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If $G=G_1\times G_2$ does it follow that $H\leq G \implies H=H_1\times H_2$ where $H_1\leq G_1$ and $H_2\leq G_2$. I am thinking that it does, being that the projection of $H$ into $G_1$ and $G_2$ is a homomorphism, but now that I am writing this it might be that $H\subset H_1\times H_2$ rather than equality.

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marked as duplicate by Derek Holt group-theory Sep 14 '15 at 17:02

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    $\begingroup$ With different letters the same question was asked here. $\endgroup$ – j.p. Sep 14 '15 at 16:48
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No. Consider the case $G_1=G_2$ and $H=\{\,(g,g):g\in G\,\}$.

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