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

1. ## [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

$(Ha)(Hb) = Hc \implies gHg^{-1} = H$ or $gH = Hg$ for all $g \in G$.

2. Originally Posted by davismj

I've looked at this a few different ways. I don't really get it.

I don't see how

$(Ha)(Hb) = Hc \implies gHg^{-1} = H$ or $gH = Hg$ for all $g \in G$.
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 $HaHb=Hc$ we may conclude from the partitioning of cosets that $c=ab$. Thus, we may define a group $G/H$ to be the set $\left\{gH:g\in G\right\}$ with $(Hg)(Hg')=H(gg')$. Clearly $He=H$ is the identity element, it's associativity is associated from $G$ (why)? It is easy to prove that $\left(Hg\right)^{-1}=Hg^{-1}$ etc. Then, one can prove that $\theta:G\to G/H:g\mapsto gH$ is a well define homomorphism and $\ker\theta=H$. So?

3. Originally Posted by davismj

I've looked at this a few different ways. I don't really get it.

I don't see how

$(Ha)(Hb) = Hc \implies gHg^{-1} = H$ or $gH = Hg$ for all $g \in G$.
In particular, $HgHg^{-1}$ is some right coset $Hc$. Now we have $1 = 1g1g^{-1} \in HgHg^{-1}$, so $1 \in Hc$, i.e. $c \in H$. Therefore $HgHg^{-1}=H$ and $H$ is normal.

4. 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 $HaHb=Hc$ we may conclude from the partitioning of cosets that $c=ab$.
No! If $HaHb=Hc$ implies $ab=c$ then, since we also have $H(ha)Hb=Hc$ for any $h \in H$, it would follow that both $ab=c$ and $hab=c$...

5. Originally Posted by Bruno J.
No! If $HaHb=Hc$ implies $ab=c$ then, since we also have $H(ha)Hb=Hc$ for any $h \in H$, it would follow that both $ab=c$ and $hab=c$...
Sorry. Too late at night for group theory. What I meant to say is that $HaHb=Hc$ then $Hc=Hab$ since evidently $e\in H\implies a\in aH,b\in bH$ and thus $ab\in HaHb$ from where it follows by the partitioning that $HaHb=Hab$. What I said is clearly wrong since given any subgroup of cardinality greater than two we have that $Hh_1=Hh_2$ but $h_1\ne h_2$