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I am not a mathematics major, and I encountered this problem while self-studying group theory (because it is an important tool in my research field). Here is the problem:

Let $A$ be a subgroup of a group $G$ with finite index. Prove that there exist elements $g_{1}$, $g_{2}$, $\cdots$ in $G$ that can serve as representatives for both the right cosets and the left cosets of $A$ in $G$.

I attempted to solve this problem: I know that if $R$ is a set of right representatives, then $R^{-1}$ is a set of left representatives. Therefore, I tried to prove the existence of such $R$ that forms a group, but I seem to struggle in proving it. In other words, I found it difficult to find suitable representatives such that both themselves and their inverses are in the set of representatives. I also attempted to abandon this idea and directly prove the existence of such a set of elements, but I couldn't do that when more than $\frac{\left | G \right |}{2}$ representatives already exist. In summary, I have no means to progress with my current approach. Therefore, I want to know how I should approach solving these problems. What tools do I need?

Shaun
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Lin Han
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    Unless you have a normal subgroup, the set of cosets will not form a group. – calc ll Mar 02 '24 at 13:14
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    The usual proof goes via Hall's Marriage Theorem. See the answers to this question – lulu Mar 02 '24 at 13:15
  • Note: this question has nothing to do with linear algebra. – lulu Mar 02 '24 at 13:15
  • @Nothingspecial: Your procedure gives no guarantee that the resulting set serves simultaneously as representatives for both right cosets and left cosets of $A$ in $G$. – Lee Mosher Mar 02 '24 at 15:50
  • @lulu Thank you for your response! I've checked out the post you mentioned and looked into Hall's marriage theorem, but it seems like there's still some distance from this theorem to the solution of the problem. According to another post's answer, I did find a proof in Hall's book, but it's so lengthy and involves some concepts I haven't encountered before (laughs). I have to consider if I've gone too far, so I think it's best to leave the answer here for now and temporarily abandon this problem. I'll revisit it if needed in the future. Anyway, thank you for your help. – Lin Han Mar 04 '24 at 08:58
  • Here'a an answer: Hall, Marshall, Jr. Combinatorial theory. Theorem 5.1.7, page 51. – Lin Han Mar 04 '24 at 08:58

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