Proof: Cosets equal or disjoint

[Leb]

Homework Statement

Two left cosets aH, bH of H in G are equal if and only
if a^{−1}b ∈ H. This is also equivalent to the statement b ∈ aH.
Proof:
Suppose that aH = bH. Then e ∈ H. So, b = be ∈ bH. If
aH = bH then b ∈ aH. So, b = ah for some h ∈ H. But, solving for
h, we get h = a
−1
b ∈ H

Homework Equations

Let G be a group, H a subgroup. a,b ∈ G.
Def: Left coset aH={ah |h ∈ H}

Attempt:
I never managed to understand this simple concept of cosets, but maybe understanding this will help. I do not see why when saying that aH = bH implies that b ∈ aH, which implies b = ah for some h ∈ H. That is how {ah |h ∈ H} = {bh |h ∈ H} implies b ∈ aH and b=ah, and not bh ∈ aH and bh =ah ? The definition of the coset only gives information on h, not on b...

 

[HallsofIvy]

Homework Statement

Two left cosets aH, bH of H in G are equal if and only
if a^{−1}b ∈ H. This is also equivalent to the statement b ∈ aH.
Proof:
Suppose that aH = bH. Then e ∈ H. So, b = be ∈ bH. If
aH = bH then b ∈ aH. So, b = ah for some h ∈ H. But, solving for
h, we get h = a
−1
b ∈ H

Homework Equations

Let G be a group, H a subgroup. a,b ∈ G.
Def: Left coset aH={ah |h ∈ H}

Attempt:
I never managed to understand this simple concept of cosets, but maybe understanding this will help. I do not see why when saying that aH = bH implies that b ∈ aH, which implies b = ah for some h ∈ H.

It is the same as you have in the first proof above. H is a subgroup so the identity, e, is in H and therefore bH contains b. Since we are given that aH= bH, b is in aH. Any member of aH is of the form ah so we have b= ah for some h in H.

That is how {ah |h ∈ H} = {bh |h ∈ H} implies b ∈ aH and b=ah, and not bh ∈aH[/quote]
What h are you talking about? Certainly it is true that bh∈ aH for some h in H- specifically h= e.

and bh =ah ?

If that were true then we could multiply both sides of the equation, on the right, by h^-1 and get b= a. Saying that bh= a or b= ah', possibly with h= h' or h\ne h', does not lead to ah= bh. Any member of

The definition of the coset only gives information on h, not on b...

Yes, of course, b is "given" and does not change. You don't need to know any thing more about b. The only information on h you need (or have) is that it is a member of H.

 

[Leb]

Thanks for the reply!
I think I figured it out just after I posted the message, I think my problem was that I missed the "Suppose that aH = bH" part, which was crucial for the proof to work.