S is a system of representatives left cosets -> S^-1 system of representatives right cosets

Question

Let G be a group and let H is a subgroup of G. Prove the following:

If S is a system of representatives for the left cosets of H in G, then $S^{-1}=\{s^{-1}:s \in S\}$ is a system of representatives for the right cosets of H in G

How I tried to prove it:

s belongs to S -> $s=a*h_{1}$ for some $h_{1}$ belongs to H and for an unique a belongs to G. $s=a*h_{1} -> h_{1}^{-1}*a^{-1}=s^{-1}$

a is unique -> a^-1 is unique, total amount of different a^-1=total amount of different a's=$[G:H]$(=number of left and right cosets). Thus proven that each s^-1 belongs to an unique right coset of G (since $h^{-1}_{1}\in H$ and $a^{-1} \in G$)

Is my proof ok? And if not, what can I improve about it? Thanks in advance

Answer 1

Disregarding any of the mathematical content, usually we write proofs in $\LaTeX$ (or mathtype, in this case) and do not overuse mathematical symbols like $\forall$, $\exists$, $\Rightarrow$. This means that we write in full sentences and try to explain what we're doing with words. In your case, although the first part of your proof is logically correct, it is hard to digest for a reader. You could improve it by cleaning up some of the parts:

If $s \in S$, then we may write $s = ah_1$ for some $h_1 \in H$ and an unique $a \in G$. Taking the inverse of this equation gives us $h_1^{-1}a^{-1} = s^{-1}$.

And this applies to your entire proof.

In the next part, though, your logic is not clear. The number of unique $a$'s is not equal to $[G:H]$. If $s = ah_1$, then $sh_1^{-1} = a$. So $a$ is uniquely determined by our choice of $h_1 \in H$, and therefore we actually have $|H|$ choices for $a$. At this point, your proof breaks down.

Hint? To prove this, I would prove that $gH \mapsto Hg^{-1}$ is a well-defined bijection between between the left and right cosets of $H$ in $G$.