Can the centralizer of Inn(G) in Aut(G), where G is preferably any non-abelian finite one, equal to Inn(G) itself? Clearly, such centralizer contains all $f$ in Aut(G) where $f(g)g^{-1}$ are in Z(G).

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    $\begingroup$ Such a group would either be trivial, cyclic of order 2, or nilpotent of class exactly two. I'll work on it a bit later today. $\endgroup$ Commented Mar 31, 2011 at 13:25
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    $\begingroup$ For p-groups, it is a necessary and sufficient condition that $Z(G)=G'$ and is cyclic. Thus, extra-special groups satisfy your question. For general groups, $G/Z(G)$ would be abelian, hence $G$ is nilpotent of class $\le2$, and so we need the above condition for p-groups to hold in each Sylow. $\endgroup$
    – user641
    Commented Mar 31, 2011 at 14:00
  • $\begingroup$ @Steve: I'm sure this is something I should be able to do, but I confess I'm stuck (I can get sufficiency, but only via what I suspect is a ridiculously complicated argument). Could you sketch (or post) the equivalence for $p$-groups? $\endgroup$ Commented Apr 1, 2011 at 15:03
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    $\begingroup$ No, the argument is fairly involved :-). I didn't want to post the link because you said you wanted to work on it. The relevant paper is www.maths.otago.ac.nz/home/downloads/john_curran/CommAlg.pdf $\endgroup$
    – user641
    Commented Apr 1, 2011 at 16:50
  • $\begingroup$ @Steve: Ah, well, that explains it. (-: Interestingly, I think my argument for the sufficiency may be somewhat different from theirs; it came down to the Chinese Remainder Theorem... $\endgroup$ Commented Apr 1, 2011 at 19:19

3 Answers 3


To summarize (thanks to Steve D. for the article pointer):

If $G$ is a group, an automorphism $\sigma\in\mathrm{Aut}(G)$ is said to be a "central automorphism" if $\sigma$ commutes with every automorphism in $\mathrm{Inn}(G)$, or equivalently if $\sigma(g)g^{-1}\in Z(G)$ for all $g\in G$. The central automorphisms form a normal subgroup of $\mathrm{Aut}(G)$, denoted by $\mathrm{Aut}_c(G)$.

You are asking for examples with $\mathrm{Inn}(G)=\mathrm{Aut}_c(G)$. In particular, this requires $\mathrm{Inn}(G)\cong G/Z(G)$ to be abelian, hence $G$ is nilpotent of class at most $2$.

Case 1. $G$ abelian.

If $G$ is abelian, then $\mathrm{Inn}(G)=\{1\}$ and $\mathrm{Aut}_c(G)=\mathrm{Aut}(G)$. For the equality between the group of central automorphisms and inner automorphisms to hold, we would need the group of automorphisms to be trivial. If $G$ is finite, the only possibilities are $|G|=1$ and $|G|=2$. As was recently discussed in this very forum, whether there are any infinite abelian groups with this property depends on your Set Theory. If we assume the Axiom of Choice (in fact, weaker axioms are sufficient here), then there are no infinite abelian examples. But there are models of $ZF$ without choice where there are such infinite groups, as Asaf Karagila showed in that question.

Case 2. Finite nonabelian groups.

Moving on to groups of nilpotency class exactly $2$. A finite nilpotent group is the products of its Sylow $p$-subgroups, which are characteristic and have no nontrivial homomorphisms between them; so the group of automorphisms of a finite nilpotent group naturally breaks up into a product of automorphisms groups of $p$-groups. Thus, the problem reduces to characterizing the finite $p$-groups of nilpotency class two in which every central automorphism is inner. A theorem of M.J. Curran and D.J. McCaughan (Central automorphisms that are almost inner, Comm. Algebra 29 (no. 5), 2081-2087, 2001) completely solves the problem then:

Theorem. If $G$ if a finite nonabelian p-group, then $\mathrm{Aut}_c(G)=\mathrm{Inn}(G)$ if and only if $G'=Z(G)$ and $Z(G)$ is cyclic.

There are many examples of such groups: all nonabelian groups of order $p^3$, all extra-special $p$-groups, etc.

So a finite group satisfies the condition if and only:

  1. $G$ is trivial; or
  2. $G$ is a direct product of its Sylow $p$-subgroups, $Z(G)=G'$ and $Z(G)$ is cyclic; or
  3. $G\cong C_2\times H$, where $C_2$ is the cyclic group of order two, and $H$ is a group of odd order of types 1 or 2.

