# 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]

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

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