Question

Do the same for Theorem 9.3.

Short answer

It has been proved that Theorem 9.3 is false if$N$is not normal.

Step-by-step solution

Step-By-Step SolutionStep 1: State the theorem

Theorem 9.3 If $M$and $N$ are normal subgroups of a group $G$such that $G=MN$and $M\cap N=⟨e⟩$, then $G=M\times N$.

Take D4 as example

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

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

This group follows the relation as:

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

Consider its subgroups $M=\{1,r,r^2,r^3\}$ and $N=\{1,t\}$

Here, role="math" localid="1652784354355" $G=MN$and $M\cap N=\left\{e\right\}$ but $G\ne M\times N$since $rt\in G$.

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

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

Hence, its image$M\times N$ in must be$\left(1,1\right)$ but this is not the case as$(r,t)(r,t)=(r^2,t^2)$ implies$(r,t)(r,t)=(r^2,1)$ .

Thus, it can be concluded that if$N$ is not normal then, $G\ne M\times N$.