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Suppose $H_i < G$ where $G$ is a group, and $H_i$ is a subgroup of $G$. Then it is true that $\bigcap_{i=1}^{\infty} H_i$ still a subgroup?

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Have you tried to prove it is? – Tara B Feb 20 '13 at 23:45
As a general rule, the intersection of a(ny) family of substructures turns out to be a substructure. – Pete L. Clark Feb 20 '13 at 23:46
It's easier for finite intersection, not sure if I can apply the same technique for infinite – love Feb 20 '13 at 23:48
@love: Well, when you're not sure, just try, and see if it works. =] – Tara B Feb 20 '13 at 23:52
up vote 6 down vote accepted

Sure it is. It contains $1$, and then if $a, b \in \bigcap_{i=1}^{\infty} H_i$, then $a, b \in H_i$ for each $i$, so $a b^{-1} \in H_i$ for each $i$, so $a b^{-1} \in \bigcap_{i=1}^{\infty} H_i$.

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"Infinite intersection" need not be countable... – GEdgar Feb 21 '13 at 1:55
@GEdgar, I answered the question as formulated in the question body, of course the very same argument works for an arbitrary intersection. – Andreas Caranti Feb 21 '13 at 9:17

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