# Aut(G) = $C_3$, G = ?

Is there a group G such that Aut(G) = $C_3$? What if we replace 3 with a prime number p?

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Dear Greg, your question is a common homework problem in a 1st course on group theory, and not appropriate for MO. – Ryan Budney Nov 5 '10 at 16:04
I agree with Ryan and voted to close. Most probably it is a homework question. Anybody who knows the definition of Aut and Inn should be able to answer it. – Mark Sapir Nov 5 '10 at 16:08
I personally felt this was closed too abruptly. Seeing that the OP possesses a PhD in mathematics, it is probably not a homework question. I would have voted to re-open at meta, except for some reason I'm unable to sign in. As far as I can tell, while the problem is not hard, it's not a complete triviality either. I have in the meantime mailed the OP my own solution. – Todd Trimble Nov 6 '10 at 4:26
I've just cast the last reopen vote. I would urge the questioner to add some motivation: what led you to ask this question, why are you interested in replacing 3 by an arbitrary prime? Is that extra just a casual add-on or is it something you're really interested in in your research? – Loop Space Nov 6 '10 at 12:40
The question was more curiosity since I've had the chance to play with gap recently. It just occurred to me that although many of the small groups also appear as automorphism groups, C3 never seemed to. As others have clearly indicated, the idea that p be prime is not necessary, only that p be odd. Although my dissertation was in Lie algebras I've since retrained and switched to modeling biological systems, something a number of people here do. Having not had the chance to teach abstract algebra even at the undergraduate level, I've not seen group theory in 15 years. – Greg Gibson Nov 7 '10 at 19:14

There is no group $G$ (finite or infinite) for which $Aut(G) \cong C_p$ (the cyclic group of order $p$), if $p > 1$ is an odd number.

Suppose otherwise. The inner automorphism group $Inn(G)$ is a subgroup, also cyclic, and a well-known exercise in group theory is that if $Inn(G) \cong G/Z(G)$ is cyclic, then $G$ is abelian.

An abelian group $G$ has an involution given by inversion. Unless inversion is trivial, we get an element of order 2 in $Aut(G) = C_p$, contradiction.

If inversion is trivial, then the abelian group $G$ becomes a vector space over $\mathbb{F}_2$. In that case it is an easy to prove that either $Aut(G)$ is trivial or has an element of order 2; either way we get a contradiction.

Edit: After listening to some comments about this at meta, I amended my answer so that it gives less away or leaves a bit more to the imagination, or so I hope.

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I can't see what's wrong with reading other people's proofs. I do that all the time. – Franz Lemmermeyer Nov 6 '10 at 15:57
This proof isn't that elementary, because the axiom of choice is invoked! Is it consistent with ZF that there exists a vector space over F_2 admitting no nontrivial involutions? Worse yet, is it consistent with ZF that there exists a vector space V over F_2 with Aut(V) = C_3?? – Jared Weinstein Nov 6 '10 at 18:47
Perhaps the MO rule should be, undergraduate homework no, graduate homework yes? Outside our own specialties, we are all at the level of graduate students. – Gerry Myerson Nov 6 '10 at 21:50
@Gerry: that's pretty close to how I feel. There are some problems in Hartshorne that I never figured out. @Mark: sigh. Unless there's another way, one has to hit upon the idea of using inversion, there is the reformulation in terms of vector spaces, there is AC. Perhaps it all seems very trivial to you. It may not be trivial for everyone. – Todd Trimble Nov 6 '10 at 22:12
I agree with Gerry Myerson that outside of our own specialties, our knowledge level is the same as that of graduate students: slightly familiar with the terminology, but unsure of its applications. I actually have considered this to be the raison d'etre for Math-Overflow $-$ what may appear to be a tall hard-to-climb-over step or fence from the point of view of one mathematical or scientific specialty may be similar or equivalent in another specialty to a very easy problem with a well-known canonical solution. Asking here allows us to let other experts help us when we're stalled. :) – sleepless in beantown Nov 7 '10 at 5:45

The original question has already been answered, but Jared Weinstein asked in a comment about what happens if we don't assume the axiom of choice. I've convinced myself that it's consistent with ZF to have a vector space over $\mathbb{F}_2$ with automorphism group $C_3$. In case any set theorists (other than me) are looking at this question, here's the model I have in mind. (It's a permutation model, using atoms, but the Jech-Sochor theorem suffices to convert it into a ZF-model.) Start with the full universe $V$ built from a countable set $A$ of atoms (and satisfying AC). In $V$, give $A$ the structure of a $\mathbb{F}_4$-vector space, obviously of dimension $\aleph_0$. (The relevance of the 4-element field $\mathbb{F}_4$ is that the two elements that are not in the 2-element subfield are cube roots of 1, so multiplication by either of them gives an automorphism of order 3.) Let $G$ be the group of automorphisms of this vector space, and let $M$ be the Fraenkel-Mostowski-Specker permutation submodel of $V$ determined by the group $G$ with finite subsets of $A$ as supports. In $M$, $A$ is an $\mathbb{F}_4$-vector-space. Multiplication by the elements of $\mathbb{F}_4\setminus\mathbb{F}_2$ gives a $C_3$-action on the underlying abelian group. Fairly easy calculations (admittedly not yet written down) convince me that this abelian group has no automorphisms in $M$ beyond this copy of $C_3$.

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Thanks! This is a good example where I see an advantage to the strategy you suggest (a permutation model, and then the Jech-Sochor embedding), than directly using a symmetric model or something like that. – Andrés E. Caicedo Nov 8 '10 at 0:49
I'll second those thanks. So the trivial homework problem can't be settled within ZF, and involves something more than just playing with definitions of Aut and Inn. Learn something new every day. – Todd Trimble Nov 8 '10 at 14:13
@Todd: Indeed, there's something more involved, something that depends on the axiom of choice, but in this case the relevant "something" is only that vector spaces have bases, and that permutations of a basis induce automorphisms of the vector space. – Andreas Blass Nov 8 '10 at 16:30
@Andreas: I know that; indeed I wrote that in my answer before I revised it. I'm really referring back to one of Mark's comments which I take issue with. – Todd Trimble Nov 8 '10 at 17:49