# Property of minimal normal subgroups

So, I'm having some problems with understanding one part of the proof of the theorem which states :

Let $G$ be a nontrivial finite group. If $K$ is a minimal normal subgroup of $G$, and $L$ is any normal subgroup of $G$, then either $K \leq L$ or $\langle K,L\rangle$=$K \times L$.

Proof: Since $K \cap L \vartriangleleft G$ the minimality of $K$ shows that either $K \leq L$ or $K \cap L = 1$. In the latter case $\langle K,L \rangle = KL = K \times L$ because both $K$ and $L$ are normal.

I don't understand the last equality which states that $KL = K \times L$. If somebody could give me a hint or something...

Thanks!!

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This follows from a standard property:

Lemma. If $M$ and $N$ are normal subgroups of $G$ and $M\cap N=\{1\}$, then $nm=mn$ for all $n\in N$ and $m\in M$.

Proof. Let $n\in N$ and $m\in M$. Consider $x=n^{-1}m^{-1}nm$. Writing it as $(n^{-1}m^{-1}n)m$ it follows from normality of $M$ that $x\in M$. Writing it as $n^{-1}(m^{-1}nm)$ it follows from normality of $N$ that $x\in N$. Since $N\cap M=\{1\}$, then $x=1$. But $n^{-1}m^{-1}nm=1$ is equivalent to $nm=mn$. $\Box$

The isomorphism $KL\cong K\times L$ (or equality if you think of the product as an internal direct product) now follows by mapping $(k,\ell)\in K\times L$ to $k\ell\in KL$.

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Thanks, I got it... –  aldo Jan 9 '12 at 18:25