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Problem of the Week #23 - September 3rd, 2012

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Chris L T521

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Jan 26, 2012
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Again, sorry for posting this late. 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 $G$. Show that $H\cap K$ is a subgroup of $G$. Furthermore, show that this is true for any arbitrary intersection of subgroups of $G$; i.e. if $\{H_{\alpha}\}$ is a collection of subgroups of $G$, then $\bigcap_{\alpha} H_{\alpha}$ is also a subgroup of $G$.

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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.

Theorem: Let \(G\) be a group. The non-empty subset \(H\) of \(G\) is a subgroup of \(G\) if and only if \(ab^{-1}\in H\,\forall\,a,b\in H\).

(Reference: Elementary group theory - Wikipedia, the free encyclopedia)


Let \(H\) and \(K\) be subgroups of \(G\). Take any two elements, \(a,b\in H\cap K\). Then,


\[a,b\in H\mbox{ and }a,b\in K\]


Since \(H\) and \(K\) are subgroups of \(G\), \(ab^{-1}\in H\mbox{ and }ab^{-1}\in K\). Therefore,


\[ab^{-1}\in H\cap K\,\forall\,a,b\in H\cap K\]


\[\Rightarrow H\cap K\leq G\]


Q.E.D.


Take any two elements, \(a,b\in \bigcap_{\alpha} H_{\alpha}\mbox{ where }\{H_{\alpha}\}\) is an arbitrary collection of subgroups of \(G\). Then,


\[ab^{-1}\in H_{\alpha}\mbox{ for each }H_{\alpha}\in\{H_{\alpha}\}\]


\[\therefore ab^{-1}\in\bigcap_{\alpha} H_{\alpha}\,\forall\,a,b\in \bigcap_{\alpha} H_{\alpha}\]


\[\Rightarrow\bigcap_{\alpha} H_{\alpha}\leq G\]


Q.E.D.
 
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