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.

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

Yes, this is correct.

I'm posting this CW answer so that users who confidently concur have something to vote on, and so this question doesn't stagnate in the Unanswered Questions Queue. If however anyone would like to write a more substantial response to the question, please downvote this answer and post your own.


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