Algebra Homework 2

Proof. We consider the two cases: either g ∈ H or g ∈ G−H. Case 1 Suppose g ∈ H. Then for any gh ∈ gH, gh ∈ H. This implies gH ⊆ H. Now, let h be any element of H. Then h = (gg−1)h = g(g−1h), and g−1h ∈ H because g−1 and h are both in the subgroup H. It follows that h ∈ gH and H ⊆ gH. Therefore, gH = H. It can be similarly shown that Hg = H. This means gH = Hg and H £ G by Lemma 2.5.1 from class. Case 2 Suppose g ∈ G−H. Then for any gh ∈ gH, gh ∈ G−H because it cannot be the case that gh ∈ H. (If gh were in H, then ghh−1 = g would have to be in H by Lemma 2.4.3, which is clearly false.) Hence, we know gH ⊆ G −H. Now, since |G/H| = 2 we know by the Lagrange Theorem that |G| = 2|H|, and by Lemma 2.4.6, |aH| = |H| for any a ∈ G. Thus, in this case, |G| = |gH|+ |H|. This allows us to say that |gH| = |G−H|. Since gH ⊆ G−H and their order is equal, it follows that gH = G −H. It can be similarly shown that Hg = G − H. This means gH = Hg and H £ G. Having demonstrated both possible cases, we can conclude in general that H £ G whenever H ≤ G and |G/H| = 2. ¤