Discriminating and finite iff trivial

Statement

The following are equivalent for a group:

1. It is both a discriminating group and a finite group.
2. It is the trivial group.

Proof

(2) implies (1)

This is obvious.

(1) implies (2)

It is easier to prove the result in a somewhat different form: a finite nontrivial group cannot be discriminating.

Suppose $G$ is a finite nontrivial group. Suppose $n$ is the order of $G$. Consider $n+1$ letters $x_{1},x_{2},\dots ,x_{n+1}$, and the following $n(n+1)/2<math>words,parametrizedbyorderedpairs<math>(i,j)$ with $1\leq i<j\leq n+1$:

$w_{i,j}(x_{1},x_{2},\dots ,x_{n},x_{n+1})=x_{i}x_{j}^{-1}$

We note that:

• For any tuple $(g_{1},g_{2},\dots ,g_{n},g_{n+1})\in G^{n+1}$, size considerations force that there exist $(i,j)$ with $1\leq i<j\leq n+1$ such that $g_{i}=g_{j}$, so that $w_{i,j}$ is trivial.
• However, for any particular word $w_{i,j}$, we can choose a tuple with $g_{i}\neq g_{j}$ so that $w_{i,j}$ is not universally satisfied (this is where we use nontriviality).

Thus, $G$ is not discriminating.