Question

Show that $\mathit{G}\mathbf{/}\mathit{M}$ is not isomorphic to $\mathit{G}\mathbf{/}\mathit{N}$ (the operation table for $\mathit{G}\mathbf{/}\mathit{N}$is in example 4). Thus, for normal subgroups $\mathit{M}$ and $\mathit{N}$ the fact that $\mathit{M}\mathbf{\cong }\mathit{N}$ does not imply that $\mathit{G}\mathbf{/}\mathit{M}$ is not isomorphic to$\mathit{G}\mathbf{/}\mathit{N}$.

Short answer

The group $G/M$ is not isomorphic to $G/N$.

Step-by-step solution

Normal subgroup and Quotient group 

Let $N$ be the normal subgroup of $G$. Then

1. $G/N$ is a group under the operation defined by $\left(Na\right)\left(Nc\right)=Nac$.

2. If $G$ is finite, then the order of$G/N$ is role="math" localid="1654500118450" $\left|G\right|/\left|N\right|$.

3. If $G$ is an abelian group, then so is $G/N$.

The group$G/N$ is called the quotient group or factor group of $G$ by$N$ .

Group G and their subgroups

Let $\mathbf{G}\mathbf{=}{\mathbf{\mathbb{Z}}}_{\mathbf{2}}\mathbf{\times }{\mathbf{\mathbb{Z}}}_{\mathbf{4}}$ be the group of order 8 and M be the subgroup generated by $(0,2)$.

The elements in M are $\left\{(0,2)(0,0)\right\}$, which is of order 2; then the quotient group $G/M$and$G/N$ has order 4.

G/M is not isomorphic to G/N  

The element $N+(0,1)$ has order 4 in $G/N$ , but there are no elements of order 4 in $G/M$.

Therefore, the two groups $G/M$ and $G/N$ are not isomorphic.