Question

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

Short answer

It is proved that every element of $G$has finite order.

Step-by-step solution

Consider the elements

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

We have to show that every element of $G$has finite order.

Assume $g\in G$be an arbitrary element.

As every element of $G/N$ has finite order, this implies that $GN$ has a finite order in $G/N$, that is, there exists $k\in N$ such that $g^k\in N$.

Prove the result

Since every element of $G$ has finite order, this means that $g^k$ must have finite order.

Consider that there exists localid="1654520951573" $n\in \mathbb{N}$ such that ${\left(g^k\right)}^n=e$, where localid="1654520989040" $e$is the identity element of localid="1654520970431" $G$.

This implies that $g$ has order $nk$, that is, $G$has finite order.

Hence, it is proved that every element $G$ has finite order.