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

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Chris L T521

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Jan 26, 2012
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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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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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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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