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.
By Infinite Cyclic Group is Isomorphic to Integers:
- $G \cong \struct {\Z, +}$
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.
From Infinite Cyclic Group is Isomorphic to Integers:
- $H \cong \struct {\Z, +}$
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.
Sources
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.9$: Theorem $15$
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.4$: Cyclic groups: Theorem $1$
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $5$: Cosets and Lagrange's Theorem: Example $5.7$