So, i have encountered two ways to introduce the multiplication of cosets, and i want to understand exactly what is happening in each, specifically in light of the multiplication of cosets being well-defined or not.
Given a group $G, K$ a subgroup of $G$, $a,b,c,d,u,v$ elements of $G …$
1 – We can introduce the multiplication of cosets of K by defining uK o vK = (uv)K. And then we can go on to prove it's well-defined ( if aK=cK and bK=dK then abK = cdK, for all pais (a,c) and (b,d) of representatives in each coset ).
So, i want to understand … Here we defined the function in a way to yield closure of the Quotient Group, but in turn we had to prove the funcion not only yields closure but is also well-defined. Is that correct ? Now, i have another way to introduce multiplication of cosets and i want to relate to this one.
2 – We can introduce multiplication of cosets of K, based on the product of group subsets ( http://en.wikipedia.org/wiki/Product_of_group_subsets ) , which gives us some results like KK=K and from that we can easily prove that if K is normal then the multiplication of two cosets of K yields another coset of K :
(xK)(vK) = xKvK = x(Kv)K = x(vK)K = xvK.
Now, my main doubt. Here, did we only proved closure of the function ( in the Quotient Group )and do we still have to prove it's well-defined ? Or in this approach, we don't need to prove the function is well-defined, because this is already embedded in the definition ?
Help me clarify : In the first introduction, we assumed multiplication of cosets was closed in the Quotient Group, and had to prove it was also well-defined in case K is normal. In the second definition, we didn't assume it was closed, proved it's closed if K is normal but now we don't need to prove it's well-defined ?
Thanks a lot !
Best Answer
Yes, take cosets $A=aK$, $B=bK$, then the first definition $$ A\cdot B := (ab)K $$ is a coset again, by definition, but we have to check that the choice of representatives $a\in A$ and $b\in B$ is irrelevant.
For the second definition, $$ A\cdot B := AB = \{\, gh : g\in A, h\in B\,\}, $$ you don't have to check any independence of choice of representatives, because the definition doesn't invole representatives. Instead, you have to check that what you get is a coset again, and as it turns out, you get $AB=(ab)K$, so both definitions agree.
Be aware that you need normality of $K$ in both cases to get something that is well-defined and a coset again.