Case 3. Infinite nonabelian groups. I can't answer it completely, except to say that examples certainly exist.

This one seems more complicated; I'll take a look at the Curran-MacCaughan paper later on, but there are certainly examples with $G'=Z(G)$ cyclic. For example, consider the relatively free nilpotent group of class $2$. This is $$F = \Bigl\langle x,y\Bigm| [x,y,x]=[x,y,y]=1\Bigr\rangle,$$ and is isomorphic to the group of $3\times 3$ upper triangular matrices with integer coefficients and $1$s in the diagonal, by the map that sends $x$ to $I+E_{12}$ and $y$ to $I+E_{21}$. Every element has normal form $x^ay^b[y,x]^c$ with $a,b,c\in\mathbb{Z}$, and multiplication given by $$(x^ay^b[y,c]^c)(x^{\alpha}y^{\beta}|[y,x]^{\delta}) = x^{a+\alpha}y^{b+\beta}[y,x]^{c+\delta+\alpha b}.$$

Any two elements of $F$ determine an endomorphism by mapping $x$ to the first element and $y$ to the second; thus, the automorphisms of $F$ correspond to maps of the form $$\begin{align*} x &\longmapsto x^a y^b [y,x]^{k}\\ y &\longmapsto x^c y^d [y,x]^{\ell} \end{align*}$$ with $ad-bc = \pm 1$. An inner automorphism given by conjugation by $x^ry^s$ maps $x$ to $x[x,x^ry^s] = x[y,x]^{-s}$ and $y$ to $y[y,x^ry^s]=y[y,x]^r$. Thus, inner automorphisms correspond to $b=c=0$ and $a=d=1$.

If an automorphism of $F$ is central, then $(x^ay^b[y,x]^k)x^{-1} = x^{a-1}y^b[y,x]^{k-b}$ and $(x^cy^d[y,x]^{\ell})y^{-1}=x^cy^{d-1}[y,x]^{\ell}$ are both central; since the center is the commutator subgroup, this requires $a-1=b=c=d-1=0$, showing that the automorphism is inner. Hence $\mathrm{Inn}(F)=\mathrm{Aut}_c(F)$.

I think that the conditions $G'=Z(G)$ and $Z(G)$ cyclic will still be at least sufficient in the infinite case.

  • $\begingroup$ The finite group conclusion is a tiny bit more complicated: If G has finite odd order, then 2×G works iff G works. Irritatingly, Z(2×G) = 2×Z(G) = 2×G′ ≠ (2×G)' = 1×G′. $\endgroup$ Commented Apr 1, 2011 at 21:28
  • $\begingroup$ @Jack: Ah, good catch. Thanks. The theorem should also say "nonabelian", since $C_2$ does not satisfy the condition. $\endgroup$ Commented Apr 1, 2011 at 21:40
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    $\begingroup$ @Babak, @Jack: I'm pretty sure I can prove that if $G/Z(G)$ is a direct sum of cyclic groups, and $G'=Z(G)$ is cyclic, then the condition holds for $G$. Don't know what happens if $G/Z(G)$ is more complicated, and I haven't thought about the converse yet. Babak: you never said if your interest was mainly finite groups (in which case we are done), or any group; if infinite groups hold particular interest, I'll make a point of double-checking and posting. If not... then I'll do it if I get a chance or a breakthrough. (-: $\endgroup$ Commented Apr 2, 2011 at 18:20

I was fiddling with this, but never had time to finish. Maybe it is interesting to someone.

We define two bijections that help to describe the centralizer of the inner automorphism group in the full automorphism group.

  • exp : Hom(G,Z(G)) → CAut(G)(Inn(G)) : ζ ↦ ( g ↦ gζ(g) )
  • log : CAut(G)(Inn(G)) → Hom(G,Z(G)) : φ ↦ ( g ↦ g−1 φ(g) )

We need to show these are well-defined (that is they have the specified ranges).

