Suppose that H is a Normal Subgroup of G . a, b in G. Prove that (aH)(bH)=abH
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
I have attached my solution please let me know if it is right or how can I fix it? CCI04202013_00001.bmp
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.