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**Statement:**

Let $G$ be a group and let $a \in G$. Then $H= \{a^{n} \ | \ n \in \mathbb{Z}\}$ is a subgroup of $G$ and is the smallest subgroup of $G$ that contains $a$, that is, every subgroup containing $a$ contains $H$.

**Proof:**

We check the conditions given in Theorem 5.14 for a subset of a group to give a subgroup. Since $a^{r}a^{s} = a^{r+s}$ for $r,s \in \mathbb{Z}$, we see that the product in $G$ of two elements of $H$ is again in $H$. Thus $H$ is closed under the group operation of $G$. Also $a^{0} = e$, so $e \in H$, and for $a^{r} \in H$, $a^{-r} \in H$ and $a^{-r}a^{r} = e$. Hence all the conditions are satisfied, and $H \leq G$

Now, let $K \leq G$ with $a \in K$, and suppose that $\mid K \mid < \mid H \mid$.

Then there exists some element in $H$ that is not an element of $K$. Without loss of generality, let this element be denoted $a^n$ with $n \in \mathbb{Z}$.

Additionally, $a^{-1} \in K$ by definition of subgroup, since $a \in K$.

Then $n$ cannot be less than zero, since $a^{-1} \cdot a^{-1} \cdot ... \cdot a^{-1}$ with $n$ factors of $a^{-1}$ must be an element of $K$ by the requirement that subgroups be closed under the operation.

However, $n$ cannot be greater than zero, since $a^{1} \cdot a^{1} \cdot ... \cdot a^{1}$ with $n$ factors of $a^{1}$ must also be an element of $K$ by the requirement that subgroups be closed under the operation.

Thus, it must be the case that $n = 0$, by process of elimination.

But this is a contradiction of our assumption that $K \leq G$, since $a^{0} = e$, which must be an element of $K$ by definition of subgroup.

Therefore, $\mid K \mid \not < \mid H \mid$.

It follows that $H$ is the smallest subgroup of $G$, since it was shown that no smaller subset can be a group.