Fraction of tuples satisfying groupy relation in subgroup is at least as much as in whole group

Statement
Suppose $$G$$ is a finite group. Suppose $$R \subseteq G^n$$ is a $$n$$-ary relation on $$G$$ with the property that if we consider a $$n$$-tuple $$(g_1,g_2, \dots, g_i,\dots, g_n)$$, and we fix all coordinates except the $$i^{th}$$ coordinate, the set of possibilities for $$g_i$$ forms a subgroup of $$G$$.

Suppose $$H$$ is a subgroup of $$G$$. Then, we have:

$$\! \frac{|H^n \cap R|}{|H^n|} \ge \frac{|R|}{|G^n|}$$

In other words, the fraction of $$n$$-tuples from $$H$$ satisfying $$R$$ is at least as much as the fraction of $$n$$-tuples in $$G$$ satisfying $$R$$.

Applications

 * Commuting fraction in subgroup is at least as much as in whole group
 * Commuting fraction in subring of finite non-associative ring is at least as much as in whole ring
 * Commuting fraction in subring of finite Lie ring is at least as much as in whole ring
 * Associating fraction in subring of finite non-associative ring is at least as much as in whole ring
 * Fraction of tuples for iterated Lie bracket word in subring of finite Lie ring is at least as much as in whole ring

Facts used

 * 1) uses::Index satisfies transfer inequality: This states that if $$H,K \le G$$, then $$[H:H \cap K] \le [G:K]$$. However, the form in which we use it is the conditional probability formulation, which states that:

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

(Note: The letters $$H,K$$ have interchanged roles compared to the formulation of the result on its original page.)

Proof
Given: Finite group $$G$$, groupy $$n$$-ary relation $$R$$ on $$G$$. Subgroup $$H$$ of $$G$$.

To prove: $$\! \frac{|H^n \cap R|}{|H^n|} \ge \frac{|R|}{|G^n|}$$

Proof: We prove the statement by induction on $$n$$. The proof for the base case $$n = 1$$ and the induction step are similar, but we give both proofs.

Proof of base case: In the base case, the relation has only one element, which means that it is just a subset $$R$$ of $$G$$. The groupiness now says that this subset is a subgroup.Then, fact (1) yields that:

$$\! \frac{|H|}{|H \cap R|} \le \frac{|G|}{|R|}$$

Inverting both sides and flipping the inequality sign yields:

$$\! \frac{|H \cap R|}{|H|} \ge \frac{|R|}{|G|}$$

Proof of induction step: Suppose we have shown the result for all $$(n-1)$$-ary relations. We now need to show it for the $$n$$-ary relation $$R$$.

Define $$R_{(g_1,g_2,\dots,g_{n-1})}$$ as the subset of $$G$$ comprising those values of $$g_n$$ such that $$(g_1,g_2,\dots,g_{n-1},g_n) \in R$$. Each such $$R_{(g_1,g_2,\dots,g_{n-1})}$$ is a subgroup, and:

$$\! |R| = \sum_{(g_1,g_2,\dots,g_{n-1}) \in G^{n-1}} |R_{(g_1,g_2,\dots,g_{n-1})}|\qquad (1)$$

Also, we have:

$$\! |R \cap (G^{n-1} \times H)| = \sum_{(g_1,g_2,\dots,g_{n-1}) \in G^{n-1}} |R_{(g_1,g_2,\dots,g_{n-1})} \cap H| \qquad(2)$$

Dividing both sides of (1) by $$|G|$$ and both sides of (2) by $$|H|$$, we get respectively:

$$\! \frac{|R|}{|G|} = \sum_{(g_1,g_2,\dots,g_{n-1}) \in G^{n-1}} \frac{|R_{(g_1,g_2,\dots,g_{n-1})}|}{|G|} \qquad (3)$$

$$\! \frac{|R \cap (G^{n-1} \times H)|}{|H|} = \sum_{(g_1,g_2,\dots,g_{n-1}) \in G^{n-1}} \frac{|R_{(g_1,g_2,\dots,g_{n-1})} \cap H|}{|H|} \qquad(4)$$

By fact (1), we have $$[H:R_{(g_1,g_2,\dots,g_{n-1})} \cap H] \le [G:R_{(g_1,g_2,\dots,g_{n-1})}]$$, yielding:

$$\! \frac{|R_{(g_1,g_2,\dots,g_{n-1})} \cap H|}{|H|} \ge \frac{|R_{(g_1,g_2,\dots,g_{n-1})}|}{|G|}$$

Thus, each of the summands to (4) is bigger than the corresponding summand to (3), and thus we get:

$$\! \frac{|R \cap (G^{n-1} \times H)|}{|H|} \ge \frac{|R|}{|G|}$$

Multiplying both denominators by $$|G^{n-1}|$$, we get:

$$\! \frac{|R \cap (G^{n-1} \times H)|}{|G^{n-1} \times H|} \ge \frac{|R|}{|G^n|} \qquad (5)$$

For $$g \in G$$, define $$R_{g,n}$$ as the relation on $$G^{n-1}$$ induced by $$R$$ with the last coordinate $$g$$. Then:

$$\! |R \cap (G^{n-1} \cap H)| = \sum_{h \in H} |R_{h,n}| \qquad (6)$$

Finally, we have:

$$\! |R \cap H^n| = \sum_{h \in H} | R_{h,n} \cap H^{n-1}| \qquad (7)$$

Dividing (6) and (7) by $$|G^{n-1} \times H|$$ and $$|H^n|$$ respectively, we get:

$$\! \frac{|R \cap (G^{n-1} \cap H)|}{|G^{n-1} \times H|} = \frac{1}{|H|}\sum_{h \in H} \frac{|R_{h,n}|}{|G^{n-1}|}\qquad(8)$$

and:

$$\! \frac{|R \cap H^n|}{|H^n|} = \frac{1}{|H|} \sum_{h \in H} \frac{|R_{h,n} \cap H^{n-1}|}{|H^{n-1}|}\qquad (9)$$

Each $$R_{h,n}$$ is groupy. Thus, by induction, each $$R_{h,n}$$ satisfies:

$$\! \frac{|R_{h,n} \cap H^{n-1}|}{|H^{n-1}|}\ge \frac{|R_{h,n}|}{|G^{n-1}|} \qquad (10)$$

Plugging (10) into the summations on the right sides of (8) and (9), we get:

$$\! \frac{|R \cap H^n|}{|H^n|} \ge \frac{|R \cap (G^{n-1} \cap H)|}{|G^{n-1} \times H|} \qquad (11)$$

Combining (5) and (11) yields the desired results.