# 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

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 $n(n+1)/2[itex] words, parametrized by ordered pairs [itex](i,j)$ with $1 \le i < j \le n + 1$:

$w_{i,j}(x_1,x_2,\dots,x_n,x_{n+1}) = x_ix_j^{-1}$

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 $1 \le i < j \le n + 1$ 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 $g_i \ne g_j$ so that $w_{i,j}$ is not universally satisfied (this is where we use nontriviality).

Thus, $G$ is not discriminating.