# Module function on automorphisms of discrete locally compact group

Let $G$ be a discrete locally compact group and let $\alpha: G \to G$ be an automorphism. Show that the module $\rm{mod}_{G}(\alpha)$ is 1.

In the case of a locally compact field $k$ and $\alpha$ being multiplication by an element $a$ with $\rm{mod}_{k}(a) < 1$, it is proven by considering the sequence $a^{n}$ which has $\rm{mod}_{k}(a^{n}) \to 0$, hence $a^{n} \to 0$, so $\{0\}$ is not open. For this reason, I was attempting to consider the sequence $\rm{mod}_{G}(\alpha^{n})$ with $\rm{mod}_{G}(\alpha) < 1$, but I wasn't sure what to conclude.

• Isn't Haar measure on a discrete group just counting measure, so this follows immediately from the definition of the modulus of an automorphism of $G$ as the ratio of the measure of a set to the measure of its image under the automorphism? – Keenan Kidwell Jul 18 '14 at 19:02
• @KeenanKidwell Is there anything special about a discrete group that allows the counting measure or is it just that one point sets usually aren't open, hence not necessarily Borel? – Jacob Bond Jul 18 '14 at 19:14
• A discrete group is automatically locally compact, and as (left) Haar measure on a locally compact group is unique up to scaling, and counting measure on a discrete group is a Haar measure, it is "the" Haar measure. Also, locally compact fields need not be discrete, and the modulus of multiplication by an element is the normalized absolute value of that element, not usually $1$, so that line of thought is irrelevant. – Keenan Kidwell Jul 18 '14 at 19:24
• @KeenanKidwell I got the measure part of that, but why ISN'T the module of an automorphism on a nondiscrete group always 1? Where does the reasoning break down? – Jacob Bond Jul 18 '14 at 20:00
• @jaebond If it were always $1$ it would be pretty useless. – egreg Jul 18 '14 at 20:07

The Haar measure on a discrete group is counting measure, so $\mu(E)=\#E$ for any Borel set $E$ (where $\#E$ means $\infty$ if $E$ is infinite). It's clear that an automorphism $\alpha$ of $G$ preserves cardinality (in the naive since that it takes infinite sets to infinite sets and it takes sets of finite cardinality to sets of the same finite cardinality), so for any Borel set $E$, $\mu(\alpha(E))=\mu(E)$, hence the modulos of $\alpha$ is $1$.