TITLE: Prove that $G$ is a group
QUESTION [7 upvotes]: Let $G$ be a subset of $\mathbb Z$. 
$G$ has at least one positive and one negative element and if $a, b \in G$ then $a+b \in G$.  How to show that $G$ is group under + operation?

REPLY [5 votes]: The subgroup criterion gives us the claim, if we can show that $G$ is closed under negation. 
Let $x\in G$ be arbitrary. Assume that $x>0$ (the other case is handled by making the obvious changes to this argument). There is a negative number $m<0$ in $G$. Applying the assumption $x-1$ times gives us that $2m=m+m, 3m=2m+m,\ldots, xm=(x-1)m+m$ are all elements of $G$. 
Then we get $mx+x=(m+1)x\in G$, $mx+2x=(m+2)x\in G,\ldots$ and eventually $(-1)x\in G$, because from $m$ we get to $-1$ by adding $+1$ finitely many times.