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

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- April 19th 2013, 12:19 PMmohammedlabeebAbstract Algebra (left cosets)
Suppose that H is a Normal Subgroup of G . a, b in G. Prove that (aH)(bH)=abH

- April 19th 2013, 12:42 PMemakarovRe: Abstract Algebra (left cosets)
This follows directly from the definition of a normal subgroup.

- April 19th 2013, 02:41 PMmohammedlabeebRe: 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 - April 19th 2013, 03:02 PMemakarovRe: Abstract Algebra (left cosets)
- April 19th 2013, 05:08 PMmohammedlabeebRe: Abstract Algebra (left cosets)
What is my next step?

If I start with the LHS: (aH)(bH)

Is it right if I say ah1 bh2 ? - April 19th 2013, 09:30 PMjohngRe: Abstract Algebra (left cosets)
Hi,

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

Attachment 28040 - April 19th 2013, 10:19 PMmohammedlabeebRe: Abstract Algebra (left cosets)
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- April 19th 2013, 10:21 PMmohammedlabeebRe: Abstract Algebra (left cosets)
- April 20th 2013, 10:22 AMmohammedlabeebRe: 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

- April 20th 2013, 02:22 PMemakarovRe: Abstract Algebra (left cosets)
The solution is OK, but you proved the two key facts several times:

(1) ∀a ∈ G ∀h_{1}∈ H ∃h_{2}∈ H, ah_{1}= h_{2}a

(2) ∀a ∈ G ∀h_{1}∈ H ∃h_{2}∈ H, h_{1}a = ah_{2}.

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.