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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.

Seirios
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tmpys
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  • 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 07:13
  • 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 07:40
  • 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 08:10
  • 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 08:39
  • I have HK={hk|h in H, k in K} – tmpys Apr 25 '14 at 09:11
  • For it to be called a direct product the condition $HK=G$ is not sufficient. That set, $HK={hk\mid h\in H, k\in K}$, is not always even a subgroup - consider $K\le S_3$ as above and $H=\langle(13)\rangle$. It is a subgroup, if one of $H$ or $K$ is normal. It is a direct product, if $H\cap K={e}$ and both are normal. – Jyrki Lahtonen Apr 25 '14 at 09:34
  • you are the only person that has said the words direct product. – tmpys Apr 25 '14 at 19:06
  • Look at the title of your question. Admittedly the point has become moot a long time ago. Sorry about the noise :-) – Jyrki Lahtonen Apr 25 '14 at 19:09
  • Ahh lol, ok im sorry. I forgot I put that in the header.... I am very sorry, you are right. – tmpys Apr 25 '14 at 23:18

1 Answers1

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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!

Amitesh Datta
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