Index satisfies transfer inequality

From Groupprops
Jump to: navigation, search

Statement

In terms of index

Suppose G is a group and H, K are subgroups of G. Then:

[K:H \cap K] \le [G:H].

In terms of conditional probability

This formulation is valid for finite groups. It says that if G is a group and H, K are subgroups, then:

\frac{|H \cap K|}{|K|} \ge \frac{|H|}{|G|}

In other words, what it says is that, for a uniform distribution on a finite group, knowing that a particular element is in the subgroup K either increases or keeps the same the probability that the element is in the subgroup H.

Related facts

Applications

The formulation in terms of conditional probability is particularly useful to prove results on the fractions of tuples satisfying a groupy relation. See, for instance:

Facts used

  1. Product formula: if H, K \le G are subgroups, there is a natural bijection between the left cosets of H \cap K in K and the left cosets of H in HK.

Proof

Given: A group G with subgroups H, K.

To prove: [K:H \cap K] \le [G:H].

Proof: By fact (1), the number of left cosets of H in HK equals the number of left cosets of H \cap K in K. Thus, the number of left cosets of H in G is at least as much as the number of left cosets of H \cap K in k, yielding the desired inequality.