# Abelian Group proof

• Mar 9th 2009, 08:50 PM
jzellt
Abelian Group proof
Suppose a,b e G, where G is an abelian group under +. Show -(a + b) = (-a) + (-b).

Thanks for any help.
• Mar 9th 2009, 09:10 PM
Chris L T521
Quote:

Originally Posted by jzellt
Suppose a,b e G, where G is an abelian group under +. Show -(a + b) = (-a) + (-b).

Thanks for any help.

By a property of inverses, $\displaystyle -(a+b) = (-b) + (-a)$. Since $\displaystyle a,b\in G$ and G is Abelian, then $\displaystyle (-b)+(-a)=(-a) + (-b)$
• Mar 9th 2009, 09:14 PM
jzellt
How do we know that -(a + b) = (-b) + (-a)?

Isn't that something I would have to show while giving a solid proof?
• Mar 9th 2009, 09:17 PM
Chris L T521
Quote:

Originally Posted by jzellt
How do we know that -(a + b) = (-b) + (-a)?

Isn't that something I would have to show while giving a solid proof?

You need to keep in mind that if $\displaystyle \left(G,*\right)$ is a group $\displaystyle a,b\in G$, then $\displaystyle \left(a*b\right)^{-1}=b^{-1}*a^{-1}$. In additive notation, this would be the same as $\displaystyle -(a*b)=(-b)*(-a)$, where $\displaystyle *$ is a binary operation. In this case, that binary operation is $\displaystyle +$.

Does this clarify things?