# Randomized black-box group algorithm for normality testing has false positive rate of at most three-fourths

## Statement

Suppose $G$ is a finite group and $H$ is a subgroup of $G$ that is not normal in $G$. In other words, there exist $g \in G$ and $h \in H$ such that $ghg^{-1} \notin H$. Then, suppose $C$ is the set: $C = \{(g,h) \mid g \in G, h \in H, ghg^{-1} \in H \}$

Then we have: $\! |C| \le \frac{3}{4} |G||H|$

In other words, the probability that a randomly selected pair from $G \times H$ is in $C$ is at most $3/4$.

## Related facts

### Applications

The fact has applications for the normality testing problem, and forms the basis of the randomized black-box group algorithm for normality testing.

## Facts used

1. Lagrange's theorem in the special form subgroup of size more than half is whole group

## Proof

Given: A finite group $G$, subgroup $H$ that is not normal in $G$. $C = \{(g,h) \mid g \in G, h \in H, ghg^{-1} \in H \}$

To prove: $|C| \le \frac{3}{4}|G||H|$

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Let $A$ be the set of elements in $G$ that do not normalize $H$. Then, $|A| \ge \frac{1}{2}|G|$. Fact (1) $H$ is not normal in $G$, $G$ is finite The normalizer $N_G(H)$ is a subgroup of $G$. Since $H$ is not normal, it is a proper subgroup of $G$. Hence, by Fact (1), it has size at most half the size of $G$. Therefore, its complement, $A$, has size at least half the size of $G$.
2 For $g \in A$, let $D(g)$ be the set of elements $h \in H$ such that $ghg^{-1} \notin H$. Then, $|D(g)| \ge \frac{1}{2}|H|$. Fact (1) Step (1) Since $g$ does not normalize $H$, we have that $g^{-1}Hg$ is not contained in $H$. Thus, $g^{-1}Hg \cap H$ is a proper subgroup of $H$. By Fact (1), it has size at most half the size of $H$. Therefore, its complement in $H$, which is $D(g)$, has size at least half the size of $H$.
3 $\sum_{g \in A} |D(g)| \ge \frac{1}{4} |G||H|$ Steps (1), (2) Direct combination of steps
4 $|C| \le \frac{3}{4}|G||H|$ Steps (1), (2), (3) The complement of $C$ in $G \times H$ is precisely the set of pairs $(g,h)$ where $g \in A$ and $h \in D(g)$. By Step (3), the number of such pairs is at least $\frac{1}{4}|G||H|$, so the size of the complement is at most $\frac{3}{4}|G||H|$.