# Injective mapping onto a subgroup of a finite group.

Question: Let $$G$$ be a finite group and $$H$$ be a subgroup of $$G$$ such that $$\gcd(m,o(H))=1$$. Is $$\phi:H \to H$$ defined by $$\phi(x)=x^m$$ an injective mapping?

Attempt: $$\phi(x)=\phi(y) \implies x^m=y^m \implies x^my^{-m}=e \implies (y^{-1}x)^m=e$$. Now, by closure, $$y^{-1}x \in H$$ and if $$y^{-1}x\neq e$$, then we must have $$o(y^{-1}x) \mid m$$ and $$o(y^{-1}x) \mid o(H) \implies \gcd(m,o(H))>1$$, a contradiction.

Injection on a finite set into itself is bijective. Hence we can regard it a permutation on $$H$$.

Is this correct? Kindly verify.

• Looks correct. Of course at the end you meant $o(y^{-1}x)|o(H)$. – Mark Aug 31 '19 at 14:05
• @Mark, would you please consider checking math.stackexchange.com/questions/3339914/… ? – Subhasis Biswas Aug 31 '19 at 14:07
• Sorry, I saw the comment only now. If you assume $\gcd(m,o(H))=1$ then your solution becomes correct. – Mark Aug 31 '19 at 16:50
• @Mark , it is regarding the link in the comment? Well, $m = o(G/N)$, so $\gcd$ is always $1$. – Subhasis Biswas Aug 31 '19 at 17:07
• Yes, then it is correct. – Mark Aug 31 '19 at 17:09