## 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.