A group $G$ can have an element $g$ for which every automorphism of the group fixes $g$. Obviously, the identity is one such element, and one can easily find order-2 examples: the unique order-2 element in $C_{2n}$, or $-1$ in the quaternion group.

My question is whether one can find a group $G$ with an element $g$ of order at least 3 which is fixed by every automorphism of $G$.

One might suspect that there is always an automorphism taking $g$ to $g^{-1}$, presenting an obvious obstacle to the high-order case, but this is false. For example, take $G$ to be the unique nonabelian group of order $21$, realized as a semidirect product of $C_7$ and $C_3$. Then any of the $14$ elements of order $3$ cannot be sent to their inverses. However, none of these elements are fixed by every automorphism; the $14$ elements fall into two conjugacy classes of size $7$.

Restricting our attention to those automorphisms given by conjugation, we see that $g\in Z(G)$. However, I haven't found a way to strengthen this restriction into a proof of impossibility.

Edit: This post originally contained the sentence

Intuitively, one can think of such an element as "unique" in the sense that it has group-theoretic properties not shared by any other element.

but I have moved it to the bottom to avoid confusion. By this, I just mean that automorphisms exchange elements that in some sense "serve the same role" as each other in the group; an element which is fixed by every automorphism can be thought of as having no such counterparts. I do not mean to search for a group in which only one element has the property of being fixed by every automorphism.

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    $\begingroup$ If $g$ is fixed by every automorphism, then so is $g^{-1}$. If you want $g$ to be the only nonidentity element that is fixed by every automorphism, then this forces $g=g^{-1}$, so $g$ must have order $2$. $\endgroup$ Oct 21 '20 at 15:29
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    $\begingroup$ @ArturoMagidin The OP did not actually ask for there to be a unique such element, they were just using "unique" as a synonym for the property of being fixed by every automorphism. In short, the question is simply "can an element of order at least 3 be fixed by every automorphism". I don't think your argument actually answers this. $\endgroup$
    – verret
    Oct 21 '20 at 20:14
  • $\begingroup$ @verret: I agree my argument does not answer that question, but I disagree that what you describe is the question that was asked. The first paragraph says they want a "group-theoretic propert[y] not shared by any other element". $\endgroup$ Oct 21 '20 at 20:18
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    $\begingroup$ Yes, I think that is how they are trying to describe informally the property of being fixed by every automorphism (and it's not a bad way to describe it). The fact that they are putting unique in quotation marks, saying "in the sense of" and then using colons also suggests that what follows the colons is the definition of unique. Anyway, rather than try to divine the words, we can just wait for OP to clarify... $\endgroup$
    – verret
    Oct 21 '20 at 20:27
  • $\begingroup$ Apologies for the ambiguity. I meant for the colon to indicate a restatement of the previous sentence, not a description of the specific property desired. That is, I meant that “fixed by every automorphism” essentially means “has unique group-theoretic properties”; at least in the finite case, I believe it literally is equivalent to the existence of a logical formula $\phi$ about the group operations with a unique element $g$ such that $\phi(g)$ holds. I've edited the post in a way that I hope makes the meaning clearer. $\endgroup$ Oct 21 '20 at 20:30

I think the question is asking for a finite group having an element of order at least 3 that is fixed by every automorphism. The smallest such group has order 63 and is the unique non-trivial semidirect product $C_7\rtimes C_9$.

Here's a sketch of why it works. Let $G=C_7\rtimes C_9$. First, it's not too hard to see that the center of $G$ has order $3$. We claim that elements of the center are fixed by every automorphism. Now, since $Z(G)$ is characteristic in $G$, $\mathrm{Aut}(G)$ has an induced action on $G/Z(G)$, which is the non-abelian group of order $21$. As the OP pointed out, in this group, an element of order $3$ and its inverse are in different orbits of the automorphism group. It then follows that, if we pull back the action to the full $G$, the inverse pair of elements of order $3$ in $Z(G)$ are also in different orbits.

  • $\begingroup$ Ah, thanks! After playing around with the group structure a bit, I think I have a sense for how these elements "look" and why they ought to be unique in the group. As a very slight generalization, I think $C_7 \rtimes C_{3n}$ ought to produce such elements of order $n$, though this might only work if $n$ is prime? $\endgroup$ Oct 21 '20 at 21:41
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    $\begingroup$ No, your generalisation does not work. Suppose that $n$ is coprime to $6$, for example. Then we get $C_7\rtimes (C_{3n})\cong C_7\rtimes (C_3\times C_n)\cong(C_7\rtimes C_3)\times C_n$, but then we get additional automorphisms which act on the $C_n$ factor. $\endgroup$
    – verret
    Oct 22 '20 at 2:58
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    $\begingroup$ If you want to generalise to different primes, the easiest is probably to take $C_n\rtimes C_{p^2}$, where $n-1$ is divisible by $p$ but not $p^2$. $\endgroup$
    – verret
    Oct 22 '20 at 4:02
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    $\begingroup$ Here's another source of examples. It is known that "almost all" $p$-groups have automorphism group a $p$-group. (Just google it.) Now, if a $p$-group acts on another $p$-group, it must centralise a non-trivial subgroup. So in fact, "almost all" $p$-groups have the property you want. $\endgroup$
    – verret
    Oct 22 '20 at 4:06

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