# Problem of the Week #24 - September 10th, 2012

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#### Chris L T521

##### Well-known member
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Thanks to those who participated in last week's POTW!! Here's this week's problem!

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Problem: Let $H$ and $K$ be subgroups of a group $G$. Define $HK = \{hk : h\in H, k\in K\}$. Show that $HK$ is a subgroup of $G$ if and only if $HK=KH$.

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#### Chris L T521

##### Well-known member
Staff member
This week's question was correctly answered by Sudharaka. You can find his solution below:

Let $$HK\leq G$$. Take any element, $$kh\in KH$$. Then,

$kh=(h^{-1}k^{-1})^{-1}$

$$h^{-1}k^{-1}\in HK$$ and since $$HK$$ is a group $$kh=(h^{-1}k^{-1})^{-1}\in HK$$.

$\therefore KH\subseteq HK~~~~~~~~(1)$

Take any element, $$hk\in HK$$. Since $$HK$$ is a group there exist, $$(hk)^{-1}\in HK$$ such that,

$(hk)(hk)^{-1}=1$

$\Rightarrow (hk)^{-1}=k^{-1} h^{-1}\in KH$

That is each element of $$HK$$ has its' inverse in $$KH$$. But since $$KH\subseteq HK$$ each element of $$KH$$ is also an element in $$HK$$. Therefore it follows that each element of $$KH$$ has its inverse in $$KH$$. Now take any element $$hk\in HK$$.

$hk=(k^{-1}h^{-1})^{-1}\in KH$

$\therefore HK\subseteq KH~~~~~~~~~~~~~(2)$

By (1) and (2),

$HK=KH$

Conversely let us assume that, $$HK=KH$$. Take any element $$h_{1}k_{1},\,h_{2}k_{2}\in HK$$. Then,

$h_{1}k_{1}(h_{2}k_{2})^{-1}=h_{1}k_{1}k_{2}^{-1}h_{2}^{-1}$

Since $$HK=KH$$ there exist $$h_{3}\in H\mbox{ and }k_{3}\in K$$ such that, $$k_{1}k_{2}^{-1}h_{2}^{-1}=h_{3}k_{3}$$. Therefore,

$h_{1}k_{1}(h_{2}k_{2})^{-1}=h_{1}k_{1}k_{2}^{-1}h_{2}^{-1}=h_{1}h_{3}k_{3}\in HK$

That is,

$h_{1}k_{1}(h_{2}k_{2})^{-1}\in HK\,\forall\,h_{1}k_{1},\,h_{2}k_{2}\in HK$

$\therefore HK\leq G$

Q.E.D.

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