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

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 $\\{A_i \times B_i \\}_{i=0}^{t=s}$.