11

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.

RavenclawPrefect
  • 17,299
  • 10
    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 $2$. – Arturo Magidin Oct 21 '20 at 15:29
  • 1
    @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. – verret Oct 21 '20 at 20:14
  • @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". – Arturo Magidin Oct 21 '20 at 20:18
  • 1
    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... – verret Oct 21 '20 at 20:27
  • 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. – RavenclawPrefect Oct 21 '20 at 20:30
  • @RavenclawPrefect Can you edit your question so that, at some point, you write exactly what is your question formally without the "intuitive" part? For example, if your question is "does there exists a finite group with an element of order at least 3 fixed by every group automorphism" then say that explicitly. If you mean something else, then say that. – verret Oct 21 '20 at 20:32
  • When you say “has a property not shared by any other group element” you create exactly the same impression/ambiguity as you had before. You want the element to be fixed by every automorphism and no other condition, fine, say that; but don’t tell us this property is “not shared by any other element” because then you are implying that you want it to be the unique element of the group with that property. It’s a lousy turn of phrase to use there. Just say “groups with nonidentity elements that are fixed by every automorphism”, instead of introducing fuzzy language. – Arturo Magidin Oct 21 '20 at 20:37
  • I agree with Arturo. Even though I understood what you meant, you are not explaining it very well. While it's encouraged to have a motivation and some background in these questions, you should still write things in a way that the actual question can be understood with as little background and interpretation as possible. I gave a suggested phrasing twice already, I'm not sure why you are not using it. – verret Oct 21 '20 at 20:40
  • And “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” is again saying you want your $g$ to be the unique element of the group that has the given property. So you are not saying what you mean, either in the comments, or in the message. Just state exactly what you mean, explicitly. Drop the scare quotes and intuition when you write down the specific question you are asking. Add them later to explain context. – Arturo Magidin Oct 21 '20 at 20:44
  • Edited further. I confess I'm not sure why the previous iteration was not clear enough (it opened with a formal description of the desired property and asked if any such elements existed with order $\ge3$), but the sentence starting with "My question is" should be fine now, I hope. – RavenclawPrefect Oct 21 '20 at 21:02
  • Arturo Magidin: I'm not sure I follow your comment about logical formulas. In the quaternion group $Q$, the formula $\phi(g): (g\ne e) \wedge (g^2=e)$ is satisfied by exactly one $g\in Q$. In general, if one can state such a formula in terms of the language of set theory and the group operation (from which one can in turn define the identity), and exactly one element satisfies that formula, it must be fixed under every automorphism. I believe the converse is true in all finite groups. I don't mean that $\phi$ is a formula expressing the property of being fixed under all automorphisms. – RavenclawPrefect Oct 21 '20 at 21:07
  • @RavenclawPrefect: There are two distinct questions you could ask, and your wording continues to confuse them. You could ask: "Is there a group $G$ with an element $g$ that is fixed by every automorphism of $G$, and $g$ has order strictly greater than $2$?" That's the question verret interpreted, answered, and you endorsed. You could also be asking "Is there a group $G$, and an element $g\in G$ such that (i) $g$ is fixed by every automorphism of $G$; (ii) the order of $g$ is greater than $2$; and (iii) $g$ is the only element of $G$ satisfying (i) and (ii)?" (cont) – Arturo Magidin Oct 21 '20 at 21:11
  • @RavenclawPrefect: If you want to ask the first question, then please ask it explicitly and clearly. If you also mean to ask whether "If $g$ is fixed by every automorphism of $G$, then there exists a sentence/formula $\phi$ in the language of group theory such that $\phi(x)$ holds for $x\in G$ if and only if $x=g$", then you can ask that and phrase it clearly, instead of fuzzily with scare quotes and confusing language. We don't read minds, we read words. – Arturo Magidin Oct 21 '20 at 21:15
  • The first question is the one I mean. I see why the original wording was so unclear now, thanks for your patience. All I meant by my talk of uniqueness was that the sentence "$g$ is fixed by every automorphism of $G$" can be thought of as meaning "$g$ serves a unique role in $G$, and cannot be substituted with another element in a way that preserves $G$'s structure". I don't mean that $g$ should be the only element to possess this property, though I see how the original post can be read that way. – RavenclawPrefect Oct 21 '20 at 21:26

1 Answers1

9

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.

verret
  • 6,691
  • 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? – RavenclawPrefect Oct 21 '20 at 21:41
  • 1
    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. – verret Oct 22 '20 at 02:58
  • 1
    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 $p^2$. – verret Oct 22 '20 at 04:02
  • 3
    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. – verret Oct 22 '20 at 04:06