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Assume that $G$ is a group, and that $H$ is a (not necessarily normal) subgroup of $G$ having finite index $r=[G:H]$. A subset $\{x_1,\cdots,x_r\}\subset{G}$ is called a left transversal of $H$ in $G$ provided that $\{x_1{H},\cdots,x_r{H}\}$ is complete set of $r$ distinct left cosets of $H$ in $G$. Similarly, a subset $\{y_1,\cdots,y_r\}\subset{G}$ is called a right transversal of $H$ in $G$ provided that $\{{H}y_1,\cdots,{H}y_r\}$ is a complete set of $r$ distinct right cosets of $H$ in $G$. Prove that there exists a distinguished subset $\{x_1,\cdots,x_r\}\subset{G}$ that is simultaneously (both) a left and a right transversal of $H$ in $G$; i.e. such that $$G={\bigcup}^{r}_{j=1}{x_j}H = {\bigcup}^{r}_{j=1} H{x_j}.$$ Not sure how to approach this. I felt that that the conjugates of $H$ might play a role, and proved that any subgroup of the form $gHg^{-1}$ must also have index $r$. But am not able to use this fact! This problem seems to have been discussed on math stack exchange but I cannot find a solution. As respondents observed, this is a consequence of Hall's "marriage theorem". Is there a direct proof of this fact which does not use the marriage theorem?

bof
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student
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  • By definition we have $r$ cosets, so that $G$ is a union of these cosets. You don't have to do much. Did you cover the standard properties of cosets? – Dietrich Burde Jan 25 '20 at 09:15
  • @DietrichBurde It's a standard application of Hall's theorem (the so-called marriage theorem) but I wouldn't say it's exactly trivial. – bof Jan 25 '20 at 09:45
  • See this answer https://math.stackexchange.com/q/178458 – bof Jan 25 '20 at 09:51
  • @DietrichBurde Did you read the question carefully? It is not assumed that $H$ is a normal subgroup. – bof Jan 25 '20 at 09:55
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    Do you know how to prove it if $G$ is a finite group? The case of a finite index subgroup of an infinite group can be reduced to the case of a finite group. By the way, it's also true that a common transversal exists if $H$ is a finite subgroup of infinite index, But there are counterexamples when $H$ is an infinite subgroup of infinite index. – bof Jan 25 '20 at 11:00
  • Folks! Just saw your responses. The problem does not assume that $G$ is a finite group, or that $H$ is a normal subgroup of $G$. Of course, if $H$ is a normal subgroup of $G$, the result is immediate, since $gH=Hg$ for every $g\in{G}$. – student Jan 25 '20 at 16:44
  • I edited the question and posed it (hopefully) with greater clarity after the (useful) comments responding to my question. – student Jan 25 '20 at 17:52

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