Abelian Group proof

**jzellt** · March 9th 2009, 08:50 PM

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

Thanks for any help.

**Chris L T521** · March 9th 2009, 09:10 PM

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, $-(a+b) = (-b) + (-a)$ . Since $a,b\in G$ and G is Abelian, then $(-b)+(-a)=(-a) + (-b)$

**jzellt** · March 9th 2009, 09:14 PM

How do we know that -(a + b) = (-b) + (-a)?

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

**Chris L T521** · March 9th 2009, 09:17 PM

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

Does this clarify things?

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