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

Question

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.

Answer 1

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!