# Solvability of a direct product of solvable groups [duplicate]

I can prove that a direct product of cyclic groups is not necessarily a cyclic group. Also it is easy to show that a direct product of abelian groups is abelian. I am curious about the next question.

Is a direct product of solvable groups is solvable?

I do not know if I should show that there is a subnormal series whose factors are abelian, or try to show that such a series does not exist.

There is an easy way to show that a direct product of solvable groups is a solvable group.

Suppose that $$N\unlhd G$$, $$N$$ is solvable and $$G/N$$ is solvable. Then $$G$$ is solvable.

In your case $$H\times K = G$$, where $$H, K$$ are solvable groups. We have $$G/K\simeq H$$ is solvable, so $$G$$ is solvable by the previous statement.

Another way is to construct a subnormal series with abelian quotients. Since $$H$$ and $$K$$ are solvable, there are series $$H\unlhd H_1\unlhd\ldots\unlhd H_{n-1}\unlhd\{e\}$$ and $$K\unlhd K_1\unlhd\ldots\unlhd K_{m-1}\unlhd\{e\}$$, whose quotients are abelian.

Let us construct such a series for $$H\times K$$.

$$H\times K\unlhd H_1\times K\unlhd\ldots H_{n-1}\times K\unlhd \{e\}\times K\unlhd\{e\}\times K_1\unlhd\ldots\unlhd\{e\}\times K_{m-1}\unlhd\{e\}\times\{e\}$$

By the definition of direct product it is easy to see that first $$n-1$$ quotients are isomorphic to quotients of $$H$$ and the others are isomorphic to quotients of $$K$$ (so all quotients are abelian). We constructed a subnormal series with abelian quotients, so $$H\times K$$ is solvable.

A group $$G$$ is solvable if and only if there exists a normal subgroup $$N$$ such that $$N$$ and $$G/N$$ are both solvable. Apply this and induction, and you obtain that a direct product of solvable groups is solvable.

Intuitively, the idea is to take a composition series. Start with a composition series for one direct factor. Then the composition series for the second direct factor can be glued on top of the first. Now continue with this construction.