How to prove that in an abelian group $-(-a) = a$?

Question

I have to prove that in an abelian group $-(-a) = a$, and the only hint given is that the inverses are unique.

My attempt is as follows:

$-(-a) = a$ is equivalent to $-(-a) - a = 0$, but I don't know how to continue. Some ideas?

Answer 1

By definition of inverse (of $-a$ and $a$, respectively), $-(-a)+(-a)=0$ and $a+(-a)=0$.

Therefore $-(-a)$ and $a$ are both inverses of $-a$, by definition of inverse.

Using the hint that inverses are unique, therefore, $-(-a)=a.\quad$ QED

Answer 2

First, let us recall some defintions:

1. $0$ is an element such that for every element $b$, $$b + 0 = 0 + b = b$$

2. For any element $b$, its inverse $-b$ is an element such that $$b + (-b) = (-b) + b = 0$$

Start from these, we have:

$$\begin{array}{rll} -(-a) &= -(-a) + 0 & \color{blue}{\text{ definition of } 0}\\ &= -(-a) + ((-a) + a) & \color{blue}{\text{ definition of } -a}\\ &= (-(-a) + (-a)) + a & \color{blue}{\text{ addition is associative }}\\ &= 0 + a & \color{blue}{\text{ definition of } -(-a)}\\ &= a & \color{blue}{\text{ definition of } 0} \end{array}$$