Subgroup of Free Group is Free Group

Theorem

Let $G$ be a free group.

Let $H \le G$ be a subgroup of $G$ such that $H$ is not the trivial group.

Then $H$ is also a free group.

Proof

Let $x \in H$ such that $x$ is not the identity element.

Because $H$ is not the trivial group, such an $x$ always exists.

As $H$ is a subgroup of $G$:

$x \in G$

where, by hypothesis, $G$ is a free group.

By Element of Free Group can be Expressed Uniquely as Finite Product, $x$ can be expressed uniquely in the form:

$a^\alpha b^\beta \dotsm r^\sigma$

where:

adjacent elements $a, b, \ldots, r$ of $a^\alpha b^\beta \dotsm r^\sigma$ are distinct elements of $G$

$\alpha, \beta, \ldots, \sigma$ are non-zero integers.

Hence, by definition, $H$ is a free group.

$\blacksquare$

Sources

• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): free group
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): free group