Answer to Question #223549 in Abstract Algebra for Prathibha Rose

Question

Let a be an element of.order n in a group and k be a positive integer .prove that |a^k| = n/(gcd (n,k))

Answer 1

Let $a$ be an element of order $n$ in a group $G$ with identity $e,$ and $k$ be a positive integer. Let $n=dn_1, k=dk_1,$ where $d=gcd(n,k).$ Then $(n_1, k_1)=1.$ Let $|a^k|=t.$ If $(a^k)^t=e,$ then $a^{kt}=e,$ and hence $n=dn_1$ is a divisor of $kt=dk_1t.$ Since $(n_1, k_1)=1,$ we conclude that $n_1$ divides $t.$ Taking into account that $(a^k)^{n_1}=a^{dk_1n_1}=a^{nk_1}=(a^n)^{k_1}=e$ and $n_1$ divides $t,$ we conclude that $t=n_1,$ and hence $|a^k|=n_1=\frac{n}{d}=\frac{n}{gcd(n,k)}.$