This follows directly from the definition of a normal subgroup.
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
(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.