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 !

  • $\begingroup$ I don't get it: the bottom line is that in "both" definitions we get, after all, the very same result, right? So what's the problem here? In either way, $\;xK\cdot yK=(xy)K\;$ ...if this is clear, now show this is well definied iff $\;K\lhd G\;$ . $\endgroup$
    – DonAntonio
    Apr 15, 2014 at 10:22
  • $\begingroup$ In the second case, before we say that K is a normal subgroup, its not clear that xK.yK = (xy)K. its only clear that (xK)(yK) = xKyK .... $\endgroup$
    – nerdy
    Apr 15, 2014 at 12:58

1 Answer 1


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.

  • $\begingroup$ Man, this clarified everything.Could i ask you something else ? Is it always the case that if we want a function to be well-defined, we will want it to behave the same way even if the same input is expressed in different forms ( different representatives for a single set for example ) Do you know in what context ( field of mathematics, site, etc )i can read more about functions being well-defined, and specifically congruence relations that i'm confused about ? $\endgroup$
    – nerdy
    Apr 15, 2014 at 13:17
  • $\begingroup$ @nerdy I might not be an expert on this, but I think what happens is that you typically define a relation before checking that its a function (ie for every input, you have gotten precisely 1 output). In order for it to be a $\textit{group}$, this function also has to be have the closure property. In the display where you introduce it as the product of group subsets, the work is already done on proving that the multiplication of group subsets is a function. So that this way all that's left to do in proving its a valid group operation is closure. $\endgroup$ Apr 24, 2022 at 1:53
  • $\begingroup$ hold on I need to go, but remind me about this comment and I will come back to finish it. $\endgroup$ Apr 24, 2022 at 2:27

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