I've looked at this a few different ways. I don't really get it.
I don't see how
$\displaystyle (Ha)(Hb) = Hc \implies gHg^{-1} = H$ or $\displaystyle gH = Hg$ for all $\displaystyle g \in G$.
I've looked at this a few different ways. I don't really get it.
I don't see how
$\displaystyle (Ha)(Hb) = Hc \implies gHg^{-1} = H$ or $\displaystyle gH = Hg$ for all $\displaystyle g \in G$.
It's late, and I don't remember if there is an easier manipulation way..but you'll have to do this eventually might as well start now.
If $\displaystyle HaHb=Hc$ we may conclude from the partitioning of cosets that $\displaystyle c=ab$. Thus, we may define a group $\displaystyle G/H$ to be the set $\displaystyle \left\{gH:g\in G\right\}$ with $\displaystyle (Hg)(Hg')=H(gg')$. Clearly $\displaystyle He=H$ is the identity element, it's associativity is associated from $\displaystyle G$ (why)? It is easy to prove that $\displaystyle \left(Hg\right)^{-1}=Hg^{-1}$ etc. Then, one can prove that $\displaystyle \theta:G\to G/H:g\mapsto gH$ is a well define homomorphism and $\displaystyle \ker\theta=H$. So?
Sorry. Too late at night for group theory. What I meant to say is that $\displaystyle HaHb=Hc$ then $\displaystyle Hc=Hab$ since evidently $\displaystyle e\in H\implies a\in aH,b\in bH$ and thus $\displaystyle ab\in HaHb$ from where it follows by the partitioning that $\displaystyle HaHb=Hab$. What I said is clearly wrong since given any subgroup of cardinality greater than two we have that $\displaystyle Hh_1=Hh_2$ but $\displaystyle h_1\ne h_2$