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It is very tempting to perform this step, but I feel like it is not true. I couldn't come up with a counterexample though.

user103461
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3 Answers3

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No: $H = \mathbb{Z}_4$, $K = \langle 2 \rangle$.

More generally: $H = $ your favorite cyclic group, $K = $ your favorite non-trivial subgroup of $H$. As Dan Shved points out below, this doesn't always work - but it does give infinitely many counterexamples.

  • Thanks. I'll post another question, since this step came up in a specific proof. I am guessing there is some reason it holds in that particular case. – user103461 Oct 27 '13 at 05:25
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    The idea after "more generally" doesn't always work. If $|K|$ and $|H/K|$ are coprime, then $H \simeq G \times K$. – Dan Shved Oct 27 '13 at 05:34
  • @DanShved Why is that true? If $|K|$ and $|H/J|$ are coprime, then $H\cong K\unlhd_\varphi G$ for some $\varphi$. – Alex Youcis Oct 27 '13 at 06:37
  • @Alex What does $\unlhd_\varphi$ mean? Is it the semidirect product? If so, remember that in the construction above $H$ is cyclic, hence abelian, so $\varphi$ must be a trivial homomorphism. – Dan Shved Oct 27 '13 at 09:01
  • @DanShved It was supposed to be $\rtimes_\varphi$. I thought you were referring to some general setup, I didn't see that we were only considering abelian groups. – Alex Youcis Oct 27 '13 at 09:12
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Let $S_4$ be symmetric group with $4$ symbols and $A_4\leq S_4$ is alternative group. As you know, $A_4$ is a normal subgroup because $[S_4:A_4]=2$.

Suppose that your statement is true. Therefore, we should have $S_4\cong A_4\times \mathbb Z_2$ as, $\dfrac{S_4}{A_4}\cong \mathbb Z_2$.

It is not hard to see that the group $A_4\times \mathbb Z_2$ has no element with order $4$. However, element $(1\,2\,3\,4)\in S_4$ is order $4$.

Bobby
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$$H=\mathbb Z \,;\, K =n \mathbb Z$$

N. S.
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