Injective mapping onto a subgroup of a finite group.

Question

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.

Answer 1

Yes, this is correct.

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