Suppose $H$ and $K$ are subgroups of a finite group $G$ where $|H||K|=|G|$. Show that $H\cap K=\{e\}$ iff $G=HK$


Suppose $|H|=m$.

Let $H=\{h_0,h_1,h_2,....,h_{m-1}\}$.

Since $h_iK=h_jK$ iff $h_i^{-1}h_j\in K$ iff $h_i^{-1}h_j=e$ iff $h_j=h_i$, we have that $h_0K\cap h_1K \cap ..... \cap h_{m-1}K=\emptyset$ and $|h_0K\cup h_1K\cup ....\cup h_{m-1}K|=m|K|=|H||K|$.

This shows we have $|H||K|=|G|$ distinct elements of the form $hk$ where $h\in H \text{ and } k\in K$. So $G\subset HK$ and for sure $HK\subset G$ So $G=HK$.

I am having the problem with the other direction...

$G=HK\implies H\cap K =\{e\}$

Any help with this would be great. Thanks in advance.

  • $\begingroup$ Hmm maybe I just use the same idea, a (not e) in the intersection implies a is in H and a is in K. So now I look at cosets and I will not get the entire group in the union of cosets? $\endgroup$ – tmpys Apr 25 '14 at 7:13
  • $\begingroup$ I'm a bit troubled by the use of direct product in the title. For example the subgroups $H=\langle(123)\rangle$ and $K=\langle(12)\rangle$ of $G=S_3$ satisfy everything here, but yet $G$ is not their direct product. $\endgroup$ – Jyrki Lahtonen Apr 25 '14 at 7:40
  • $\begingroup$ In this situation you speak of G is HK H = <(123)>= {e,(123),(132)} K = <(12)>= {e,(12)} HK = {e*e, e(123), e(132), (12)e, (12)(123), (12)(132)} = {e, (123), (132), (12), (23),(13) } = G = S_3 $\endgroup$ – tmpys Apr 25 '14 at 8:10
  • $\begingroup$ Yes, all that is correct. But it is not a direct product because $H$ and $K$ don't commute. That was my point. $\endgroup$ – Jyrki Lahtonen Apr 25 '14 at 8:39
  • $\begingroup$ I have HK={hk|h in H, k in K} $\endgroup$ – tmpys Apr 25 '14 at 9:11

Do you know the formula $\left|HK\right|=\frac{\left|H\right|\left|K\right|}{\left|H\cap K\right|}$ for subgroups $H$ and $K$ of the finite group $G$? (I think you have proved this formula in the case $H\cap K=\{e\}$ in your attempt at the solution above - if you are interested, then try to see if you can make the minor modifications necessary to prove this in general.)

Using the formula, the result is almost immediate ... (Can you see how?)

Hope this helps!

  • $\begingroup$ I guess that would make $|H\cap K|=1$ thus it must be $e$ $\endgroup$ – tmpys Apr 25 '14 at 7:22
  • $\begingroup$ I am guessing there is a way to do the missing direction without this though? $\endgroup$ – tmpys Apr 25 '14 at 7:24
  • $\begingroup$ Dear @tmpys, yes, there is a way to do this without using this formula but it would probably be less elegant and at least partly be redoing the proof of this formula $\endgroup$ – Amitesh Datta Apr 25 '14 at 7:35
  • $\begingroup$ do you have a proof of this proposition? $\endgroup$ – tmpys Apr 25 '14 at 7:41
  • $\begingroup$ Dear @tmpys, please see: math.stackexchange.com/questions/168942/… $\endgroup$ – Amitesh Datta Apr 25 '14 at 8:01

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