# common left and right coset representatives for a subgroup of finite index

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?

• 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? Jan 25, 2020 at 9:15
• @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, 2020 at 9:45
– bof
Jan 25, 2020 at 9:51
• @DietrichBurde Did you read the question carefully? It is not assumed that $H$ is a normal subgroup.
– bof
Jan 25, 2020 at 9:55
• 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, 2020 at 11:00