2
$\begingroup$

This question already has an answer here:

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

$\endgroup$

marked as duplicate by Lord Shark the Unknown, onurcanbektas, Brahadeesh, José Carlos Santos, Namaste group-theory 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.

  • $\begingroup$ This has been covered many times on this site - just type "finite index" into the search box and pick your favourite! $\endgroup$ – user1729 Nov 14 '12 at 17:21
2
$\begingroup$

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)

$\endgroup$
  • $\begingroup$ DonAntonio, please check: Is my answer linked here correct? $\endgroup$ – user198044 Oct 16 '18 at 3:49
2
$\begingroup$

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.