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I am interested in the following question about finite groups. Consider a finite group $G$ and a function which takes group elements and returns their $n$-th power, equivalently $g \mapsto g^n$. For certain finite groups, e.g., cyclic groups of prime order $p$ and $n \perp p$ this map is invertible on all elements, while for the alternating group on $4$ elements and $m = 2$, the map is only invertible on cycle types of the form $(1, 0, 1, 0)$ (e.g., $(1,2,3)(4)$).

What are methods to determine the size of the pre-image for a given $h = g^m$, and conditions under which this pre-image is uniquely $g$?

pedalferrous
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  • So you want to know, for all finite groups $G$, and all elements $g$ of $G$, and for all $n$, how many $n$th roots $g$ has in $G$. That's a lot to ask. – Gerry Myerson Nov 29 '20 at 05:01
  • For example, in groups of order $16$, the number of solutions of $x^2=1$ can be $2,4,6,8,10,12$ or $16$. – Gerry Myerson Nov 29 '20 at 05:04
  • There is a cute formula for the size of the preimage in terms of characters: https://mathoverflow.net/questions/42653/number-of-n-th-roots-of-elements-in-a-finite-group-and-higher-frobenius-schur-in – Qiaochu Yuan Nov 29 '20 at 05:14
  • Another reference is Huppert's book "Character Theory of Finite Groups", there is a chapter named "On the number of solutions to $g^n = 1$" or something similar. I don't have the book at hand right now, so I unfortunately can't give the precise page numbers. – Marktmeister Nov 29 '20 at 09:26
  • https://math.stackexchange.com/questions/1132336/when-is-a-power-map-a-homomorphism – tryst with freedom Mar 10 '22 at 00:40

1 Answers1

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Proposition Let $G$ be a finite group, $n$ a non-zero integer, then the map $f: G \rightarrow G$ defined by $f(g)=g^n$ is a bijection if and only if gcd$(n,|G|)=1$.

Proof Assume $f$ is bijection and let $p$ be a prime dividing both $|G|$ and $n$. By Cauchy's Theorem, one can find a $g \in G$ of order $p$. But $f(g)=g^n=(g^p)^{\frac{n}{p}}=1$, so $g=1$ as $f$ is bijective, a contradiction.
If $|G|$ and $n$ are relatively prime, then by Bézout's Theorem, we can find $a,b \in \mathbb{Z}$ with $1=a|G|+bn$. Hence for each $g \in G$, $g=g^1=g^{a|G|+bn}=(g^{|G|})^a \cdot (g^b)^n=(g^b)^n$, showing that $f$ is surjective. Since $G$ is finite, it follows that $f$ is injective.

If $n$ and $|G|$ are not relatively prime, then one can use character theory of finite groups over $\mathbb{C}$ to say something about "$\sqrt[n]{g}$". One can define, assuming $n$ is positive, $\vartheta_n(g)=\#\{h \in G: h^n=g\}$. This is a class function on $G$ (constant on conjugacy classes) and hence one can write $\vartheta_n=\sum_{\chi \in Irr(G)}\nu_n(\chi)\chi$, where $\nu_n(\chi) \in \mathbb{C}$ are uniquely determined. So if you know these values of $\nu_n$ for each $\chi \in Irr(G)$, you know the numbers $\vartheta_n$. More on this theory can be found in many textbooks, for example that of I.M. Isaacs, Character Theory of Finite Groups, Chapter $4$, Products of Characters, page $49$ and following. Or see Y.G. Berkovich, L.S. Kazarin, E.M. Zhmud Characters of Finite Groups, page $152$, IV.5-7.

Nicky Hekster
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  • Thank you for a response to this question; while I agree with your analysis I think my original concern was not whether the map $f$ you've defined is a bijection, but rather if it is invertible on any non-trivial subset of its image. E.g., within $S_4$, squares of elements generate a subgroup ($A_4$), but this map is only invertible on three-cycles. – pedalferrous Dec 10 '20 at 16:22
  • Excellent work. – tryst with freedom Mar 10 '22 at 00:42