Question

Suppose that His a subgroup of a group G and that$\mathit{a}\mathbf{\in }\mathit{G}$ has order n. If ${\mathit{a}}^{\mathbf{k}}\mathbf{\in }\mathit{H}$and$\left(k,n\right)\mathbf{=}\mathbf{1}$ , prove that $\mathit{a}\mathbf{\in }\mathit{H}$.

Short answer

It is proven that $a\in H$

Step-by-step solution

Theorem 1.2

Let a and b be integers, and let d be their greatest common divisor. Then, there exist integers u and v such that $d=au+bv$.

Prove that a∈H

It is given that $\left(k,n\right)=1$. So, from Theorem 1.2:

$ck+bn=1$

Hence, it can be written as:

$$\begin{gathered} a^1=a^{ck+bn} \\ =a^{ck}{a}^{bn} \\ ={\left(a^k\right)}^c{\left(a^n\right)}^b \\ ={\left(a^k\right)}^c\in H \end{gathered}$$

Therefore, $H$satisfies the axiom of closure since $H$is a subgroup.

Hence, it is proved that $a\in H$.