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Suppose I have a group $G$ which is the (inner) direct product of subgroups $H$ and $K$.

That is, $G = HK$, $H\cap K = \{1\}$, and $H$ and $K$ are both normal subgroups of $G$.

Then for a subgroup $G'$ of $G$, I want to write (in fact a proof I'm trying to follow uses this fact) that $G' = (G'\cap H)(G'\cap K)$.

This looks like it should be true in general. $(\supset)$ is clear, but $(\subset)$ won't work out for me.

Suppose $m\in G'$. Then write $m = hk$ for some $h\in H$, $k\in K$. If $h\in G'$, then so is $k$. Similarly if $k\in G'$, then so is $h$. The case left to deal with is if $h,k$ are both not in $G'$. I cannot rule this out.

Extra hypothesis (in the context of the theorem) that might help: $G'$ is a maximal subgroup of $G$ and $H$ and $K$ are both Sylow subgroups of $G$. $G$ is a finite group.

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At least some of the extra hypotheses are needed. For example, if $H=K$ and $G=H\times K$, then the diagonal $G'=\{(h,h)\mid h\in H\}$ is a subgroup with $G'\cap H=G'\cap K=1$. (And apparently finiteness is not the decicive condition) – Hagen von Eitzen Oct 11 '12 at 17:26
up vote 1 down vote accepted

I shall use the additional properties that $G$ is finite and $H,K$ are Sylow.

As $H$, $K$ are Sylow subgroups of orders $p^n$ and $q^m$, respectively, we see that $G$ has $p^nq^m$ elements. This implies $p\ne q$ (or one of the groups $H,K$ is trivial; in that case the claim is trivial as well).

Then $G'\cap H$ is a $p$-group, hence contained in a $p$-Sylow group of $G'$. This again is contained in a $p$-Sylow $H'$ of $G$, which must be conjugate to $H$. Since $H$ is normal, we conclude $H=H'$. In other words: $G\cap H$ is a $p$-Sylow subgroup of $G'$, of order $p^r$ say. Similarly, $G\cap K$ is a $q$-Sylow subgroup of $G'$, of order $q^s$ say. Then the order of $G'$ must be $p^rq^s$, i.e. $G'=(G\cap H)(G\cap J)$.

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I'm assuming you mean $G'\cap H$ is a $p$-Sylow subgroup of $G'$, but how are you concluding this? All the previous line establishes is that $G'\cap H$ is contained in some Sylow $p$-subgroup of $G'$. – roo Oct 11 '12 at 19:01

I got this using the first Sylow Theorem, with an idea sparked by Hagen's answer above.

Since there exists a conjugate $H'$ of $H$ such that $G'\cap H'$ is a Sylow $p$-subgroup of $G'$. Since $H$ is normal (as Hagen pointed out above) we have $G'\cap H = G'\cap H'$ is a Sylow subgroup of $G'$. And the rest of the argument is what Hagen outlined above.

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