# Intuition for why a group can fail to have an automorphism sending a particular element to its inverse.

It is known that there are groups $$G$$ containing an element $$g$$ which is not mapped to $$g^{-1}$$ by an automorphism of $$G$$, but I find this counterintuitive; when I visualize the symmetries of some object in $$\mathbb{R}^2$$ or $$\mathbb{R}^3$$, it seems evident to me that doing a rotation one way vs its inverse are performing the same "role" in the group, and hence should be mapped to each other by an automorphism. I'm also aware that every (finite) group can be viewed as the symmetries of some object in $$\mathbb{R^n}$$

Is there an example (ideally geometric) where my intuition can accept that this can happen? Doing a computation in some semi-direct product isn't satisfying to me.

• I think it would be better if you studied the counterexample in the answer that you cite and then asked again for help if you failed to understand it. We can't usefully speculate about what your intuition can accept or on what reasoning is acceptable to you. Commented Oct 13, 2020 at 0:02
• I believe I understand the example, but I don't find it enlightening; it'd probably be difficult to construct an object in some higher dimensional space with that symmetry group and visualize it well enough, I'm just wondering if there's some nicer way Commented Oct 13, 2020 at 0:14
• An interesting aside is that for any group $G$, and any element $x\in G$, there is always a group $K$, $G\leq K$, where $x$ is conjugate to $x^{-1}$. This can be achieved through HNN extensions. Commented Oct 20, 2020 at 2:07
• The best I can come up with is the following. You are absolutely right, that rotation in one direction is the 'opposite' of rotation in the other direction. The automorphism that maps rotation to its inverse is a reflection in the appropriate hyperplane. Why should that rotation preserve the other elements of the group? Commented Oct 20, 2020 at 22:14
• I think the problem lies in the fact that it is simply impossible, and very noninstructive, to base your intuition for groups solely on subgroups of $O(\Bbb{R}^2)$ and $O(\Bbb{R^3})$. Almost nothing happens there. Visualising $S_5$ this way is painful, and you can't even get to $S_6$! Commented Oct 21, 2020 at 20:26

Below is an arguably geometric example, it may not be what you're looking for.

For a vector space $$V$$ over a field $$F$$, let $$A(V)$$ be the group of affine maps on $$V$$ which are compositions of translations and scaling maps, i.e. the affine maps of the form $$x \mapsto ax + v$$, for $$a \in F \setminus \{0\}$$ and $$v \in V$$. Let $$T(V)$$ be the subgroup of $$A(V)$$ consisting of translations, i.e. the maps $$x \mapsto x + v$$. Note that $$T(V) \cong V$$ as groups. We include the translations because they encode additional structure of the action of scaling maps on $$V$$; if we only considered the group of scaling maps, this would just be $$F \setminus \{0\}$$ under multiplication, which is abelian, and thus has automorphism $$g \mapsto g^{-1}$$.

We will now prove that when $$F = \mathbb{Q}$$ or $$\mathbb{F}_p$$, every automorphism $$\phi$$ of $$A(V)$$ preserves scaling factor, meaning if $$f \in A(V)$$ is of the form $$x \mapsto ax + v$$, then $$\phi(f)$$ is of the form $$x \mapsto ax + v'$$. In particular, this will mean that every element of $$A(V)$$ with scaling factor $$a \neq \pm 1$$ is not sent to its inverse by any automorphism of $$A(V)$$.

Lemma 1: Let $$m, n$$ be integers which are not zero in $$F$$. Then for $$f \in A(V)$$, $$f$$ has the property that $$fg^n = g^m f$$ for all $$g \in T(V)$$ if and only if $$f$$ has scaling factor $$m/n$$, meaning $$f$$ is of the form $$x \mapsto (m/n)x + v$$.

Proof: If $$f$$ is of the given form, then for $$g(x) = x + u$$, clearly $$f(g^n(x)) = (m/n)(x + nu) + v = (m/n)x + v + mu = g^m(f(x))$$ so $$fg^n = g^m f$$. In the other direction, if $$f$$ satisfies $$fg^n = g^m f$$ for all $$g \in T(V)$$, then writing $$f(x) = ax + v$$, and taking $$g(x) = x + u$$ for some $$u \neq 0$$, we have $$a(x + nu) + v = ax + v + mu$$, giving $$anu = mu$$, hence $$a = m/n$$ as desired.

Lemma 2: If $$F = \mathbb{Q}$$ or $$\mathbb{F}_p$$, then every automorphism $$\phi$$ of $$A(V)$$ preserves $$T(V)$$, meaning $$\phi(T(V)) = T(V)$$.

