# Group with finite outer automorphism group and large center

Does there exist a finitely generated group $$G$$ with outer automorphism group $$\mathrm{Out}(G)$$ finite, whose center contains infinitely many elements of order $$p$$ for some prime $$p$$?

A motivation is that the existence of such a group would answer the question in this 2011 MO post. Indeed such a group would be an example of a finitely generated group with finite Out but with finite index in a group with infinite Out (namely $$G\times\mathbf{Z}/p\mathbf{Z}$$).

Edit: here is an example when we drop the finite generation assumption: the universal central extension $$G$$ of $$\mathrm{SL}_n(\mathbf{Q})$$ when $$n\ge 5$$.

Indeed, the central kernel is isomorphic (by standard stability results for $$K_2$$ of fields) to $$K_2(\mathbf{Q})$$, which is isomorphic to $$C_2\oplus\bigoplus_{p>2}C_{p-1}$$, where $$p$$ ranges over odd primes (see Milnor K-theory book, Chapter 11). So $$K_2(\mathbf{Q})$$ contains infinitely many elements of order 2 (or even of any other prime order, by Dirichlet's theorem of primes in an arithmetic progression). On the other hand, by Schreier-Van der Waerden, $$\mathrm{Aut}(\mathrm{(P)SL}_n(K))$$, for any field $$K$$, is generated by inner automorphisms, the inverse-by-transposition involution, and automorphisms induced by $$\mathrm{Aut}_{\mathrm{field}}(K)$$ acting entry-wise. Since here the field has a trivial field automorphism group, we obtain that $$\mathrm{Out((P)SL}_n(\mathbf{Q}))$$ has two elements only. It is immediate that the canonical map from $$\mathrm{Aut}(G)$$ to $$\mathrm{Aut}(G/Z(G))$$ is injective, and it follows that $$\mathrm{Out}(G)$$ is finite.

Actually I expect that central extensions of $$\mathrm{SL}_n$$ of some well-chosen finitely generated commutative ring should be a source of finitely generated exemples, but it sounds harder.

• Yves, "transposition" should read contrgradient ($X\mapsto (X^t)^{-1}$). Also, $\mbox{Aut}(Z(G))$ should read $\mbox{Aut}(G/Z(G))$. – Anton Klyachko Jul 2 '15 at 20:45
• @AntonKlyachko: Oddly, it's contragredient and not contragradient. (I'm guilty of misspelling this in the past.) I don't know where this term came from; in particular I don't think that gredient has any meaning on its own. – Dan Ramras Feb 17 '18 at 20:55
• @DanRamras here international-dictionnaire.com/definitions/… they suggest it comes from "ingredient" (but do not provide source) – YCor Feb 17 '18 at 21:08

Theorem B of this paper implies that we can take any two nontrivial involution-free groups $A$ and $B$ and construct a complete simple groups $D$ with a (diagrammatically) aspherical presentation $D=A*B/\langle\!\langle w_1, w_2,\dots\rangle\!\rangle$ (though such use of this theorem is killing a mosquito with a cannon). Here, complete means naturally isomorphic to the automorphism group (i.e. centreless and without outer automorphisms). Now, suppose that $A$ and $B$ coincides with their commutator subgroups.

Acphericity implies that the centre of the free central extension $\widetilde D=A*B/[\langle\!\langle w_1, w_2,\dots\rangle\!\rangle,A*B]$ of $D$ is the free abelian group with the basis $\widetilde{w_1},\widetilde{w_2},\dots$ (see, e.g., Olshanskii's book, Section 34.4). Now, we can do whatever we want. For instance, we can take the quotient $G=\widetilde D/\langle\widetilde {w_1}^p,\widetilde{w_2}^p,\dots\rangle$ and obtain a desired group $G$. The natural map $\mbox{Aut}\,\widetilde D\to\mbox{Aut}\,G$ exists because the subgroup $\langle\widetilde {w_1}^p,\widetilde{w_2}^p,\dots\rangle$ is characteristic in $\widetilde D$ (as this subgroup consists of $p$th powers of all central elements), and this map is injective because $\widetilde D$ coincides with its commutator subgroup. So, there are no outer automorphisms of $G$.

Jul 6, 2015. More details added, as suggested by Mamuka.

(1) $G$ (as well as $D$ and $\widetilde D$) is finitely generated if $A$ and $B$ are finitely generated.

(2) Suppose that a group is a quotient of another group: $L=M/N$. Then

• there is a natural map $f\colon \mbox{Aut}\,M\to \mbox{Aut}\,L$ if $N$ is characteristic;

• $f(\mbox{Inn}\,M)=\mbox{Inn}\,L$ (i.e. $f$ sends inner automorphisms to inner automorphisms, and each inner automorphism of $L$ has at least one inner preimage);

• $f$ is injective if $N$ is central and $M$ coincides with its commutator subgroup.

Now, take $M=G$ and $L=D$ (and $N=\langle\widetilde {w_1},\widetilde{w_2},\dots\rangle$). Then $\mbox{Aut}\,L=\mbox{Inn}\,L$ and, hence, $\mbox{Aut}\,M=\mbox{Inn}\,M$ by (2), i.e. all automorphisms of $M=G$ are inner.

• Sorry, I cannot figure out how do you deduce the last sentence, and why can one arrange for a finitely generated $G$. Could you please elaborate a little bit? – მამუკა ჯიბლაძე Jul 6 '15 at 6:08
• I added some details. – Anton Klyachko Jul 6 '15 at 10:21
• Thank you for expanding it. And sorry for asking about (1) - by some reason I thought $w_i$ were some "new" generators... – მამუკა ჯიბლაძე Jul 9 '15 at 4:10