Let $H$ be a subgroup of the group $G$ and let $T$ be a set of representatives for the distinct right cosets of $H$ in $G$. In particular, if $t\in T$ and $g \in G$ then $tg$ belongs to a unique coset of the form $Ht'$ for some $t'\in T$. Write $t'=t\cdot g$. Prove that if $S\subseteq G$ generates $G$, then the set $\{ts(t\cdot s)^{-1}|t\in T, s\in S\}$ generates $H$. Suggestion: If $K$ denotes the subgroup generated by this set, prove the stronger assertion that $KT=G$. Start by showing that $KT$ is stable under right multiplication by elements of $G$.
The above is a question from a list of qualifying exam questions I've been trying to work on in prep for my examination, however, this question had kept me stumped for a few weeks. Would anyone happen to have an idea for how to obtain the result of the suggestion?
As the suggestion says, let $K$ be generated by $R=\{ts(t\cdot s)^{-1}\mid t\in T, s\in S\}$. We want to prove that $KT$ is stable under the right multiplication action of $G$. To that end, since $S$ generates $G$, it is enough to prove that $KTS\subset KT$. Also if $k_1, k_2\in K, t\in T$ and $s\in S$, if $k_1ts\in KT$, then so is $k_2k_1ts$ obviously. Therefore it is enough to prove that given $t,t'\in T$, $s,s'\in S$ then $ts(t\cdot s)^{-1}t's'\in KT$. But $$ ts(t\cdot s)^{-1}t's'= \underbrace{[ts(t\cdot s)^{-1}]}_{\in R}\underbrace{[t's'(t'\cdot s')^{-1}]}_{\in R}\underbrace{(t'\cdot s')}_{\in T} $$ so we are done. This takes care of the second part of the suggestion. Now let $g=kt\in KT$. Then $1=gg^{-1}\in KTG\subset KT$ due to stability. Again due to stability then $G=KT$.
Now note that $ts=h_1 (t\cdot s)$ for some $h_1$ by definition. So $ts(t\cdot s)^{-1}=h_1\in H$, meaning $R\subset H$ and in turn $K\subset H$. The fact that $G=KT=HT$ now shows that $H=K$.