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

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#### Chris L T521

##### Well-known member
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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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#### Chris L T521

##### Well-known member
Staff member
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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