# Maximal implies modular

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., maximal subgroup) must also satisfy the second subgroup property (i.e., modular subgroup)
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## Statement

### Verbal statement

Any maximal subgroup of a group is a modular subgroup.

### Statement with symbols

Suppose $A$ is a maximal subgroup of $G$. Then, if $B,C$ are subgroups of $G$ such that $A \le C$, we have:

$\langle A, B \cap C \rangle = \langle A,B \rangle \cap C$.

## Proof

Given: A group $G$, a maximal subgroup $A$.

To prove: If $B,C$ are subgroups of $G$ such that $A \le C$, then:

$\langle A, B \cap C \rangle = \langle A,B \rangle \cap C$.

Proof: Since $A$ is maximal, either $C = A$ or $C = G$. We consider both cases.

If $A = C$, then $B \cap C \le A$, so the left side is $A$. $\langle A, B \rangle$ contains $A$, so the right side is $A$. Thus, the identity holds for $A = C$.

If $C = G$, then both sides are $\langle A, B$, and again the identity holds.