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For any subgroup $H\leq G$, we have $|gH| = |H|$, so there is a bijection between the elements of H and the elements of any coset. This makes me wonder: is there any canonical map of sets between $H$ and a given coset $gH$? In particular, there would be a distinguished element in each coset corresponding to $1$.

Although a coset of a proper subgroup is not closed under multiplication, it seems that any sort of canonical map would in a sense translate information about the structure of $H$ to each coset. I tried experimenting with this idea, seeing if there were any interplay between cosets that preserved some sense of the structure of $H$, in a way that ended up resembling grading in a ring. I know this is vague, but in any case I failed to discover anything conclusive or even interesting.

I suspect that there is no such canonical map, but what is the case, and why?

Is there a difference between the cases where $G$ is a finite or an infinite group?

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You need to define «canonical» for your question to become a real question. What requirements, exactly, do you want on the bijection? –  Mariano Suárez-Alvarez Aug 5 '12 at 0:13
Some ways of representing groups naturally produce a well-ordering of the group, and the canonical or distinguished coset representative is the least element. This is the case with permutation groups, polycyclically presented groups, and Coxeter groups. In the case of Coxeter groups and parabolic subgroups $H$, there is some sense in which the structure is translated. In the other cases, the canonical representative is just a convenient marker. –  Jack Schmidt Aug 5 '12 at 16:49

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You have not made clear what you mean by canonical, but I suspect that in any event the answer is no. Let $G$ be a group, $H$ a subgroup, $C$ a coset of $H$ in $G$, $x$ any element of $C$. Then $h\to xh$ is a bijection between $H$ and $C$, and I don't see why there should be a "canonical" choice of $x$.

Let's look at an example. Let $G$ be the integers, let $H$ be the even integers. What is there to make one odd number more "canonical" than all the others?

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Now I see that my question was very ignorant. Thank you for answering. –  PrimeRibeyeDeal Aug 5 '12 at 16:11

Canonical is in the eye of the beholder. There's usually no group-theoretic reason for singling out any of the representatives, but you can apply other criteria. For instance, if you have a well-ordering on the group (which you can always define if the group is finite), you can choose the least element from each coset. The most interesting case is when you have no explicit formula at all to make the choice, for instance if you want to choose representatives for $\mathbb R/\mathbb Q$. Then you need the axiom of choice to even claim that there is a set of representatives, and you can't explicitly construct one.

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