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Let $G$ be a group and $H$ a subgroup of $G$. If $G$ is finite, then according to Philip Hall's "marriage theorem" there is a left transversal $T$ of $H$ in $G$ (that is, $T$ contains precisely one element from each left coset of $H$) such that $T$ is also a right transversal of $H$.

Does this theorem generalize to infinite groups? The original theorem treats only the case where $|H| < \infty$ and $[G:H] < \infty$.

What can we say when $|H| = \infty$ and $[G:H] = \infty$? What about $|H| < \infty$ and $[G:H] = \infty$? Also $|H| = \infty$ and $[G:H] < \infty$?

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This is just to observe that if $|G:H|$ is finite then $G$ has a normal subgroup of finite index contained in $H$, so the problem reduces to the case when $G$ is finite, and a common transversal exists. All other cases are covered by Geoff Robinson's answer. –  Derek Holt Jan 1 '13 at 9:26

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We can (and indeed must be able to if it can be done at all) break up the problem of a simultaneous set of right and left coset representatives for $H$ according to $(H,H)$-double cosets. For a given $(H,H)$-double coset $HxH,$ we seek a single indexing set $I$ and collections of elements of $H$ $\{ s_{i}: i\in I \}$ and $\{t_{i}:i \in I \}$ such that $HxH = \cup_{i \in I} Hs_{i}xt_{i}H,$ and the union is disjoint, and so that the elements of $ \{s_{i}xt_{i}: i \in I\}$ are all in different right cosets of $H$ and all in different left cosets of $H.$

Now $(s_{i}xt_{i})(t_{j}^{-1}x^{-1}s_{j}^{-1}) \in H$ if and only if $t_{i}t_{j}^{-1} \in H \cap x^{-1}Hx$. Hence we require that the elements $t_{i} :i \in I$ all lie in different right cosets of $H \cap x^{-1}Hx$ in$ H$. Similarly, we require that the elements $ s_{i} : i \in I $ all lie in different left cosets of $H \cap H \cap xHx^{-1}$ in $H$. In fact, this tells us when we can solve the problem. The double coset $HxH$ is a union of a collection of right cosets of $H$ in $G,$ and union of left cosets of $H$ in $G$. If there is a bijection between the set of right cosets of $H$ in $HxH$ and the set of left cosets of $H$ in $HxH,$ then we can take to be an indexing set of the common cardinality of these two sets. Then we can take $\{ s_{i}: i \in I\}$ to be a full set of representatives for the left cosets of $H \cap H \cap xHx^{-1}$ in $H$ and $\{ t_{i}: i \in I\}$ to be a full set of representatives for the right cosets of $H \cap H \cap x^{-1}Hx$ in $H.$ Then $HxH = \cup_{i \in I} Hxt_{i} = \cup_{i \in I} s_{i}xH$ (disjoint) and the elements $\{ s_{i}xt_{i} : i \in I \}$ are simultaneously a set of representatives for the right and left cosets of $H$ contained within $HxH.$

Analyzing the argument, we can find a simultaneous right and left tranversal for $H$ in $G$ if and only if for every $x \in G,$ we have $[H:H \cap xHx^{-1}] = [H : H \cap x^{-1}Hx]$ (note that if $Y$ is a fixed subgroup of a group $X,$ then inversion gives a bijection between the set of right cosets of $Y$ in $X$ and the set of left cosets of $Y$ in $X$, and we denote the common cardinality of these sets as $[X:Y]).$ When $H$ is finite, these two indices are always equal, since the subgroups $x^{-1}Hx \cap H$ and $xHx^{-1} \cap H$ are conjugate, hence of the same size,while $H$ itself is finite.

Note that $[G:H]$ seems to play no role in this condition: however, even when $H$ is countable, I see no obvious way to exclude the possibility that $H \cap xHx^{-1}$ is countably infinite, but $[H : H \cap x H x^{-1}]$ is finite, while $[H : H \cap x^{-1}Hx]$ is countably infinite.

Later edit: Indeed, Derek Holt has provided in the comments an example to illustrate that the situation suggested above does occur "in nature". In view of this, it may be that the most general easily stated condition which guarantees a simultaneous left and right transversal to $H$ in $G$ is that for each $ x \in G \backslash N_{G}(H)$, the subgroup $H \cap xHx^{-1}$ has finite order. Even later edit: Just to make it clear, the example provided by Derek gives a case where there is no simultaneous right and left transversal.

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Let $H = \langle a,b \rangle$ be free, and let $G$ be the HNN extension $\langle a,b,x \mid x^{-1}ax=a^2, x^{-1}bx=b^2 \rangle$ of $H$. Then $H \cap x^{-1}Hx = \langle a^2,b^2 \rangle$, which has infinite index in $H$, but $(xax^{-1})^2=a$, $(xbx^{-1})^2=b$, so $xHx^{-1}$ contains $H$. –  Derek Holt Dec 31 '12 at 19:53
    
Thanks Derek. I did think that there might be some such counterexample using HNN extensions, since they usually provide counterintuitive examples for conjugacy questions, but I had not got around to trying to construct such an example. –  Geoff Robinson Dec 31 '12 at 20:06
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So putting this answer and the comments of Derek together, am I right in saying that the only case where the mutual transversal might not exist is the one where $|H| = \infty$ and $[G:H] = \infty$? –  spin Jan 1 '13 at 16:45
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Yes that's right! And there do exist such examples in which there is no mutual transversal. –  Derek Holt Jan 1 '13 at 16:55
    
That is right, but I think you need a little argument when $[G:H]$ is finite. For then we have $[H :H \cap xHx^{-1}] \leq [G:xHx^{-1}] = [G:H] < \infty$ for all $x \in G,$ and we have $[G:H \cap xHx^{-1}] =[G: H \cap x^{-1}Hx]$ for all $x\in G.$ –  Geoff Robinson Jan 1 '13 at 17:10

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