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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?

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    $\begingroup$ If $N$ is a normal subgroup of $G$, then $G$ is solvable if and only if both $N$ and $G/N$ are solvable. $\endgroup$ – Daniel Fischer Dec 4 '14 at 16:48
  • $\begingroup$ I know that two groups are normal subgroups of their direct product but I don't know what I can say about the quotient groups. $\endgroup$ – Marc Dec 4 '14 at 16:51
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    $\begingroup$ You know e.g. $(G_1\times G_2)/(G_1\times \{1\})$ up to isomorphism. $\endgroup$ – Daniel Fischer Dec 4 '14 at 16:52
  • $\begingroup$ Ah, this is $G_2$. So since $G_1$ and $G_2$ are normal in the direct product so are the quotient groups. So the direct product is always solvable. $\endgroup$ – Marc Dec 4 '14 at 17:12
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    $\begingroup$ Right, a direct product of (finitely many) solvable groups is solvable. $\endgroup$ – Daniel Fischer Dec 4 '14 at 17:13
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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 $G_1\times G_2$.

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