Question

If $\mathit{N}$is a finite normal subgroup of a group $\mathit{G}$ and if $\mathit{G}\mathbf{/}\mathit{N}$ contains an element of order $\mathit{n}$, prove that $\mathit{G}$ contains an element of order $\mathit{n}$.

Short answer

It is proved that $G$contains an element of order $n$.

Step-by-step solution

Consider the elements

Consider $G$ to be a group and $N$ be a finite normal subgroup of $G$. Consider that $G/N$has an element of order $n$.

We have to show that $G$ has an element of order $n$.

Assume $gN\in G/N$ be an arbitrary element of order $n$. This implies that we have $g^n\in N$.

Prove the result

Since N is a finite group, the order of each element of $G/N$ has finite order. Since $g^n$has finite order; this implies that:

$\begin{array}{c}{\left(g^n\right)}^k=g^{nk}\\ =e\end{array}$

This implies that $g^k$ is the element of order $n$.

Hence, it is proved that $G$ contains an element of order $n$.