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Let $G$ be a finite group, and $H,K$ be proper non-trivial subgroups such that $H\cap K=1$, and $HK=G$. Is it necessary that one of these subgroups is normal in $G$?

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No, neither of the subgroups needs to be normal.

Take for example the alternating group $A_5$ which has a subgroup of order 12 (a copy of $A_4$) and one of order 5 (the 5-Sylow). Now clearly, these intersect trivially, as their orders are coprime. But since the product of their orders is exactly $60 = |A_5|$ this means that their product is all of $A_5$. Since $A_5$ is simple, neither of these can be normal.

For more information, see the question Has this "generalized semidirect product" been studied? where I got some very nice answers about this particular phenomenon.

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I thought I was the only person who remembered my answers! I'm glad other people don't just read them and forget... – user1729 Sep 17 '12 at 13:59

Please have a look at this link, There, @GeoffRobinson presented another neat example in his first comment. :)

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:-) $\quad +1\quad$ – amWhy Mar 21 '13 at 1:47

I added to the link given by Tobias the following:

"These are also called matched pairs of groups. I just did a google search on this term and it gave over 11,400 results, often related to Hopf algebras. The n-fold case is discussed in arXiv:1201.0059 . But I was not aware of the older references given above [i.e. in the link], which are nice to see."

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