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I have some doubt in the following statement: Let $H$ and $K$ be two finite Groups and let $G$ denote their direct product. If $S＜G$ , then does $S$ must be of the form $H'\times K'$ for some subgroups $H'$ and $K'$ of $H$ and $K$ respectively?

This is my idea: for all $(h, k)\in S$ , we must have $$ (h_1, k_1)(h_2, k_2)^{-1}=(h_1h_2^{-1}, k_1k_2^{-1})\in S\ , $$ as desired.

Arturo Magidin
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Quay Chern
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    Think of the case $H = K = \mathbf{Z}/2$. – ronno Sep 19 '23 at 01:38
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    No, this is not true, and your argument does not say anything about whether it's true or not. See Goursat's lemma: https://en.wikipedia.org/wiki/Goursat%27s_lemma – Qiaochu Yuan Sep 19 '23 at 01:40
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    If $|H|$ and $|K|$ are not relatively prime, then just find elements $h\in H$ and $k\in K$ of the same order, and look at $\langle (h,k)\rangle$.The result holds for finite groups if and only if $\gcd(|H|,|K|)=1$. – Arturo Magidin Sep 19 '23 at 01:53
  • @ronno Thanks for your example, I have realized it. – Quay Chern Sep 19 '23 at 02:38
  • Yes, my argument just show that the first component forms a subgroup of $H$ and the second one forms a subgroup of $K$ , but not necessarily the direct product of these two subgroups. Thanks. @QiaochuYuan – Quay Chern Sep 19 '23 at 03:03
  • Thanks for giving the general case of this problem. @ArturoMagidin – Quay Chern Sep 19 '23 at 03:05
  • @ArturoMagidin Is that any proof of your latter statement, I need to use this result, thanks. – Quay Chern Mar 13 '24 at 01:58
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    @DianWei https://math.stackexchange.com/q/4212538/742 – Arturo Magidin Mar 13 '24 at 02:09

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