Question

For any group G, let Δ(G) = {g ∈ G: [G : CG(g)] < ∞}, and Δ+(G) = {g ∈ Δ(G) : g has finite order}.
Show that Δ+(G) is a characteristic subgroup of G and that Δ+(G) is the union of all finite normal subgroups of G.

Answer 1

$\Delta^{*}(G)$ is clearly closed under inverses. To see that it is closed under multiplication, consider $a_1, a_2 \in \Delta^{*}(G)$, and the subgroup $H$ they generate in $G$. Since each $a_i$ has finite order and finitely many conjugates in $G$, implies that $H$ is finite. Therefore $a_1a_2 \in H$ also has finite order, and we have $a_1a_2 \in \Delta^{*}(G)$. Any automorphism of $G$ induces an automorphism of $\Delta(G)$, which in turn induces an automorphism of $\Delta^{*}(G)$. Therefore, $\Delta^{*}(G)$ is a characteristic subgroup of $G$.

If $K$ is any finite normal subgroup of $G$, we have clearly $K \subseteq \Delta^{*}(G)$. Conversely, if $a \in \Delta^{*}(G)$, let $A = \{a_{1}, \ldots, a_{n}\}$ be the set of all conjugates of $a$. Then $A \subseteq \Delta^{*}(G)$, and $A$ generates a finite normal subgroup of $\Delta^{*}(G)$ containing $a$.