N3: Prove that the intersection of an arbitrary nonempty collection of subgroups of G is again a subgroup of G. (do not assume that the collection is countable)

Question

anonymous · Answer

Whoa there integer!  You in undergrad class?  Cause this forum is for high school math...lol 
Some Calc 1

anonymous · Answer

The answer to this question is in the definition of a subset.  Which is what I think you meant instead of subgroup.

anonymous · Answer

no it means subgroup.

anonymous · Answer

what do you have to show?   everything will work out right .
the identity is in it because the identity is in each subgroup, so it is in their intersection

anonymous · Answer

oh i see it is the countable part that is not obvious.  otherwise you could just show it by doing it for two subgroups and then use induction.  damn

anonymous · Answer

oh no it makes no difference.

anonymous · Answer

what are you using to index your subgroups?

anonymous · Answer

this is impossible for me to type for some reason but it is one line basically. for  if a and b belong to the intersection then for each say 
$$\lambda \in \Lambda$$ you know 
$$a,b\in H_{\lambda}$$ for all 
$$\lambda$$so 
$$ab^{-1}\in H_{\lambda}$$ and so 
$$ab^{-1}\in H$$