Proof: In the case $$F = \mathbb{Q}$$, we can identify the subgroup $$T(V)$$ of $$A(V)$$ as the set of "divisible" elements, namely those elements $$g \in A(V)$$ for which, for any positive integer $$n$$, there is an element $$h \in A(V)$$ with $$h^n = g$$. Elements $$g \in A(V)$$ of the form $$x \mapsto ax + v$$ for $$a \neq 1$$ do not have this property, since there can be an $$h$$ with $$h^n = g$$ only if $$a$$ is an $$n$$-th power in $$\mathbb{Q}$$, and the only nonzero element of $$\mathbb{Q}$$ which is an $$n$$-th power for each $$n$$ is $$a = 1$$. It is clear that any automorphism maps divisible elements to divisible elements, and non-divisible elements to non-divisible elements, so any $$\phi$$ has $$\phi(T(V)) = T(V)$$.

In the case $$F = \mathbb{F}_p$$, we can identify $$T(V)$$ as the set of elements of order $$p$$ (together with the identity). For an element $$g$$ of the form $$x \mapsto ax + v$$ for $$a \neq 1$$, $$g$$ is conjugate to the map $$x \mapsto ax$$, and thus $$g$$ has order dividing $$p-1$$, since this latter map has order dividing $$p-1$$. Automorphisms preserve order, so again any $$\phi$$ has $$\phi(T(V)) = T(V)$$. [end proof]

Now, let $$\phi$$ be an automorphism of $$A(V)$$, and let $$f \in A(V)$$, so $$f(x) = ax + v$$ for some $$a, v$$. Since $$F = \mathbb{Q}$$ or $$\mathbb{F}_p$$, $$a$$ is of the form $$m/n$$ for some integers $$m, n$$ which are not zero in $$F$$, and thus by Lemma 1, $$fg^n = g^m f$$ for any $$g \in T(V)$$. By Lemma 2, $$\phi$$ preserves $$T(V)$$, so $$\phi(f)$$ also has the property that $$\phi(f) g^n = g^m \phi(f)$$ for all $$g \in T(V)$$. Then, again by Lemma 1, this means that $$\phi(f)$$ has scaling factor $$m/n = a$$, so $$\phi$$ preserves the scaling factor of $$f$$.

• This is very nice! While not quite geometric in the sense I was thinking, I do find this a satisfying example. Slightly better would be an example like this, using the reals instead of the rationals (just to make things feel more geometric/no number theory) Commented Oct 23, 2020 at 22:40

Suppose you have an automorphism $$\phi:G\to G$$ such that $$g\mapsto g^{-1}$$ for every $$g\in G$$. Now let $$g,h\in G$$ and consider $$\phi(gh)$$. We have $$\phi(gh)=(gh)^{-1}=h^{-1}g^{-1}$$. On the other hand $$\phi(gh)=\phi(g)\phi(h)=g^{-1}h^{-1}$$. So it is necessary to $$g^{-1}h^{-1}=h^{-1}g^{-1}$$ for every $$g,h\in G$$. So $$G$$ must be abelian. Therefore you can take any group that is not abelian as your example. Any free group is fine, but if you want geometric example take dihedral group $$D_2$$ (isometries of a square). For $$g$$ take reflection with respect to vertical axis and for $$h$$ a reflection with respect to horizontal axis.
Edit: Considering comment below: the element which cannot be mapped to its iverse in this situation is $$gh$$ - a rotation by angle $$\pi/2$$.

• The question is not about automorphisms which invert every element. Commented Oct 14, 2020 at 9:08
• The question is about example of a group with element, that is not mapped to its inverse by an automorphism. Author of the question find this counterintuitive so i put myself in his shoes and supposed contrary, that there is not such element in some group. Therefore every element must be mapped to its inverse. I showed that this implies, that the group must be abelian. So if our group was not abelian we have contradiction. So we need to look for examples among groups that are not abelian, and i provided such examples. Commented Oct 14, 2020 at 9:18
• Your "example" won't do, elements of order $2$ are all mapped to their inverses by the trivial automorphism. With respect I don't think you understand the question. Commented Oct 14, 2020 at 9:25
• The negation of the statement "there is some $g\in G$ such that no automorphism maps $g$ to $g^{-1}$" is "for every $g\in G$ there is an automorphism $\phi$ such that $\phi(g)=g^{-1}$". But your answer talks about the stronger property: "there is an automorphism $\phi$ such that $\phi(g)=g^{-1}$ for all $g\in G$". Commented Oct 14, 2020 at 10:56
• So I agree that this doesn’t answer the question. However, I don’t think it’s irrelevant (and I didn’t downvote). But perhaps it would be better as a comment that $G$ is abelian iff $x\mapsto x^{-1}$ is an automorphism, and so the behavior the OP describes can only happen in non-abelian groups. Commented Oct 14, 2020 at 10:57