Denote by $K_n(G)$ the $n$-th group in the lower central series, why $K_n(A \times B) = K_n(A) \times K_n(B)$? Let $G$ be a group, then
$$
 K_n(G) := \langle [g_1, \ldots, g_n] : g_i \in G \rangle
$$
where $[g_1, \ldots, g_n] := [[g_1, \ldots, g_{n-1}], g_n]$ is defined recursively. It is possible to show that
$$
 K_1(G) = G, \quad K_{i+1} = [K_i(G), G]
$$
where $[g,h] = g^{-1}h^{-1}gh$ denotes the commutator, and $[A,B]$ the subgroup generated by all commutators $[a,b]$ with $a \in A, b \in B$.
Why do we have 
$$
 K_n(A \times B) = K_n(A) \times K_n(B)?
$$
Of course the inclusion $K_n(A)K_n(B) \le K_n(A \times B)$ is trivial, and as these groups are characteristic and have trivial intersection as subsets of $A$ and $B$ we have $K_n(A)K_n(B) = K_n(A) \times K_n(B)$. But the other inclusion I do not see immediately?
 A: I guess I found an answer, but shorter, more elegant or alternative answers would be appreciated. We use the following lemma:
Lemma: Let $A,B,C,D \unlhd G$ be normal subgroups, then
$$
 [AB,CD] = [A,C][A,D][B,C][B,D].
$$
Proof: For $[ab,cd] \in [AB,CD]$ we have
\begin{align*}
 [ab,cd] & = [a,cd]^b[b,cd] \\
         & = ([a,d][a,c]^d)^b[b,d][b,c]^d \\
         & = [a^b, d^b][a^{db}, c^{db}][b,d][b^d,c^d] \le [A,D][A,C][B,D][B,C]
\end{align*}
as all subgroups are normal. Further as the commutator subgroups are also normal, they commute pairwise and $[A,D][A,C][B,D][B,C] = [A,C][A,D][B,C][B,D]$
and their product forms a group. Hence every finite product of commutators of the above form lies in it and we have
$$
 [AB,CD] \le [A,C][A,D][B,C][B,D].
$$
The other inclusion follows as $[A,C], [A,D], [B,C], [B,D]$ are all contained in $[AB,CD]$. $\square$
Now for the original claim. If $n = 2$ we have
$$
 K_2(A\times B) = [AB, AB] = [A,A][A,B][B,A][B,B]
$$
and as $[A,B] \le A\cap B = 1$ and similar $[B,A] = 1$ and also as $[A,A] \le A, [B,B] \le B$ their intersection is trivial and
$$
 [A,A][A,B][B,A][B,B] = [A,A] \times [B,B] = K_2(A) \times K_2(B).
$$
Suppose inductively $n > 2$, then
\begin{align*}
 K_n(A \times B) & = [K_{n-1}(A\times B), A\times B] \\
                 & = [K_{n-1}(A) \times K_{n-1}(B), A \times B] \\
                 & = [K_{n-1}(A),A][K_{n-1}(A),B][K_{n-1}(B), A][K_{n-1}(B), B] \\
 & = [K_{n-1}(A), A] \times [K_{n-1}(B), B] \\
 & = K_n(A) \times K_n(B)
\end{align*}
with a similar reasoning as above.
