{p} = H$. Since $H$ is a $p$-group, $a^{p} \in H$. Since $a \in G$, $a^{p} \in G$. So $G$ has prime order. Is this correct so far?

Thank you very much!

  • $\begingroup$ You showed an arbitrary element $a\in G$ has order a power of $p$. This does indeed show $G$ is a $p$-group by contradiction if you invoke Cauchy's theorem (if $G$ had order divisible by a different prime $\ell$, then there would be an element of order $\ell$, a contradiction). But wouldn't it just be easier to use $|G|=|G/H||H|$? $\endgroup$ – arctic tern Nov 27 '16 at 23:57
  • $\begingroup$ So I used the contradiction route, would this be okay so far? $\endgroup$ – user211962 Nov 28 '16 at 2:00

All you need is that $\vert G \vert = \vert G \vert \vert G/H \vert$. Then since both $\vert G \vert$ and $\vert G/H \vert$ are powers of $p$, so is $\vert G \vert$.


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