8
$\begingroup$

Is there a finite non-abelian group $G$ of prime exponent such that the full automorphism group of $G$ is of odd order?

$\endgroup$
6
  • 2
    $\begingroup$ I added the Lie algebra tag, because a $p$-group of exponent $p$ and nilpotency length $<p$ can be viewed as a Lie algebra by Malcev correspondence. $\endgroup$
    – YCor
    Jan 18, 2016 at 11:56
  • $\begingroup$ Did you do any computer verification (in view of my answer, I'd be interested in a checking for groups of order $3^4$, $3^5$, $3^6$, $5^6$ expecting there are no examples, and would expect an example in order $p^7$ or $p^8$ for some $p$)? $\endgroup$
    – YCor
    Jan 19, 2016 at 11:51
  • $\begingroup$ @YCor. I have done it for all non-abelian groups of exponent $3$ of order at most $3^6$. There is no such example among such groups. $\endgroup$ Jan 19, 2016 at 12:08
  • $\begingroup$ @YCor. There is also no such group of order $5^6$. Note that all groups of exponent $3$ and of order at most $3^6$ is of class at most $2$. $\endgroup$ Jan 19, 2016 at 12:16
  • 1
    $\begingroup$ Also, 32000 groups of order $5^7$ is tested and there is not such example. Only one group with odd order automorphism exist, but the exponent of this group is $125$. $\endgroup$
    – Shahrooz
    Jan 20, 2016 at 15:58

1 Answer 1

13
$\begingroup$

Yes:

for every prime $p\ge 7$, there's a finite group of exponent $p$ whose automorphism group is a $p$-group.

Initial answer (Jan 18' 2016) Start from any (finite-dimensional) complex nilpotent Lie algebra $\mathfrak{g}$ that is defined over $\mathbf{Q}$ and has a unipotent automorphism group (see e.g. Luks' Lie algebra top of p14 in (Ancochea Campoamor survey) for a 16-dimensional example), see also first edit (Jan 19' 16) below.

Fix a rational structure, so we can defined $\mathfrak{g}_p$ as the corresponding Lie algebra modulo $p$ for $p$ large enough (i.e., not dividing any of the denominators). (Formally speaking, this means writing $\mathbf{g}=\mathfrak{h}\otimes_{\mathbf{Q}}\mathbf{C}$ for some Lie algebra $\mathfrak{h}$ over $\mathbf{Q}$.)

Then for $p$ large enough, the automorphism group of $\mathfrak{g}_p$ is a $p$-group. Indeed otherwise, if for a growing subsequence $(p_i)$ this is not the case, say $\mathfrak{g}_{p_i}$ has an automorphism of prime order $\neq p_i$, passing to the algebraic closure $K_i$ of $\mathbf{F}_{p_i}$ and then to a ultralimit over $i$ we get that the ultraproduct, which is a Lie algebra over the ultraproduct $K$ of the $K_i$ has a nontrivial semisimple automorphism. But this ultraproduct is isomorphic to $\mathfrak{h}\otimes_{\mathbf{Q}}K$. This contradicts that $\mathfrak{g}$ has a unipotent automorphism group.

Note that this provides a group of order $p^{16}$ with some well-controlled nilpotency length (I haven't made the computation for Luks' Lie algebra but it's easy to do), but gives no control of $p$, although for small $p$ a computer computation is doable).


We can expect a computer computation to yield explicit examples.

Anyway no example is possible for $p$-group of nilpotency length 2 (and odd exponent $p$): indeed such a group admits an action by automorphisms of the multiplicative group modulo $p$ (of order $p-1$), because the corresponding Lie algebra is Carnot. Classification along with Malcev correspondence also shows that no group of order $p^{\le 6}$ for prime $p\ge 7$ can work, because the corresponding Lie algebras have nontrivial integral gradings according to the classification by de Graaf in here (J. Algebra 2007) [this also works in order $5^{\le 5}$, and in order $5^6$ except for nilpotency length $=5$].


Edit: (Jan 19' 2016) I have checked that the 7-dimensional (with nilpotency length 6) Lie algebra given page 17 line 8 in (Ancochea Campoamor survey), viewed with coefficients in an arbitrary field, has only unipotent automorphisms. If one takes it in the field on $p$ elements for any prime $p\ge 7$, the Baker-Campbell-Hausdorff (which realizes Malcev's correspondence) provides a group of order $p^7$ whose automorphism group is a $p$-group (thus of odd order).

According to what's above and computer check by Alireza for $p=3,5$, for $p$ odd any nontrivial $p$-group of order $\le p^6$ has an automorphism of order 2. So $p^7$ is the smallest possible order for such exceptions, for $p=3,5$ possibly one has to go a little higher since Alireza checks that every group of order $3^7$ has an automorphism of order 2. (At the moment this gives no example for $p=3,5$ although they certainly exist.)


Edit: (Jan 25' 2018) Thinking a little about the remaining cases $p=3,5$, they seem different and deserve separate additional comments:

first recall by above remark that for any odd $p$, the nilpotency length of any example should be $\ge 3$.

  • for $p=5$, Malcev correspondence is valid in nilpotency length $\le 4$. Therefore if one searches among nilpotent rational Lie algebras of nilpotency length in $\{3,4\}$ and unipotent automorphism groups, one can reasonably hope to find one that reduces modulo 5 and for which the reduction modulo 5 has no extra automorphisms.
  • $p=3$ is quite different, because exponent 3 implies nilpotency length $\le 3$ (in contrast, by Razmyslov's theorem, for all $p>3$ there are finite groups of exponent $p$ and arbitrary large solvability length). So in this case, Lie algebras are of no help. Finding a finite group of exponent 3 whose automorphism group is a 3-group (or of odd order) sounds a quite different question.
$\endgroup$
2
  • $\begingroup$ Is it possible to give an explicit presentation for the group of order $p^7$ in your last EDIT? I have used the same relations as the Lie algebra with the relations of 7th powers of all generators. This gives me a group of nilpotency class 5, expoenent 7 and order $7^6$ whose automorphism group is of order 242121642. Is this because my group is not obtained from BCH? $\endgroup$ Jan 20, 2016 at 6:27
  • $\begingroup$ Yes you really have to use BCH. You have a presentation with generators $x_1,\dots,x_7$, relators $x_i^7=1$ and, for $i<j$, $x_ix_jx_i^{-1}x_j^{-1}=m_{ij}$ for all $i,j$, where $m_{ij}$ is a word in $x_{j+1},\dots,x_7$. With BCH you can compute $m_{ij}$ in the Lie algebra, that is some combination $(\ell(j,j+1),\dots,\ell(j,7))$. This has to be converted into a word of the form $x_{j+1}^{k(j,j+1)}\dots x_7^{k(j,7)}$. Then $k(j,j+1)=\ell(j,j+1)$. To find the next exponents, first compute $x_{j+1}^{-k(j,j+1)}m_{ij}$; then $k(j,j+2)$ is the $j+2$-coefficient in the resulting element, and so on. $\endgroup$
    – YCor
    Jan 20, 2016 at 10:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.