Question

If $\mathit{N}$ is a normal subgroup of a group $\mathit{G}$ and if ${\mathit{x}}^{\mathbf{2}}\mathbf{\in }\mathit{N}$ for every $\mathit{x}\mathbf{\in }\mathit{G}$, prove that every nonidentity element of the quotient group $\mathit{G}\mathbf{/}\mathit{N}$ of order 2.

Short answer

Every nonidentity element of the quotient group $G/N$ is of order 2.

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 $\left|G\right|/\left|N\right|$.

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 $N$ .

G/N is of order 2

Let $N$ be a normal subgroup of a group $G$ and $x^2\in N$ for every $x\in G$.

Let $x\in G$be the non identity element in G; then $xN$is the non identity element in $G/N$, and

$$\begin{gathered} \left(xN\right)\left(xN\right)=x^{2}N \\ =N(\because x^2\in N) \end{gathered}$$

Thus, $xN$ has order 2 in $G/N$.

Therefore, every non identity element of the quotient group $G/N$ is of order 2.