I have a group $G$ and two subgroups $H,K$. Both $H$ and $K$ are normal in $G$, and $H\cap K = {id}$. Is this enough information to say that $G$ is the internal direct product of $H$ and $K$, or must I also show that $G = HK$? My advisor told me that was unnecessary, but he didn't go into detail why that was, and I don't see exactly why it's unnecessary.

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    $\begingroup$ You do need something equivalent to $HK = G$, though this might just be that $|H||K| = |G|$ for example. Without this it is clearly false (take for example both to be the trivial subgroup). $\endgroup$ Mar 13, 2016 at 19:10
  • $\begingroup$ Right, I have actually shown $|H||K| = |G|$ previously. I thought that maybe I needed something else besides their size to show it. Thank you! $\endgroup$
    – Auclair
    Mar 13, 2016 at 19:14
  • $\begingroup$ @Auclair Use the fact that if $A$ and $B$ are sets of same (finite) cardinality and $f:A\to B$ is injective, then $f$ is actually bijective (apply this to a canonical map $H\times K\to G$). $\endgroup$ Mar 13, 2016 at 19:23
  • $\begingroup$ @Auclair: In case it's not clear from the answer, the identity you want is $|HK| = |H||K| / |H \cap K|$, which is valid for any finite subgroups $H$ and $K$, whether or not $HK$ is a subgroup. In your case, you can conclude from this identity that $|G| = |HK|$, and therefore $G = HK$. $\endgroup$
    – user169852
    Mar 15, 2016 at 4:24
  • $\begingroup$ What you've stated alone is certainly not sufficient. Take $G$ to be any nontrivial group and let $H=K=id$. Then $H\cap K= id$ and both $H$ and $K$ are normal in $G$, but $G\neq H\times K$. $\endgroup$
    – Tyler
    Apr 1, 2017 at 20:49

1 Answer 1


How many different elements are there of the form $hk$ for $h\in H$ and $k\in K$? Can any of them be the same? Consider $h_1k_1=h_2k_2$. We then have $h_2^{-1}h_1=k_2k_1^{-1}$. The left is an element of $H$ and the right an element of $K$, so since these are the same the element is in the intersection.

This is enough to show there are $|H||K|$ elements in the product. As discussed in the comments, this could fail to be the whole group, in which case the group is not the direct product. Since you know in your case that $|H||K|=|G|$ you can, with this additional fact, conclude that it is the direct product.

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    $\begingroup$ This does not really seem to answer the question, since the relevant part was whether it was automatic that $HK = G$. $\endgroup$ Mar 13, 2016 at 19:24
  • $\begingroup$ @Tobias thanks, fixed. $\endgroup$ Mar 13, 2016 at 19:28

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