Discriminating and finite iff trivial

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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, parametrized by ordered pairs <math>(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.