Suppose $G$ has odd order, show the function $f:G\rightarrow G$ defined by $f(x)=x^2$ is injective. This proposition is easily provable if we assume $G$ is Abelian, but I don't know how to start this without the assumption of being Abelian.

  • $\begingroup$ More general question asked later. $\endgroup$
    – PinkyWay
    Commented Oct 9, 2023 at 17:30

3 Answers 3


Let $|G|=2n-1$. Assume $a^2=b^2$. Then $a=a\cdot a^{2n-1}=(a^2)^n=(b^2)^n=b$.

  • 4
    $\begingroup$ Nice...............! +1 $\endgroup$
    – DonAntonio
    Commented Oct 11, 2013 at 6:09
  • $\begingroup$ @HagenvonEitzen What about the converse of this statement. i.e. if $f(x) = x^2$ is injective, prove G has odd order? $\endgroup$
    – user67773
    Commented May 5, 2014 at 6:17
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    $\begingroup$ @Umakant If $G$ is finite and $x\mapsto x^2$ is injective, then $x\ne x^{-1}$ for all $x\ne 1$. Hence $G\setminus\{1\}$ can be grouped into pairs of inverses and is of even cardinality. $\endgroup$ Commented Jan 4, 2017 at 16:03
  • $\begingroup$ @HagenvonEitzen Thanks. $\endgroup$
    – user67773
    Commented Jan 9, 2017 at 8:39

In general, $m,n \in \mathbb{Z}_{>0}$ with $\gcd(m,n)=1$ and $G$ is a group of order $n$, then the function $f:G\rightarrow G$ defined by $f(x)=x^m$ is a bijection.

Proof. By Bézout choose $k,l \in \mathbb{Z}$ with $km+ln=1$ and proceed in a similar way as in Hagen's proof.

Following lhf's remark (thanks!), the converse is also true: assume that $f(x)=x^m$ is a bijective function. Let $p$ be a prime with $p|n$. We will show that $p$ does not divide $m$: by Cauchy's Theorem there exists an $x \in G$ with $ord(x)=p$. Since $f(1)=1^m=1$ and $x \neq 1$, $x^m$ must be a non-trivial power of $x$. Hence $ord(x^m)=p$ and this can only be the case if $p \nmid m$. It follows that $m$ and $n$ must be relatively prime $\square$.

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    $\begingroup$ The converse is also true and $x\mapsto x^m$ is a bijection iff $(m,n)=1$. $\endgroup$
    – lhf
    Commented Oct 12, 2013 at 2:20
  • $\begingroup$ Yes, thanks never realized this - I wil add it in my proof above! $\endgroup$ Commented Oct 13, 2013 at 18:21

Hagen's answer is the best. But here is another take that uses the abelian case for the general case:

Take $g\in G$ and consider the restriction $\phi$ of $f$ to $C=\langle g \rangle$, the cyclic group generated by $g$. Because $C$ is a subgroup, $\phi$ maps $C$ to $C$. Because $C$ is abelian, $\phi$ is a homomorphism $C\to C$. Since $C$ has odd order, $\ker \phi$ is trivial and $\phi$ is injective. So $\phi$ is surjective, because $C$ is finite. But this means that $f$ itself is surjective and hence injective, because $G$ is finite.

  • $\begingroup$ Adapted from math.stackexchange.com/a/57915/589. $\endgroup$
    – lhf
    Commented Oct 11, 2013 at 11:21
  • $\begingroup$ Thanks. I just forgot the function is not a homo. +1 $\endgroup$
    – Mikasa
    Commented Oct 11, 2013 at 17:17

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