8
$\begingroup$

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.

$\endgroup$
6
  • 3
    $\begingroup$ The map $g\mapsto g^k$ is an isomorphism from $G_k$ to $G$. $\endgroup$
    – ahulpke
    Commented Jul 1, 2022 at 17:36
  • $\begingroup$ @ahulpke That's unfortunate. I don't know how I miss stuff like this sometimes. $\endgroup$
    – legionwhale
    Commented Jul 1, 2022 at 17:57
  • 1
    $\begingroup$ this is a special case of a much more general phenomenon, known as transport of structure $\endgroup$
    – Atticus Stonestrom
    Commented Jul 1, 2022 at 17:58
  • 3
    $\begingroup$ 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 $\endgroup$
    – Atticus Stonestrom
    Commented Jul 1, 2022 at 18:00
  • 2
    $\begingroup$ 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 $\endgroup$
    – Atticus Stonestrom
    Commented Jul 1, 2022 at 18:06

1 Answer 1

Reset to default
8
$\begingroup$

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 $G$.)

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – legionwhale
    Commented Jul 1, 2022 at 18:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .