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.

(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 words, parametrized by ordered pairs $$(i,j)$$ 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.