# If $G$ is an infinite group, and $A,B$ subgroups of finite index in $G$, then $A \cap B$ has finite index in $G$ [duplicate]

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Question: If $G$ is an infinite group, and $A, B$ subgroups of finite index in $G$, then prove $A \cap B$ has finite index in $G$.

I'm trying to show that $A\cap B$ can not have infinite index, but I can't make contradiction. I do not see where the problem in $A \cap B$ have infinite order comes from. I thought this look easy when I first saw it, but now I'm not so sure..I'm sure its true, but do not know where the contradiction comes from.

Thank you for help if you choose to help

## marked as duplicate by Lord Shark the Unknown, onurcanbektas, Brahadeesh, José Carlos Santos, Namaste group-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 14 '18 at 10:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• This has been covered many times on this site - just type "finite index" into the search box and pick your favourite! – user1729 Nov 14 '12 at 17:21

## 2 Answers

If $\,\{a_i\}_{i\in I}\;\;,\;\{b_j\}_{j\in J}\,$ are representatives of the (left or right) cosets of $\,A\,,\,B\,$ in $\,G\,$ , show that $\,\{a_ib_j\}_{i\in I\,,\,j\in J}\,$ represent all the cosets of $\,A\cap B\,$ in $\,G\,$ (perhaps with repetitions, though)

• DonAntonio, please check: Is my answer linked here correct? – user198044 Oct 16 '18 at 3:49

Hint: Show that $g(A \cap B) = gA \cap gB$ holds for all $g \in G$.