In an informal talk I heard a statement:

"Any cyclic subgroup in a linear group is at most exponentially distorted"

with a vague reference to a work of Lubotzky with coauthors.

The works of Lubotzky that may be relevant to this quiestion and that I was able to find reference for:

Lubotzky, Alexander; Mozes, Shahar; Raghunathan, M. S. Cyclic subgroups of exponential growth and metrics on discrete groups. C. R. Acad. Sci. Paris Sér. I Math. 317 (1993), no. 8, 735–740. http://www.ams.org/mathscinet-getitem?mr=1244421

Lubotzky, Alexander; Mozes, Shahar; Raghunathan, M. S. The word and Riemannian metrics on lattices of semisimple groups. Inst. Hautes Études Sci. Publ. Math. No. 91 (2000), 5–53 (2001). http://www.ams.org/mathscinet-getitem?mr=1828742

seem to deal with lattices in Lie groups, not with arbitrary linear groups. So the question is:

Is the statement above true in such generality? If not, then for which classes of linear groups cyclic subgroups are at most exponentially distorted?

EDIT: As far as I understood, the distortion of a cyclic subgroup $H$ inside $G\subset GL(n,F)$ is meant here to be with respect to the respective word metrics on $H$ and $G$, not the underlying metric on $GL(n,F)$.

EDIT2: Sorry for the possible confusion. I guess, the distortion here is meant in the following sense:

Let $H$ be a subgroup of $G$ and fix some generating sets for $H$ and $G$. Consider a ball $B(n,1)$ of radius $n$ in the Cayley graph of $G$ centered at $1$, i.e. the set of all elements of $G$ having word length $\le n$ with respect to the generating set of $G$. Define a function $$ f(n):=\max\{|h|_H \mid h\in H\cap B(n,1)\}, $$ where $|h|_H$ is the length of the element $h$ with respect to the generating set of $H$.

Then $f(n)$ measures the distortion of subgroup $H$ inside $G$, and its order of growth is independent on the choice of the generating sets.

For example, in the group $G=\langle a,b\mid aba^{-1}=b^2\rangle$, subgroup $H=\langle b\rangle$ has exponential distortion function, as the element $h=a^mba^{-m}=b^{2^m}$ lies in the ball $B(2m+1,1)$ in $G$, but the word length of $g$ in $H$ is $2^m$.

But if we consider a group $G=\langle a,b,c\mid aba^{-1}=b^2, bcb^{-1}=c^2\rangle$ then a similar reasoning shows that $H=\langle c \rangle$ has double exponential distortion in $G$: $f(n)=2^{2^n}$.

The statement I am inquiring about stipulates that cyclic subgroups $H$ of linear groups $G\subset GL(n,F)$ cannot have distortion functions that grow faster than exponential functions.

  • $\begingroup$ Do you mean cyclically distorted with respect to an invariant metric on $GL_n(\mathbb{R})$? (If that is the case you would at least need to assume that it is discrete.) I think the main point of Lubotzky-Mozes-Raghunathan is to prove that higher-rank lattices are not distorted; a proof of "at most exponentially distorted" would likely be much simpler. For example, if $\gamma$ generates a discrete subgroup in a real Lie group then you can put it inside a $SL_2({\mathbb R})$ and I think this yields the answer in this case. $\endgroup$ Sep 27, 2013 at 7:39
  • $\begingroup$ @JeanRaimbault: As far as I understood, the distortion here is meant to be with respect to the word metrics on a cyclic subgroup $C$ and the linear group $H$ containing $C$. $\endgroup$
    – mathreader
    Sep 27, 2013 at 13:40
  • $\begingroup$ It would be very helpful if you could add a precise definition of distortion. $\endgroup$
    – Nick Gill
    Sep 27, 2013 at 13:59
  • $\begingroup$ @NickGill: I added more info, sorry for the confusion. $\endgroup$
    – mathreader
    Sep 27, 2013 at 15:15

1 Answer 1


Yes it's true. First recall the definitions:

1) if $G$ is a group (discrete), a finitely generated subgroup $H$ is at most exponentially distorted (resp. undistorted) if for some/every finite generating subset $S$ of $H$ and any finite subset $T$ of $G$ such that $H \subseteq \langle T \rangle$, there exists a function with exponential growth (resp. with linear growth) $f$ such that $|g|_S\le f(|g|_T)$ for all $g\in H$. [If $G$ is finitely generated it's enough to check with a given $T$.] Note it is trivially true if $H$ is finite.

2) A linear group is a subgroup of $\mathrm{GL}_d(K)$ for some $d$ and field $K$.

Now to answer your question:

Proposition: in a linear group, every cyclic subgroup is at most exponentially distorted.

Lemma: let $\mathbf{K}$ be a nondiscrete locally compact field, and $g\in\mathrm{GL}_d(\mathbf{K})$. Endow the latter with the word length $\ell$ with respect to some/any compact generating subset. Then exactly one of the following holds:

a) $(\ell(g^n))_{n\in\mathbf{Z}}$ is bounded. This holds iff all eigenvalues of $g$ (over a finite extension) have modulus one, and in case $\mathbf{K}$ is Archimedean you require in addition that $g$ is semisimple.

b) $(\ell(g^n))$ grows logarithmically. This holds iff $\mathbf{K}$ is Archimedean, all eigenvalues of $g$ have modulus 1, and $g$ is not semisimple.

c) $(\ell(g^n))$ grows linearly. This holds iff $g$ has an eigenvalue (over some finite extension) of norm $\neq 1$.

I leave the proof of the lemma as an instructive exercise.

Now let $G\subset\mathrm{GL}_d(K)$ be a linear group and $\langle g\rangle$ be a cyclic subgroup and let us show the result. Clearly, we can assume that $\langle g\rangle$ is infinite and $G$ is finitely generated; actually we can then assume that $K$ is a finitely generated field.

If some eigenvalue of $g$ is not a root of unity, we can embed $K$ in a nondiscrete locally compact field so that this eigenvalue has norm $\neq 1$ (this is a theorem of Tits, used in the proof of Tits' alternative). It follows that $\langle g\rangle$ is undistorted.

If otherwise every eigenvalue of $g$ is a root of unity, after passing to a power we can suppose that $g$ is unipotent. Then since $g$ has infinite order, we deduce that the characteristic of $K$ is 0 and $g$ is not semisimple. There exists an field embedding of $K$ into $\mathbf{C}$, so by the lemma the growth of $g^n$ in $\mathrm{GL}_d(\mathbf{C})$ is logarithmic. So $g$ is at most exponentially distorted.

Note that the proof also shows that in a linear group in positive characteristic, every cyclic subgroup is undistorted.

  • 1
    $\begingroup$ PS: All this is very basic and comes much before the Lubotzky-Mozes-Raghunathan theorems, in which a nontrivial result is to show that under suitable hypotheses, elements that are exponentially distorted in the ambient group actually remain exponentially distorted in the lattice. $\endgroup$
    – YCor
    Sep 27, 2013 at 15:31

Your Answer

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

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