Prove that there is no $n\in\mathbb Z$ s.t. $n+n=1$. Consider the group $(\mathbb Z,+)$. I want to prove that there is no $n\in \mathbb Z$ s.t. $$n+n=1.$$
Of course it's an obvious question if we consider the ring $(\mathbb Z,+,\cdot )$ since if $n+n=1$ then $2n=1$, then $n$ is a unit. But units are $1$ and $-1$, but neither $n=1$ nor $n=-1$ solve $2n=1$. So there is not solution.

How can I solve this question in the group $(\mathbb Z,+)$ ? What I tried is : suppose there is $n\in\mathbb Z$ s.t. $n+n=1$. Then 
$n=1-n$. But I don't see how to get a contradiction. I tried something as $$n+n=1\implies n+(1-n)=1$$
and thus $1=1$, so no contradiction. 
Any idea ?
 A: Let $G = (\mathbb{Z}, +)$.  Note that $G$ is an infinite abelian group generated by $1$.  (We use this characterization of $1$ when we say that the sum of $k$ copies of this $1$ gives the copy of $k$ in $G$.)  We will want to distinguish members of $G$ and members of the totally ordered ring $\mathbb{Z}$, generated by $1$, that are the sum of the same number of copies of the generator in each, which we will do by subscripting by either $G$, or $\mathbb{Z}$, respectively.
For the sake of contradiction, assume 
$$  n_G + n_G = 1_G  \text{.}  $$
Add $n_\mathbb{Z}$ copies of this equation to obtain
$$  n_\mathbb{Z}(n_G + n_G) = n_\mathbb{Z} 1_G  \text{.}  $$
(Here we are using the usual convention that $zg$ where $z \in \mathbb{Z}$ and $g \in G$ means the addition of $z$ copies of $g$.)  Notice that $n_\mathbb{Z}(n_G + n_G) = (2n)_\mathbb{Z}n_G$ and  $n_\mathbb{Z} 1_G = n_G$.  So we have
$$  (2n)_\mathbb{Z} n_G = n_G  \text{.}  $$
Cancelling in $G$, 
$$  (2n-1)_\mathbb{Z} n_G = 0_G  \text{.}  $$
This gives three possibilities.


*

*$n_G = 0_G$, making our assumption $0_G + 0_G = 1_G$, an impossibility, 

*$G$ has a non-$0_G$ element of finite order, but $G$ has no torsion, or

*$(2n-1)_\mathbb{Z} = 0_\mathbb{Z}$, an impossibility (since the intersection of the odd integers and the even integers is empty).


Therefore, there is no $n_G$ such that $n_G + n_G = 1_G$.
A: If $n+n=1$ then $n$ is a generator of $(\mathbb{Z}, +)$, since $1$ itself is a generator and $1=2n$ is generated by $n$. Thus
either $n=1$ and $1=2$, I think this is only possible in the trivial group
or $n=-1$ and $-2=1$, like in $\mathbb{Z}_3$, for example
A: $\mathbb Z:=\left<1\right>$. We have that $2\mathbb Z=\left<2\right>$. Suppose there is $n\in\mathbb Z$ s.t. $2n=1$. This implies that $\mathbb Z\subset 2\mathbb Z$, and thus $\mathbb Z=2\mathbb Z$. Contradiction. 
