Question

Let G be an abelian group, and let H be a subgroup of G. Show that, for every g∈G, we have gH=Hg.

Answer 1

To show that for every g ∈ G, we have gH = Hg, we need to provide a step-by-step explanation. First, we use the fact that G is an abelian group to show that every element of G commutes with every other element. Then, we show that gH ⊆ Hg and Hg ⊆ gH, which implies gH = Hg for every g ∈ G.

Step-by-step explanation:

Let's show that for every g ∈ G, we have gH = Hg.

1. Since G is an abelian group, every element of G commutes with every other element. This means that for any g ∈ G and h ∈ H, we have gh = hg.
2. Let's show that gH ⊆ Hg. Take any element gh in gH. Since g ∈ G and h ∈ H, we have gh = hg, which means that gh is also in Hg. Therefore, we can conclude that gH ⊆ Hg.
3. Now, let's show that Hg ⊆ gH. Take any element hg in Hg. Again, since g ∈ G and h ∈ H, we have gh = hg, which means that hg is also in gH. Therefore, we can conclude that Hg ⊆ gH.

From steps 2 and 3, we can deduce that gH = Hg for every g ∈ G.