Show that if $A, B$ are groups, then $A \times B$ is solvable if and only if both $A$ and $B$ are solvable?

Question

Show that if $A, B$ are groups, then $A \times B$ is solvable if and only if both $A$ and $B$ are solvable? Why is obvious that if $A$ and $B$ are solvable then $A \times B$ is solvable?

Answer 1

Suppose $A \times B$ is solvable. Since any subgroup of a solvable group is solvable we have that $A \times \{e_B\} \cong A$ and $\{e_A\} \times B \cong B$ are solvable.

Conversely, suppose $A$ and $B$ are both solvable: There exist sub-normal series

$$1=A_0 \trianglelefteq A_1 \trianglelefteq \dots \trianglelefteq A_t=A $$

$$1=B_0 \trianglelefteq B_1 \trianglelefteq \dots\trianglelefteq B_s=B $$

with abelian quotients. You can make both series have the same length by repeating one of subgroups. Consider then the subgroups $\{A_i \times B_i \}_{i=0}^{t=s}$.

Answer 2

Think to the definition of solvability of groups. If $A$ and $B$ are solvable, then since $A \times B$ has a canonical isomorphic copy of $A$ embedded within it (take $a \in A, b = 1 \in B$), then $A$ is solvable so you have a chain of normal subgroups with the property needed for solvability. The same argument holds for $B$.