# Subgroup of Infinite Cyclic Group is Infinite Cyclic Group

## Theorem

Let $G = \gen a$ be an infinite cyclic group generated by $a$, whose identity is $e$.

Let $g \in G, g \ne e: \exists k \in \Z, k \ne 0: g = a^k$.

Let $H = \gen g$.

Then $H \le G$ and $H \cong G$.

Thus, all non-trivial subgroups of an infinite cyclic group are themselves infinite cyclic groups.

A subgroup of $G = \gen a$ is denoted as follows:

$n G := \gen {a^n}$

This notation is usually used in the context of $\struct {\Z, +}$, where $n \Z$ is (informally) understood as the set of integer multiples of $n$.

## Proof

The fact that $H \le G$ follows from the definition of subgroup generator.

$G \cong \struct {\Z, +}$

Now we show that $H$ is of infinite order.

Suppose $\exists h \in H, h \ne e: \exists r \in \Z, r > 0: h^r = e$.

But:

$h \in H \implies \exists s \in \Z, s > 0: h = g^s$

where $g = a^k$.

Thus:

$e = h^r = \paren {g^s}^r = \paren {\paren {a^k}^s}^r = a^{k s r}$

and thus $a$ is of finite order.

This would mean that $G$ was also of finite order.

So $H$ must be of infinite order.

From Subgroup of Cyclic Group is Cyclic, as $G$ is cyclic, then $H$ must also be cyclic.

$H \cong \struct {\Z, +}$

Therefore, as $G \cong \struct {\Z, +}$:

$H \cong G$

$\blacksquare$

## Comment

The interesting thing to note here is that a non-trivial subgroup of an infinite group is itself isomorphic to the group of which it is a subgroup.

This can be compared with the result Infinite Set is Equivalent to Proper Subset.