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What is an example of a group $G$ with an element $g$ such that no automorphism of $G$ takes $g$ to its inverse?

Ernie060
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Joe Shipman
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  • You'd trivially want to state that the element $g$ should be nontrivial as every automorphism carries the identity to itself. – mathematics2x2life Oct 26 '13 at 02:24
  • Note that any such group must be non commutative. – N. S. Oct 26 '13 at 02:26
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    @mathematics2x2life you don't need to say that, for the same reason that if I'm asking whether there are infinitely twin primes, I don't need to ask whether there are infinitely many twin odd primes. – YCor Oct 26 '13 at 23:07

3 Answers3

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The group $G=\mathbf{Z}/7\mathbf{Z}\rtimes\mathbf{Z}/3\mathbf{Z}$, where the generator $t$ of $\mathbf{Z}/3\mathbf{Z}$ acts by multiplication by 2: $t$ is not conjugate to $t^{-1}$ by an automorphism. Indeed the full automorphism group is $H=\mathbf{Z}/7\mathbf{Z}\rtimes\mathbf{Z}/7\mathbf{Z}^*$, which contains $G$ as subgroup of index 2. So the elements of $G$ conjugates to their inverse by an automorphism are only the elements of $\mathbf{Z}/7\mathbf{Z}$.

To check the assertion on the automorphism group, observe that if we have an automorphism of $G$, then after composing by an inner automorphism, it maps the 3-Sylow $\mathbf{Z}/3\mathbf{Z}$ into itself, and after composing by an element of $\mathbf{Z}/7\mathbf{Z}^*\subset H$, it acts as the identity on $\mathbf{Z}/7\mathbf{Z}$. So it is either the identity, or acts as identity on $\mathbf{Z}/7\mathbf{Z}=\langle u\rangle$ and maps $t$ to $t^{-1}$, but the latter does not define an automorphism, because it is not compatible with the conditions $tut^{-1}=u^2\neq t^{-1}ut$.

YCor
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    PS: a similar argument works for the semidirect product $\mathbf{Z}/5\mathbf{Z}\rtimes \mathbf{Z}/5\mathbf{Z}^*$: elements of order 4 are not conjugate to their inverses by any automorphism. – YCor Oct 27 '13 at 12:41
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The free group $F(a, b)$ on two-generators works with the word $w=a^3b^4a^5b^6$. This works because every automorphism either takes $w$ to a word which, after cyclic reduction, does not contain $a^{\pm 3}$, or takes $w$ to a conjugate of one of the following words. $$\begin{align*} &a^{3}b^{4}a^{5}b^{6}\\ &a^{3}b^{-4}a^{5}b^{-6}\\ &a^{-3}b^{4}a^{-5}b^{6}\\ &a^{-3}b^{-4}a^{-5}b^{-6}\\ &b^{3}a^{4}b^{5}a^{6}\\ &b^{3}a^{-4}b^{5}a^{-6}\\ &b^{-3}a^{4}b^{-5}a^{6}\\ &b^{-3}a^{-4}b^{-5}a^{-6} \end{align*}$$ Thus, to prove that this works we need to prove that $w^{-1}=b^{-6}a^{-5}b^{-4}a^{-3}$ is conjugate to one of the above words. And it clearly is not.

This answer requires some knowledge of the automorphisms of free groups. The list of words are the orbits of $w$ under the automorphisms which fix the length of the generators. The observation about the other elements in the orbits not containing $a^{\pm3}$ follows from the paper What Does a Basis of $F(a,b)$ Look Like? by Cohen, Metzler and Zimmermann, although some work is required to make it follow.

Taking the same word in the triangle groups $\langle a, b; a^i, b^j, (ab)^k\rangle$ for $i, j, k>13$ works too, and here you only have to check finitely many automorphisms and you can do this by hand. Note that the "finitely many" is because you only need to check the finitely many outer automorphisms and then think about how inner automorphisms can act on this. If you are very careful, you can make it follow from the free group case but this uses small-cancellation theory and other technical stuff.

user1729
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    (For a general approach, look for cyclic "malnormal" subgroups of a group $G$. These are cyclic subgroups $M$ such that the following implication holds.$$g^{-1}MG\cap M\neq1\Rightarrow g\in M$$ Consider the outer automorphism group of $G$, and ensure that $g$ is not mapped to $g^{-1}$ by any representative an outer automorphism. It turns out that hyperbolic groups are very nice for doing this in.) – user1729 Oct 26 '13 at 11:31
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I found an example of this from a comment on a math overflow question. They give the example of GAP's SmallGroup(64,28) having elements not automorphic to their inverses. The restrictions mentioned in the overflow question do not hold here as you were looking for any such group with this property. The definition of the function smallgroup can be found here.

I only tried my own constructions for a short amount of time. As @N.S. stated, any such group should be noncommutative. I tried the automorphism of $Q$, the quaternion group, by sending $i$ to $j$, $j$ to $k$, and $k$ to $i$, while mapping $1$ and $-1$ to themselves. This held except for the case of $-1$ going to $-1$, which obviously maps to its own inverse. Perhaps trying some sort of dihedral group for a finite example. Something perhaps worth trying for an infinite example would be a mapping of $\mathbb{Z}$ or $\mathbb{Q}$ onto itself through a dilation or translation while keeping the identity fixed?

Community
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mathematics2x2life
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