# Thread: Abstract Algebra (left cosets)

1. ## Abstract Algebra (left cosets)

Suppose that H is a Normal Subgroup of G . a, b in G. Prove that (aH)(bH)=abH

2. ## Re: Abstract Algebra (left cosets)

This follows directly from the definition of a normal subgroup.

3. ## Re: Abstract Algebra (left cosets)

Can you show me a more detailed solution?
I know if H is normal sub group in G then hah^-1 is in H for all h in H , and all a in G
And same apply for another element in G, b in G then hbh^-1 in H for all h in H and b in G

4. ## Re: Abstract Algebra (left cosets)

Originally Posted by mohammedlabeeb
I know if H is normal sub group in G then hah^-1 is in H for all h in H , and all a in G
This is a wrong definition. The correct one is that aha-1 ∈ H for all h ∈ H and a ∈ G, which is equivalent to saying that aH = Ha for all a ∈ G.

5. ## Re: Abstract Algebra (left cosets)

What is my next step?
Is it right if I say ah1 bh2 ?

6. ## Re: Abstract Algebra (left cosets)

Hi,
I think maybe you want a detailed "set theoretic" proof. Here is one:

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8. ## Re: Abstract Algebra (left cosets)

Originally Posted by johng
Hi,
I think maybe you want a detailed "set theoretic" proof. Here is one:

you are the best

9. ## Re: Abstract Algebra (left cosets)

I have attached my solution please let me know if it is right or how can I fix it? CCI04202013_00001.bmp
Originally Posted by mohammedlabeeb
you are the best
Originally Posted by johng
Hi,
I think maybe you want a detailed "set theoretic" proof. Here is one:

10. ## Re: Abstract Algebra (left cosets)

Originally Posted by mohammedlabeeb
I have attached my solution please let me know if it is right or how can I fix it?
The solution is OK, but you proved the two key facts several times:

(1) ∀a ∈ G ∀h1 ∈ H ∃h2 ∈ H, ah1 = h2a
(2) ∀a ∈ G ∀h1 ∈ H ∃h2 ∈ H, h1a = ah2.

Note that (1) is equivalent to aH ⊆ Ha and to aHa-1 ⊆ H for all a ∈ G, and (2) is equivalent to Ha ⊆ aH and to H ⊆ aHa-1 for all a ∈ G. Thus, (1) and (2) together are equivalent to aH = Ha for all a and to aHa-1 = H for all a.

I recommend internalizing these three equivalent ways of saying that H ⊲ G.

If we use the fact that aH = Ha as the definition of H ⊲ G, then the original claim is trivial. Indeed, (aH)(bH) = a(Hb)H (by normality) = a(bH)H = (ab)HH = (ab)H. This is why I said it follows directly from the definition of a normal subgroup.