Algebra Homework 3

Proposition. Let G be a group and (Gi, i ∈ I) a family of subgroups of G. If (i) for each i, j ∈ I there exists k ∈ I with 〈Gi, Gj〉 ≤ Gk, and (ii) Gi is simple for all i ∈ I, then H = ⋃ i∈I Gi is a simple subgroup of G. Proof. Choose i, j ∈ I. Observe that (ii) implies Gi, Gj 6= ∅ because the definition of a simple group excludes this possibility. Clearly, Gi is a group and a subset of 〈Gi, Gj〉. So, together with (i), we have Gi ≤ 〈Gi, Gj〉 ≤ Gk for some k ∈ I. But, by (ii), Gk is simple. This means Gi = Gk, and similarly, Gj = Gk. Since Gi and Gj were arbitrary, it follows that every subgroup in (Gi, i ∈ I) is really just Gk. Thus, H is the union of Gk with itself, so H = Gk. Since Gk ≤ G and Gk is simple by (ii), we conclude that H is a simple subgroup of G. ¤