Suppose $G$ is a group and $H$ and $K$ are subgroups such that $G = HK$ and $H \cap K = \left\{e\right\}$, the identity element of $G$. When can we say that $HK \cong H\times K$?

I tried to set up the canonical map $(h,k) \to hk$ and worked out that this is an isomorphism if and only if $H \subset C(K)$ or $K \subset C(H)$, where $C$ denotes the centralizer, since $ h_{1}k_{1}h_{2}k_{2} = h_{1}h_{2}k_{1}k_{2}$ and therefore $hk = kh$ for each $h \in H$, $k \in K$.

Is there a better way of saying this? How do I obtain a characterization of when $HK \cong H\times K$? Here I just picked a map (albeit canonical) and worked out when it would be an isomorphism. What if some other map works?

  • 1
    $\begingroup$ This is true if and only if $H$ and $K$ are both normal subgroups of $G$. $\endgroup$ – Derek Holt Apr 17 '15 at 16:40

Your condition on centralizers is a good one. Nevertheless, the condition that both $H$ and $K$ are normal in $G$ is also equivalent to this, and is the easiest way to highlight a direct product structure.

  • $\begingroup$ Could you give some hint on why the condition is equivalent to $H$ and $K$ being normal? I tried to show it but couldn't. Maybe I'm missing something easy. $\endgroup$ – Seven Apr 18 '15 at 4:03
  • $\begingroup$ Of course. Let $K$ and $H$ be two subgroups that $<K,H>=G$ and $H\cap K$ is trivial. Then your condition on centralizers trivially implies mine. For the other implication, take $h\in H$ and $k\in K$ and use my condition about normalization to show that $[h,k]\in H\cap K$ and conclude. $\endgroup$ – Clément Guérin Apr 18 '15 at 6:06
  • $\begingroup$ Thanks for the explanation. Also about the part of choosing the canonical map to set up an isomorphism. Why should this be the only valid isomorphism? How can I prove $H \times K \cong HK$ implies $H$ and $K$ normal without setting up an explicit isomorphism between them? $\endgroup$ – Seven Apr 18 '15 at 6:23
  • $\begingroup$ Well, without the explicit isomorphism it is not necessarily true. It may be hard to state this completely in a comment but you can show that some semi direct product are actually direct product. For instance take $\mathbb{Z}_2$ acting (via $\phi$ on $\mathfrak{S}_3$ by the conjugation by $(1,2)$. Then one can show that $\mathfrak{S}_3\rtimes_{\phi}\mathbb{Z}_2=:G$ is isomorphic to $\mathfrak{S}_3\times\mathbb{Z}_2$.However the obvious copy $H=\mathbb{Z}_2$ is not normal in $G$. $\endgroup$ – Clément Guérin Apr 18 '15 at 6:51
  • $\begingroup$ My last comment may not be as enlihgtening as it should be. My point is if you want to study more carefully the link between normal subgroups, quotient groups and global group, this property as well as its semi-direct analoguous is too specific. What you want to do is studying group extensions (it admits classification). $\endgroup$ – Clément Guérin Apr 18 '15 at 7:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.