Group theory - subgroups Let $G=(\mathbb{Z}_n,+)\,$ be a group where $\,n \geq 2$ and $d \neq 0$.
Find all possible values for $d$ and $n$ so that $\{0,d\}$ is a subgroup of $G$.
I know that for subset of $G$ to be a group there needs to be a closure under the addition operation. 
So: $\,d+d = 2d\,$ belongs to $\,\{0,d\}$.
So $2d=0$.
But the problem is I'm not sure how to generalize this and to find all possible values for $\,d\,$ and $\,n\,$.
 A: $(1)$ $\mathbb{Z}_{n}$ is cyclic and has a subgroup of order $m$ only if $m$ divides $n$ (by the theorem of Lagrange).
$(2)$ Any subgroup of a cyclic group is cyclic.
$(3)$ In your problem, any subgroup of the form $D\le G = \{0, d\}$ $d\ne 0$, $d \in \mathbb{Z}_n$ clearly has order $2$ and must be cyclic, so $D = \langle d \rangle$. 
So yes, $d+d = 0$, and hence $d^{-1} = -d = d$, otherwise $|\langle d\rangle| = \{0, d\}\ne 2$.
By $(1)$, $|\langle d \rangle| = 2$ must divide $n.\;$ So $\;n = 2k\;$ where k is some positive integer. Hence $n$ must be even. 
Now, what can you say about the possible element $d$ such that $d$ generates a subgroup of order $2$, given that $|\mathbb{Z}_{n}|$ is even $n = 2k$? 


*

*$|G| = 2:\;\;$ If $G = \mathbb{Z}_2$, $D = \{0, 1\} = G$.

*$|G| > 2: \;\;k = d \;\; \implies\quad n = 2k = 2d \implies d = \dfrac{n}{2}.$
Try specific examples to get clear on what's going on here: What element in $\mathbb{Z}_4$, generates a subgroup of order $2$? Likewise, for $Z_{12}$ what is the element that generates a subgroup of order $2$? 
A: $\mathbb{Z}_{n}$ is cyclic and have a subgroup of order $l$ $\iff l\mid n$.
Moreover, this subgroup is unique.
In your case you want $l=2$ hence such a subgroup exist if and only
if $n$ is even.
You also want $H=\langle d\rangle$ hence the order of $d$ is $?$
hence $d=?$
A: HINT: You know that $2d\equiv 0\pmod n$, so $n$ has to be even, and $d$ has to be ... what?
