# Thread: Quick quotient group Q

1. ## Quick quotient group Q

I have a simple question about quotient group operations. To this end, if $\displaystyle K$ is a normal subgroup of $\displaystyle G$, then $\displaystyle G/K$ is group under the binary operation defined by $\displaystyle (g_{1}K)*(g_{2}K)=(g_{1}g_{2})K$. This is all I have in my notes.

So, my question is: if the “mother group” $\displaystyle G$ is an additive group, then do we have that $\displaystyle (g_{1}K)*(g_{2}K)=(g_{1}+g_{2})K$?

Thanks

2. Originally Posted by Danneedshelp
I have a simple question about quotient group operations. To this end, if $\displaystyle K$ is a normal subgroup of $\displaystyle G$, then $\displaystyle G/K$ is group under the binary operation defined by $\displaystyle (g_{1}K)*(g_{2}K)=(g_{1}g_{2})K$. This is all I have in my notes.

So, my question is: if the “mother group” $\displaystyle G$ is an additive group, then do we have that $\displaystyle (g_{1}K)*(g_{2}K)=(g_{1}+g_{2})K$?

Thanks
Let me say it this way. Forget our bad notation for binary operations and let $\displaystyle G$ be a group with binary operation $\displaystyle *:G\times G\to Gg_1,g_2)\mapsto *(g_1,g_2)$ then if $\displaystyle N\lhd G$ we define the binary operation $\displaystyle \star:\left(G/N\right)\times\left(G/N\right)\to G/N:\left(g_1N,g_2N\right)\mapsto *(g_1,g_2)N$. See if that answeres your question.

3. $\displaystyle (g_{1})*(Hg_{2}H)=*(g_{1},g_{2})H$ under the map you defined. So, I would think this means $\displaystyle *(g_{1},g_{2})=(g_{1}+g_{2})$, because $\displaystyle *(g_{1},g_{2})$ must be in $\displaystyle G$, so it follows that $\displaystyle *$ must be the binary operation associated with $\displaystyle G$.

Correct?

4. Originally Posted by Danneedshelp
$\displaystyle (g_{1})*(Hg_{2}H)=*(g_{1},g_{2})H$ under the map you defined. So, I would think this means $\displaystyle *(g_{1},g_{2})=(g_{1}+g_{2})$, because $\displaystyle *(g_{1},g_{2})$ must be in $\displaystyle G$, so it follows that $\displaystyle *$ must be the binary operation associated with $\displaystyle G$.

Correct?
Hmm...close enough. My point was that the notation for the binary operation is irrelevant. So, in words the binary operation on $\displaystyle G/N$ is described "The (insert binary operation here) of two cosets is the coset whose coefficient (you get what I mean) is the (insert binary operation here) of the coefficients of the two original cosets"