Question

Let $\mathbf{N}\mathbf{,}\mathbf{K}$be subgroups of a group $\mathbf{G}$, with $\mathbf{N}$normal in $\mathbf{G}$. If $\mathbf{N}$and $\mathbf{K}$are abelian groups and $\mathbf{G}\mathbf{=}\mathbf{NK}$, is $\mathbf{G}$the direct product of $\mathbf{N}$and $\mathbf{K}$?

Short answer

Answer:

No, $G$may not be the direct product of $N$and $K$.

Step-by-step solution

Claim

Given that $N,K$are subgroups of a group $G$, with $N$normal in $G$, and $N$and $K$are abelian groups and $G=NK$.

Prove that $G$may not be the direct product of $N$and $K$by providing a counter example.

Prove by counter-example

Consider the dihydral group $D_4=\left\{1,r,r^2,r^3,t,rt,r^{2}t,r^{3}t\right\}$.

Here, $\left|r\right|=4$and $\left|t\right|=2$

This group follows the relation as:

$$\begin{gathered} rt=t{r}^{-1} \\ =t{r}^3 \end{gathered}$$

Consider its subgroups $N=\left\{1,r,r^2,r^3\right\}$and $K=\left\{1,t\right\}$.

Here, $N$is normal in $G$and both $N,K$being cyclic are abelian.

But $G\ne N\times K$since $rt\in G$.

In the group $G$, $trtr$can be written as:

$$\begin{gathered} trtr=tt{r}^{3}r \\ =t^2{r}^4 \\ =1 \end{gathered}$$

Hence, its image in $N\times K$must be $\left(1,1\right)$but this is not the case as $\left(r,t\right)\left(r,t\right)=\left(r^2,t^2\right)$implies $\left(r,t\right)\left(r,t\right)=\left(r^2,1\right)$, which can be written as $\left(r,t\right)\left(r,t\right)\ne \left(1,1\right)$as order of $r$is 4.