# Prove that $G$ is a group

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?

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Note that if $a\in G$ then $na\in G$, so if $a, b\in G$ then $|a|b+|b|a\in G$, where $|c|$ denotes the absolute value. Why does this give you the identity?

One you have realised why the above line gives you the identity, can you see why it also gives you inverses?

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I see why it gives the identity. But why does it give inverses? – Mohan Dec 1 '11 at 12:03
@user774025: $b+(|a|-1)b+|b|a=0$... – user1729 Dec 1 '11 at 12:10

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.

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