Proof: If ζ is a homomorphism from G to Z(G), then exp(ζ) : G → G : g ↦ gζ(g) is indeed an automorphism, with inverse exp(−ζ) defined by −ζ, the homomorphism from G to Z(G) that takes g to ζ(g)−1. The inner automorphism defined by h takes g to h−1gh = gh. If one applies exp(ζ) and then conjugation by h, one gets ghζ(g), since ζ(g) is central and so ζ(g)h = ζ(g). On the other hand, if one conjugates by h and then applies exp(ζ), then one gets ghζ(gh), but since ζ is a homomorphism into an abelian group, ζ is constant on conjugacy classes, and ghζ(gh) = ghζ(g). Hence every exp(ζ) is in CAut(G)(Inn(G)). Conversely, if φ is an automorphism of G commuting with all inner automorphisms, then φ(x)φ(h) = φ(xh) = φ(x)h, so φ(h) and h define the same automorphism of G, and so φ(h) and h differ by some element of the center of G. Define log(φ):G→Z(G) implicitly from φ(h) = h log(φ)(h). It is not hard to check that log(φ) is a homomorphism from G to Z(G).

Notice that log(exp(ζ)) = ζ and exp(log(φ))=φ, so these are mutually inverse bijections.

Suppose then that Inn(G) = CAut(G)(Inn(G)). This means that exp(Hom(G,Z(G))) = Inn(G). This means that log : Inn(G) → Hom(G,Z(G)) is bijective! However, log(∧h) = ( g ↦ [g,h] ) has a very simple form for the inner automorphism ∧h defined by conjugation by h. In particular, [G,G] ≤ Z(G) for log(∧h) to even make sense, and so G has nilpotency class at most 2. Notice that the kernel of log(∧h) contains all of Z(G), for any h. Indeed, the kernel of log(∧h) is CG(h), and so all centralizers are normal, which is weird.

Another sick thing: log(∧hk)(g) = [g,hk] = [g,k][g,h] = [g,h][g,k] = (log(∧h) + log(∧k))(g), so in fact log is an isomorphism of abelian groups G/Z(G) ≅ Hom(G,Z(G)).

Unfortunately, I cannot find a good element of Hom(G,Z(G)). The transfer is useless, and the central series induction arguments give inequalities pointing the wrong way.

For finite groups, presumably this is taken care of by counting as in theorem 1 of:

Sanders, P. R. "The central automorphisms of a finite group." J. London Math. Soc. 44 (1969) 225–228. MR248208 DOI:10.1112/jlms/s1-44.1.225

but it didn't seem to address infinite (nilpotent) groups. The way it is done in this (very nice) paper mentioned by Steve certainly doesn't work out for infinite groups. In particular, Lemma I would be wonderful to have more concretely with homomorphisms rather than dualizing finite abelian groups a zillion times.

Curran, M. J.; McCaughan, D. J. "Central automorphisms that are almost inner." Comm. Algebra 29 (2001), no. 5, 2081–2087. MR1837963 DOI:10.1081/AGB-100002170


Yes, this can for instance happen when $Inn(G) = Aut(G)$ which is the case for the symmetric groups $S_n$ when $n\neq 6$

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    $\begingroup$ This doesn't work unless $\mathrm{Aut}(G)$ also happens to be abelian; for $G=S_n$, $n\neq 6$, you have that the centralizer of $\mathrm{Inn}(G)$ in $\mathrm{Aut}(G)$ is $Z(S_n)$, which is trivial for $n\gt 2$. $\endgroup$ Commented Mar 31, 2011 at 13:16
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    $\begingroup$ In other words: I think you misread the question as asking whether you can have $C_{\mathrm{Aut}(G)}(\mathrm{Inn}(G))$ can equal $C_{\mathrm{Inn}(G)}(\mathrm{Inn}(G)) = Z(\mathrm{Inn}(G))$, instead of the question which is whether you can have $C_{\mathrm{Aut}(G)}(\mathrm{Inn}(G)) = \mathrm{Inn}(G)$. $\endgroup$ Commented Mar 31, 2011 at 15:44
  • $\begingroup$ @Arturo: Dear Arturo, I am thinknig of Q_8 of order 8, wherein Inn(G)=V. $\endgroup$
    – Mikasa
    Commented Apr 1, 2011 at 8:30
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    $\begingroup$ @Babak: $Q_8$ satisfies the conditions given by Steve: it has center equal to the commutator subgroup and cyclic. $\endgroup$ Commented Apr 1, 2011 at 13:17
  • $\begingroup$ I don't know why your answer didnn't get any attentions. Sorry Tobias! Thanks for the time you took for the answer. $\endgroup$
    – Mikasa
    Commented Apr 15, 2022 at 11:54

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