Q:

Composition series and direct product

I have two solvable groups $G_1$ and $G_2$ and wonder whether their direct product $G_1 \times G_2$ is solvable.
I can easily write subnormal series $G_1 \times G_2 \rhd G_1 \rhd \cdots$ and $G_1 \times G_2 \rhd G_2 \rhd \cdots$ and I know that this series has to have abelian composition factors if and only if G is to be solvable. So if $(G_1 \times G_2)/G_1$ and $(G_1 \times G_2)/G_2$ are abelian, my group is solvable. I know that these factors are abelian if $(G_1 \times G_2)$ is abelian but I suspect they are not in general.
What also bugs me is that composition series have to be the same lenght. So if  $G_1 \times G_2$ is solvable and $G_1$ has more composition factors than $G_2$ it has to be somehow guaranteed that I always find enough normal subgroups to "squeeze" in between $G_1 \times G_2$ and $G_2$ in the series $G_1 \times G_2 \rhd G_2 \rhd \cdots$
Is this correct and can somebody offer some intuition here?

A:

This is an instance of the useful
Theorem: Let $G$ be a group and $N\subset G$ a normal subgroup. Then $G$ is solvable if and only if both $N$ and $G/N$ are solvable.
Here, we use $N = G_1 \times \{1\}$, and by assumption both $N$ ($\cong G_1$) and $G/N$ ($\cong G_2$) are solvable, hence $G = G_1\times G_2$ is also solvable.
We can write a composition series explicitly: If
$$G_1 = H_0 \rhd H_1 \rhd \dotsc \rhd H_m = \{1\}$$
and
$$G_2 = K_0 \rhd K_1 \rhd \dotsc \rhd K_n = \{1\}$$
are composition series of $G_1$ resp. $G_2$, then
$$G_1\times G_2 = H_0\times K_0 \rhd H_1\times K_0 \rhd \dotsc \rhd H_m\times K_0  \rhd H_m\times K_1 \rhd \dotsc \rhd H_m\times K_n = \{1\}\times\{1\}$$
is a composition series of $G_1\times G_2$.

