If $G=H\times K$ with $H,K$ solvable, show $G$ is solvable

Question

I want to prove

If $G=H\times K$ with $H,K$ solvable, then $G$ is solvable.

I'm not sure how to do this, but my idea so far is to somehow combine the normal series of the two subgroups in some way. My professor said use the fact that given $N$ normal in $G$, $G$ is solvable iff $G/N$ and $N$ are solvable. I'm not sure how the quotient group fits into the picture at all.

Thanks for the help.

Answer 1

The trick is to use the fact that $H$ and $K$ are normal. To see this, consider the projection homomorphisms: $K$ is the kernel of $\pi_1: H \times K \rightarrow H$ defined by $(h, k) \mapsto (h, e)$, and $H$ is the kernel of $\pi_2: H \times K \rightarrow K$ defined by $(h, k) \mapsto (e, k)$.

Knowing this, you can apply the useful theorem.

Answer 2

Another way that you can use is

Define $G^{(i+1)}=[G^{(i)},G^{(i)}]$
Since $H$ and $K$ is solvable, there exists $r,s\in \Bbb{N}$ such that $H^{(r)}=\{1\}$ and $K^{(s)}=\{1\}$.
Then you can verify that $(H\times K)^{(m)}=H^{(m)}\times K^{(m)}$ for any $m\geq1$.
Take $m=\max(r,s)$, you will get take $G^{(m)}=\{1\}$