# Abstract Algebra (left cosets)

• Apr 19th 2013, 12:19 PM
mohammedlabeeb
Abstract Algebra (left cosets)
Suppose that H is a Normal Subgroup of G . a, b in G. Prove that (aH)(bH)=abH
• Apr 19th 2013, 12:42 PM
emakarov
Re: Abstract Algebra (left cosets)
This follows directly from the definition of a normal subgroup.
• Apr 19th 2013, 02:41 PM
mohammedlabeeb
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
• Apr 19th 2013, 03:02 PM
emakarov
Re: Abstract Algebra (left cosets)
Quote:

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.
• Apr 19th 2013, 05:08 PM
mohammedlabeeb
Re: Abstract Algebra (left cosets)
What is my next step?
Is it right if I say ah1 bh2 ?
• Apr 19th 2013, 09:30 PM
johng
Re: Abstract Algebra (left cosets)
Hi,
I think maybe you want a detailed "set theoretic" proof. Here is one:

Attachment 28040
• Apr 19th 2013, 10:19 PM
mohammedlabeeb
Re: Abstract Algebra (left cosets)
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• Apr 19th 2013, 10:21 PM
mohammedlabeeb
Re: Abstract Algebra (left cosets)
Quote:

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

Attachment 28040

you are the best
• Apr 20th 2013, 10:22 AM
mohammedlabeeb
Re: Abstract Algebra (left cosets)
I have attached my solution please let me know if it is right or how can I fix it? Attachment 28050
Quote:

Originally Posted by mohammedlabeeb
you are the best

Quote:

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

Attachment 28040

• Apr 20th 2013, 02:22 PM
emakarov
Re: Abstract Algebra (left cosets)
Quote:

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.