Find all positive integer $n$ such that $\mathbb{Z}_{n}$ contains a subring isomorphic to $\mathbb{Z}_{2}$
A subring a subset of the ring such that it is closed under addition, contains the identity of the ring, closed under multiplication, and closed under additive inverses.
So we want a subring isomorphic to $(\mathbb{Z}_2,+)$, then $n = 2k$. The identity of $\mathbb{Z}_{2k}$ is $0$ since $a+0 = 0+a = a$. How do we describe the inverses for the subring?
Best Answer
There is some confusion in the original posting. The question asked for a subring $S$ of $R=\mathbb{Z}_n$ that included the identity. Through the comments, it appears that the OP wanted the additive identity to be included in the subring, not the multiplicative identity. I will provide discussion of both ideas here.
If $S$ must contain the multiplicative identity, $1$, then $S$ contains all sums of $1$ with itself, in other words, all the elements of $R$. Hence, if $1\in S$, then $S=R$. The only way for $S\simeq\mathbb{Z}_2$ is if $R=\mathbb{Z}_2$.
If $S$ does not need to contain the multiplicative identity, then for $S$ to be isomorphic to $\mathbb{Z}_2$, then $S$ must have two elements. It has the $0$ of $R$ because $S$ is a ring and it has one other element, which we'll call $g\not=0$. Since $S\simeq\mathbb{Z}_2$ and $S$ is a subring of $\mathbb{Z}_n$, it must he that $g+g\equiv 0\pmod n$ and $g\cdot g\equiv g\pmod n$.
Since $g+g\equiv 0\pmod n$ and $g\not=0$, we know that $n\mid 2g$ and that $2\mid n$. Therefore, $n=2k$ for some $k$. Assuming, without loss of generality that $0<g<n$, then $2g<2n$, so the only way for $n$ to divide $2g$ is if $n=2g$. In this case, we know that $g\equiv k=\frac{n}{2}\pmod n$.
Since $g\cdot g\equiv g\pmod n$, $g\not=0$, $n=2k$, and $g\equiv k\pmod n$, we consider $k^2$. If $k$ is even, then $n=2k\mid k^2$, which cannot happen because $g\cdot g\equiv g\not\equiv 0\pmod n$. Therefore, $k$ is odd. Since $k$ is odd, $k=2l+1$ for some integer $l$, then $k^2=k(2l+1)=2kl+k=nl+k\equiv k\pmod n$. Therefore, when $k$ is odd, $k^2\equiv k\pmod n$.
Combining all of this, we have that when $n=2k$ where $k$ is odd. $R=\mathbb{Z}_n$ has the subring $S=\{0,k\}$, which is isomorphic to $\mathbb{Z}_2$.