Thread: [SOLVED] If the product of two right cosets of H is a right coset of H, then H is nor

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 $\displaystyle g \in G$.

Originally Posted by davismj

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?

Originally Posted by davismj

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 $\displaystyle g \in G$.

In particular, $\displaystyle HgHg^{-1}$ is some right coset $\displaystyle Hc$. Now we have $\displaystyle 1 = 1g1g^{-1} \in HgHg^{-1}$, so $\displaystyle 1 \in Hc$, i.e. $\displaystyle c \in H$. Therefore $\displaystyle HgHg^{-1}=H$ and $\displaystyle H$ is normal.

Originally Posted by Drexel28

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 $\displaystyle c=ab$.

No! If $\displaystyle HaHb=Hc$ implies $\displaystyle ab=c$ then, since we also have $\displaystyle H(ha)Hb=Hc$ for any $\displaystyle h \in H$, it would follow that both $\displaystyle ab=c$ and $\displaystyle hab=c$...

Originally Posted by Bruno J.

No! If $\displaystyle HaHb=Hc$ implies $\displaystyle ab=c$ then, since we also have $\displaystyle H(ha)Hb=Hc$ for any $\displaystyle h \in H$, it would follow that both $\displaystyle ab=c$ and $\displaystyle hab=c$...

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 $\displaystyle h_1\ne h_2$