# Free factor implies self-normalizing or trivial

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., free factor) must also satisfy the second subgroup property (i.e., self-normalizing subgroup)

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## Statement

Suppose , i.e., is a free product of subgroups and , so is a free factor of . Suppose further that is a nontrivial group. Then, is a self-normalizing subgroup of : .

## Proof

**Given**: A free product .

**To prove**: is self-normalizing in .

**Proof**: Suppose is not self-normalizing in . Pick any . Then we can write uniquely as an alternating product of elements from and , with at least one letter from . If the first letter for the unique expression for is , we can replace by to get a new element in whose first letter is in . Similarly, if the last letter is , we can replace by to get a new element in whose last letter is in . Thus, we can, without loss of generality, assume that:

where each , and all elements are non-identity elements. Now, consider any non-identity element (we can do this because is a nontrivial group). Clearly, the word is also a reduced word, and since this word has length more than one, we see that . This contradicts the assumption that , completing the proof.