Existence of Unique Subgroup Generated by Subset/Singleton Generator

Theorem
Let $\left({G, \circ}\right)$ be a group. Let $a \in G$.

Then $H = \left\langle {a}\right\rangle = \left\{{a^n: n \in \Z}\right\}$ is the unique smallest subgroup of $G$ such that $a \in H$.

That is:
 * $K \le G: a \in K \implies H \subseteq K$

Proof
From Powers of Element form Subgroup, $H = \left\{{a^n: n \in \Z}\right\}$ is a subgroup of $G$.

Let $K \le G: a \in K$.

By Ordering on Integers is Trichotomy, there are three cases:
 * $(1) \quad n \gt 0$


 * $(2) \quad n = 0$


 * $(3) \quad n \lt 0$

Case 1
Since the Natural Numbers are Non-Negative Integers, the case where $n \gt 0$ will be proved using induction.

Basis For Induction
If $n = 1$, then $a^1 = a$ by definition of power of element.

And so $a^1 \in K$.

Induction Hypothesis
Suppose that $a^n \in K$ for some $n \in \Z_{\gt 0}$.

Induction Step
Then by the definition of power of element:
 * $a^{n+1} = a^n \circ a$

It is known that $a^n \in K$ and $a \in K$ by hypothesis.

So then by the closure of $K$ we have that:
 * $a^n \circ a \in K$

Therefore $a^{n+1} \in K$.

Case 2
Let $e$ be the identity element of $G$.

If $n = 0$, then by the definition of power of element $a^0 = e$.

By Identity of Subgroup, $e \in K$.

Therefore $a^0 \in K$.

Case 3
If $n \lt 0$, then by definition of power of element:
 * $a^n = \left({a^{-n}}\right)^{-1}$

It is also seen that:

Therefore $a^n \in K$.

It has been shown that:
 * $\forall n \in \Z: a^n \in K$

Thus, $H \subseteq K$.