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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

amWhy
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Winnie
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  • 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 Answers2

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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
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  • DonAntonio, please check: Is my answer linked here correct? –  Oct 16 '18 at 03:49
  • I am trying to restate your proof by doing a little elucidation:Given $A,B\leq G.$ Let the set of right cosets of $A,B$ be $A'={Ag_{i_1},...Ag_{i_k}},B'={Bg_{j_1},...,Bg_{j_k}}$ respectively. Now, say, $(A\cap B)g_{k_1}$ is a right coset of $A\cap B.$ Then, $Ag_{k_1}\in A'$ and $Bg_{k_1}\in B'$ as $A',B'$ are finite. Since $g_{k_1}$ is arbitary so we see $\forall (A\cap B)g_{k_1}$ $\exists i_p,j_p$ such that $Ag_{k_1}=A{i_p}\in A'$ and $Bg_{k_1}=B{j_p}\in B'$, thus, we conclude the set of cosets of $A\cap B$ finite since $A',B'$ is finite. This is what you meant, right? – Arthur Feb 06 '23 at 04:45
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Hint: Show that $g(A \cap B) = gA \cap gB$ holds for all $g \in G$.

Mikko Korhonen
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