# A question about direct products of subgroups of a finite group.

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

$\rightarrow$

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.

• 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? – tmpys Apr 25 '14 at 7:13
• 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. – Jyrki Lahtonen Apr 25 '14 at 7:40
• 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 – tmpys Apr 25 '14 at 8:10
• Yes, all that is correct. But it is not a direct product because $H$ and $K$ don't commute. That was my point. – Jyrki Lahtonen Apr 25 '14 at 8:39
• I have HK={hk|h in H, k in K} – 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.)
• I guess that would make $|H\cap K|=1$ thus it must be $e$ – tmpys Apr 25 '14 at 7:22