Direct product of two nilpotent groups is nilpotent and direct product of two solvable groups is solvable Let $G = G_{1} \times G_{2}$.
I need to prove the following two things:


*

*If $G_{i}$ is nilpotent of degree $n_{i}$, $i = 1, 2$, then $G$ is nilpotent of degree $n = \max \{ n_{1}, n_{2} \}$.

*If $G_{i}$ is solvable (some people call them soluble) of degree $n_{i}$, $i = 1,2$, then $G$ is solvable of degree $n = \max \{n_{1},n_{2} \}$.


For the first one, for $G_{1}$, letting $G_{1,0}=G_{1}$, $G_{1,i+1}=G_{1,i}/C(G_{1,i})$ for $i = 0, 1, 2, \cdots$, $G_{1}$ is nilpotent if $G_{1,n_{1}}=G_{1,n_{1}-1}/C(G_{1,n_{1}-1})=\{ e \}$ for some $n_{1}$. Likewise, for $G_{2}$, letting $G_{2,0}=G_{2}$, $G_{2,i+1}=G_{2,i}/C(G_{2,i})$ for $i = 0,1,2, \cdots$, then $G_{2}$ is nilpotent if $G_{2,n_{2}}=G_{2,n_{2}-1}/C(G_{2,n_{2}-1})= \{ e\}$ for some $n_{2}$. 
So, if $G = G_{1} \times G_{2}$ were nilpotent of degree $n = \max \{n_{1}, n_{2}\}$, then $G_{\max\{ n_{1},n_{2}\}} =G_{\max\{ n_{1},n_{2}\}-1}/C(G_{\max\{ n_{1},n_{2}\}-1}) = \{ e\} $, but I can't figure out how to put the two parts together. I'm really quite clueless on how I should proceed, if this is even the best way to prove this, or even how to prove it in general. Therefore, guidance, preferably very detailed guidance, is needed.
For the second one, I was wondering if the fact that if $G = G_{1} \times G_{2}$, then $G_{1} \trianglelefteq G$, $G_{2} \trianglelefteq G$ would be needed? Although, I'm not going to lie - I'm just as clueless on this part as I was on the first part. 
Thanks! :) 
 A: The key fact you want to use here is that $C(G_1\times G_2)=C(G_1)\times C(G_2)$ (prove this if it's not obvious to you).  And if $H_1\trianglelefteq G_1$ and $H_2\trianglelefteq G_2$, then $(G_1\times G_2)/(H_1\times H_2)\cong (G_1/H_1)\times(G_2/H_2)$.  This gives by induction on $n$ that $(G_1\times G_2)_n\cong G_{1,n}\times G_{2,n}$ for all $n$.  Since a product of groups is trivial iff each factor is trivial, this gives you the result.
For solvability the idea is very similar, but instead of looking at the operation of replacing $G$ by $G/C(G)$, you are looking at the operation of replacing $G$ by $[G,G]$.  So you want to prove that $[G_1\times G_2,G_1\times G_2]=[G_1,G_1]\times[G_2,G_2]$, and then make a similar induction argument.
A: For any groups $G_1,G_2$ and normal subgroups $H_1 \trianglelefteq G_1, H_2 \trianglelefteq G_2$, we have that
$$C(G_1 \times G_2) = C(G_1) \times C(G_2)$$
and
$$\frac{G_1 \times G_2}{H_1 \times H_2} \cong \frac{G_1}{H_1} \times \frac{G_2}{H_2}.$$
Combining these gives that
$$\frac{G_1 \times G_2}{C(G_1 \times G_2)} \cong \frac{G_1}{C(G_1)} \times \frac{G_2}{C(G_2)}.$$

Note: with the index $i$, I mean the $i$-th term in the series, not the $i$-th factor of the direct product!
Let's switch to your groups $G, G_1$ and $G_2$ now. We want to prove (by induction) that $G_{i} \cong G_{1,i} \times G_{2,i}$ or all $i = 0,1, \dots$ . The base case $i = 0$ is just $G = G_1 \times G_2$ which is exactly how we defined $G$, so nothing has to be proven. Now, assume that $G_{i} \cong G_{1,i} \times G_{2,i}$ for some $i$. We then have
$$G_{i+1} = \frac{G_i}{C(G_i)} \cong \frac{G_{1,i} \times G_{2,i}}{C(G_{1,i} \times G_{2,i})} \cong \frac{G_{1,i}}{C(G_{1,i})} \times \frac{G_{2,i}}{C(G_{2,i})} = G_{1,i+1} \times G_{2,i+1}.$$
Hence
\begin{align}G_i = \{e\} &\iff G_{1,i} \times G_{2,i} = \{e\} \times \{e\}\\
&\iff G_{1,i} = \{e\} \text{ and } G_{2,i} = \{e\}\\
&\iff i \geq n_1 \text{ and } i \geq n_2\\
&\iff i \geq \max\{n_1,n_2\}
\end{align}
thus $G$ has nilpotency class $\max\{n_1,n_2\}$.

Now, on to the solvable part. Working with general groups $G_1,G_2$ again, we can quickly show that
$$[G_1 \times G_2,G_1 \times G_2] = [G_1,G_1] \times [G_2,G_2].$$
Since a commutator subgroup is generated by the set of commutators, it suffices to note the following for $g_1,h_1 \in G_1$, $g_2,h_2 \in G_2$:
\begin{align}
[(g_1,g_2),(h_1,h_2)] &= (g_1,g_2)^{-1}(h_1,h_2)^{-1}(g_1,g_2)(h_1,h_2) \\
&= (g_1^{-1},g_2^{-1})(h_1^{-1},h_2^{-1})(g_1,g_2)(h_1,h_2)\\
&= (g_1^{-1}h_1^{-1}g_1h_1, g_2^{-1}h_2^{-1}g_2h_2)\\
&= ([g_1,h_1],[g_2,h_2]).
\end{align}

Now, back to your groups $G, G_1$ and $G_2$. Let's denote $G^{(0)} = G$ and $G^{(i+1)}  = [G^{(i)},G^{(i)}]$, and the same notation for the factors $G_1, G_2$. We want to prove by induction that $G^{(i)} = G_1^{(i)} \times G_2^{(i)}$ for all $i = 0,1 ,\dots$ . Once again the base case $i = 0$ is just the definition of $G$, so let's assume that $G^{(i)} = G_1^{(i)} \times G_2^{(i)}$ for some $i$. Then
\begin{align}
G^{(i+1)} &=  [G^{(i)},G^{(i)}]\\
&= [G_1^{(i)} \times G_2^{(i)},G_1^{(i)} \times G_2^{(i)}]\\
&= [G_1^{(i)} ,G_1^{(i)} ] \times [G_2^{(i)},G_2^{(i)}]\\
&= G_1^{(i+1)} \times G_2^{(i+1)} .
\end{align}
We finish the proof with the same argument as before:
\begin{align}G^{(i)} = \{e\} &\iff G_1^{(i)} \times G_2^{(i)} = \{e\} \times \{e\}\\
&\iff G_1^{(i)} = \{e\} \text{ and } G_2^{(i)} = \{e\}\\
&\iff i \geq n_1 \text{ and } i \geq n_2\\
&\iff i \geq \max\{n_1,n_2\}
\end{align}
thus $G$ has derived length $\max\{n_1,n_2\}$.
