I'm struggling with an assignment question on the topic of soluble groups. The question is to prove that if $G = H \times K$ is a soluble group, and $H$ and $K$ have derived lengths $n$ and $m$ respectively, then the derived length of $G$ is given by $\max\{n,m\}$.
Here, the derived length of a soluble group is the length of the derived series, given by $G \geq G^{'} \geq G^{''} \geq \cdots \geq G^{(n)} = {e}$, where $G' = \langle [x,y] : x,y \in G \rangle$ and $G^{(r)} = (G^{(r-1)})'$.
My working so far: Since $G = H \times K$, I've tried to expand out $G'$:
$G' = \langle [x,y] : x,y \in G \rangle = \langle [(h_1,k_1),(h_2,k_2)] : h_1,h_2 \in H, k_1, k_2 \in K \rangle = \langle (h_1,k_1)^{-1}(h_2,k_2)^{-1}(h_1,k_1)(h_2,k_2) : h_1,h_2 \in H, k_1, k_2 \in K \rangle = \langle (h_1^{-1}h_2^{-1}h_1h_2, k_1^{-1}k_2^{-1}k_1k_2) : h_1,h_2 \in H, k_1, k_2 \in K \rangle = \langle ([h_1, h_2], [k_1, k_2]) : h_1,h_2 \in H, k_1, k_2 \in K \rangle = (\langle [h_1, h_2] : h_1,h_2 \in H \rangle, \langle [k_1, k_2]: k_1, k_2 \in K \rangle) = H' \times K'$
Is this manipulation legitimate? If so, does it help? I can kind of see how (if $m > n$) we would have $G^{(n)} = H^{(n)} \times K^{(n)} = \{e\} \times K^{(n)}$, which would 'force' $G^{(n)} = \{e\}$, but what's to stop us from simply calculating $G^{(n+1)} = \{e\} \times K^{(n+1)}$ (since $\{e\}'=\{e\}$)?
Am I on the right track here? Any help would be greatly appreciated
You're on the right track. If $m>n$, then $H^{(m)}=K^{(m)}=\{e\}$ so $G^{(m)}=H^{(m)}\times K^{(m)} = \{e\}\times\{e\}=\{e\}$.