Is the intersection of subgroups of G always a subgroup?

[Chris L T521]

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]

This week's question was correctly answered by Sudharaka. You can find his solution below.

Spoiler

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.