Basic properties of perfect groups

A group H is called perfect if H^\prime = H. (I.e., H is its own commutator subgroup.)

  1. Prove that every nonabelian simple group is perfect.
  2. Prove that if H,K \leq G are perfect, then \langle H \cup K \rangle is perfect. Extend this to arbitrary joins of perfect groups.
  3. Prove that any conjugate of a perfect group is perfect.
  4. Prove that any group G has a unique maximal perfect subgroup and that this subgroup is normal.

  1. Let H be a nonabelian simple group. Now H^\prime is characteristic in H, hence normal. Moreover H^\prime \neq 1 since H is nonabelian. Thus H^\prime = H, and H is perfect.
  2. Let H,K \leq G be perfect subgroups. It follows from the definition of commutator that if A \leq B, then A^\prime \leq B^\prime. Thus H = H^\prime,K=K^\prime \leq \langle H \cup K \rangle^\prime. Thus \langle H \cup K \rangle \leq \langle H \cup K \rangle^\prime, so that \langle H \cup K \rangle is perfect.

    In general, \langle \bigcup_I H_i \rangle is the \leq-least subgroup containing \bigcup_I H_i. Thus if each H_i is perfect, we have H_i \leq \langle \bigcup_I H_i \rangle for each i, so that \langle \bigcup_I H_i \rangle \leq \langle \bigcup_I H_i \rangle^\prime. Hence \langle \bigcup_I H_i \rangle is perfect.

  3. Let H \leq G be a perfect subgroup. Now H = H^\prime = \langle \{ [a,b] \ |\ a,b \in H \} \rangle. Now g^{-1}Hg = g \langle \{ [a,b] \ |\ a,b \in H \} \rangle = \langle g^{-1}\{ [a,b] \ |\ a,b \in H \}g \rangle = \langle \{ g^{-1}[a,b]g \ |\ a,b \in H \} \rangle = \langle \{ [g^{-1}ag, g^{-1}bg] \ |\ a,b \in H \} \rangle = \langle \{ [a^\prime,b^\prime] \ |\ a^\prime, b^\prime \in g^{-1}Hg \} \rangle = (g^{-1}Hg)^\prime. Thus every conjugate of H is perfect. (Alternately, conjugate subgroups are isomorphic.)
  4. Let G be a group and let \mathcal{H} be the set of all perfect subgroups of G. By part 2 above, T = \langle \bigcup \mathcal{H} \rangle is a perfect subgroup of G. Moreover, by definition T is maximal, since if K \leq G is perfect then K \leq T, so that no subgroup which is strictly \leq-larger than T is perfect. Finally, T is unique since if S is some other \leq-maximal perfect subgroup then S \leq T, hence S=T.

    Now every conjugate of T is also a maximal perfect subgroup, as otherwise contradicts the maximalness of T. Thus g^{-1}Tg = T for all g \in G. Hence T is normal.

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