# Are groups always isomorphic to their image under power maps?

Let $$(G, \cdot)$$ be a finite group of order $$n$$. Consider the map:

$$G \to G, g \mapsto g^k$$

for $$k$$,$$n$$ coprime. This is injective, but generally not a homomorphism.

Define a new group $$G_k = (G,\circ)$$ by:

$$g \circ h = (g^k h^k)^{\frac{1}{k}}$$

where naturally $$g \mapsto g^{\frac{1}{k}}$$ is the (well-defined) inverse of $$g \mapsto g^k$$. It is easily verified that this gives a valid group operation.

Question: Is it true that $$G \cong G_k$$ for all $$k$$ coprime to $$n$$?

The map is clearly an isomorphism if $$G$$ is abelian (or more generally, $$k$$-abelian). Also, note that the order of any $$g \in G_k$$ is the same as the order of $$g \in G$$. Thus, $$G, G_k$$ have the same order sequence, so according to this answer, any counterexample must have $$n \ge 16$$, and I am not very familiar with these groups. (But I do not expect the statement to be true.)

I also noticed that if $$G \sim H$$ when $$H \cong G_k$$ for some $$k$$ (coprime to $$|G|$$), then $$\sim$$ is an equivalence relation.

• The map $g\mapsto g^k$ is an isomorphism from $G_k$ to $G$. Jul 1 at 17:36
• @ahulpke That's unfortunate. I don't know how I miss stuff like this sometimes. Jul 1 at 17:57
• this is a special case of a much more general phenomenon, known as transport of structure Jul 1 at 17:58
• for instance. if $(G,\cdot)$ is any group, and $f:X\to G$ is any bijection, then you can make $X$ into a group by defining $xy=f^{-1}(f(x)f(y))$ for every $x,y\in X$. (this is a good exercise to check!) the map $f$ will then give an isomorphism from $G$ to $X$ with the new multiplicative structure; the idea is that you are somehow "transporting the group structure of $G$ along $f$". your question is the special case when $X=G$ and $f$ is the map $x\mapsto x^k$. there's nothing special about groups here; you can do the same in much larger generality Jul 1 at 18:00
• the same idea applies in complete generality! :) it works with essentially any mathematical structure you may have come across. for example, you can transport structure along a bijection in the same way for rings, topological spaces, vector spaces, and so on. the idea is that, if you have a structure $Y$ and bijection $X\to Y$, all you are really doing is "relabelling" or "renaming" the elements of $Y$ by those of $X$. most of mathematics is invariant under this kind of relabelling, hence the reason we can do this Jul 1 at 18:06

Yes: the map $$f\colon G\to G_k$$ defined by $$f(g)=g^{1/k}$$ is a group isomorphism.
(Side note: one example of this is $$k=-1$$, for which $$G_k$$ is the "opposite group" of $$G$$.)
• Thanks for the answer! I was wondering then if this could be generalised to $k$ not coprime to $n$ (so that the map has some sort of "kernel") by restricting to some appropriate subset of $G$? I was curious if this could be considered as the action of $\mathbb{Z}$ (via $k \mapsto (G \to G, g \mapsto g^k)$) on all finite groups. Perhaps this is better suited as a new question, though. Jul 1 at 18:10