An exercise from Dummit and Foote, example 2.1.7
Fix some $n\in \mathbb{Z}$ with $n>1$. Find the torsion subgroup of $\mathbb{Z}\times \left( \mathbb{Z}/n\mathbb{Z}\right)$. Show that the set of elements of infinite order together with the identity is $\textit{not}$ a subgroup of this direct product.
My work;
The elements in $G=\mathbb{Z}\times \left( \mathbb{Z}/n\mathbb{Z}\right)$ are $(x,\overline{y}), x\in \mathbb{Z}, \bar{y}\in (\mathbb{Z}/n\mathbb{Z})$. The only element in $\mathbb{Z}$ that has finite order is the element $1_{\mathbb{Z}}=0$, but all elements of $(\mathbb{Z}/n\mathbb{Z})$ have finite order. Thus, elements in $H=\{(x,\overline{y})\in \mathbb{Z}\times \left( \mathbb{Z}/n\mathbb{Z}\right) \mid |(x,\overline{y})|<\infty \}$ are of the form $(0,\bar{y})$, and it is clear now that $H$ is nonempty. As inverse elements of $\mathbb{Z}/n\mathbb{Z}$ are $-\overline{y}$, then for $\overline{x}, \overline{y}\in \mathbb{Z}/n\mathbb{Z}$, $$(0,\overline{x})+(0,\overline{y})^{-1}=(0,\overline{x})+(0,-\overline{y})=(0,\overline{x}-\overline{y})=(0,\overline{x-y})\in \mathbb{Z}\times \left( \mathbb{Z}/n\mathbb{Z}\right)$$ So since $(0,\overline{x-y})$ is in $H$, $H$ is a subgroup of $G$.
So that seems pretty straight forward. Then infinite ordered elements would be of the form $(x,\overline{y})$ where $x\neq 0$. Combining this with $1_{G}=(0,\overline{0})$, we check to see if this set is a subgroup. But closure fails because for elements $(1,\overline{1}), (-1,\overline{1})$,
$$(1,\overline{1})+(-1,\overline{1})=(1-1,\overline{1}+\overline{1})=(0,\overline{2})$$
which is not of infinite order. So this subset of $G$ is not closed and thus can not be a subgroup of $G$.
Does this proof suffice?