Proof: Cosets equal or disjoint

1. Mar 14, 2012

Leb

1. The problem statement, all variables and given/known data

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

2. Relevant 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...

2. Mar 14, 2012

HallsofIvy

Staff Emeritus
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.

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

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.

3. Mar 20, 2012

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.

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