0
$\begingroup$

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.

$\endgroup$
0

2 Answers 2

Reset to default
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$