# (Abstract Algebra) proof of an equivalent relation with subgroups

• Apr 2nd 2013, 06:05 PM
mohammedlabeeb
(Abstract Algebra) proof of an equivalent relation with subgroups
Let G be a group and H,K subgroups of G. If (x,y both belong to G ), define a relation ~ on G by x~y if y=axb for some (a in H) and (b in K). Prove that ~ is an equivalence relation

I tried to solve the question and this is what I was able to say so far:

I know to show ~ equivalent relation it has to be reflexive, symmetric, and transitive
I know for reflexive {for all u in G u~u}
for symmetric {for all u, v in G whenever u~v then v~u}
for transitive { for all u, v, w in G whenever u~v and v~w then u~w}. Here I need to start by " Let u~v and v~w for all u, v, w in G and I want to show u~w"
• Apr 5th 2013, 11:01 AM
johng
Re: (Abstract Algebra) proof of an equivalent relation with subgroups
Hi,
You are describing the so called double cosets of H and K. Denote by 1 the identity of the group G.
1. x ~ x : since 1 is in both H and K and x = 1 x 1
2. if x ~ y then for some a in H , b in K, y=axb. Then $\displaystyle a^{-1}\in H\>\text{and}\> b^{-1}\in K,\> x=a^{-1}yb^{-1}$ So y ~ x.
3. transitivity - use the fact that both H and K are closed under multiplication. I leave it to you.
• Apr 7th 2013, 08:32 AM
mathguy25
Re: (Abstract Algebra) proof of an equivalent relation with subgroups
To show that ~ is reflexive, we must show that x ~ x for all x in G. Let x be in G. We need to show that x = axb for some a in H and b in K. Let e be the identity element of G. Since H and K are subgroups of G, we see that e is identity element of both H and K. Thus, e is in H and e is in K. Take a = e and b = e. Then axb = exe = (ex)e = xe = x. Thus, x ~ x and ~ is reflexive.

To show that ~ is symmetric, we must show that if x ~ y, then y ~ x for all x and y in G. Let x and y be in G. Assume that x ~ y. We know there exists a' in H and b' in K such that y = a'xb'. We need to show that y ~ x i.e. there exists a in H and b in K such that x = ayb. Since a' is in H and H is a subgroup of G, we know that a'^(-1) is in H. Similarly, b'^(-1) is in K. Take a = a'^(-1) and b = b'^(-1). Then note that ayb = a'^(-1)yb'^(-1) = a'^(-1)(a'xb')b'^(-1) = (a'^(-1)a')x(b'b'^(-1)) = exe = xe = x. Thus, y ~ x and so ~ is symmetric.

To show that ~ is transitive, we must show that if x ~ y and y ~ z, then x ~ z for all x,y,z in G. Let x,y,z be in G. Assume that x ~ y. We know there exists a', a'' in H and b', b'' in K such that y = a'xb' and z = a''yb''. We need to show that x ~ z i.e. there exists a in H and b in K such that z = axb. Note that since a',a'' is in H and H is a subgroup of G, we see that a''a' is in H. Similarly, b'b'' is in K. Take a = a''a' and b = b'b''. Then axb = a''a'xb'b'' = a''(a'xb')b'' = a''yb'' = z. Thus, x ~ z and so ~ is transitive.

Thus, ~ is an equivalence relation.

Note that a' and a'' are used as labels to represent distinct elements.