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I get the equation $a^{-1} + b^{-1} = (a + b)^{-1}$ from ordinary + operation. For ordinary + operation I mean $a^{-1} = -a$. It is also true for * of rational numbers $3^{-1}*4^{-1} = \frac{1}{3} * \frac{1}{4} = \frac{1}{12}=(3*4)^{-1}$.

I would like to know whether it is true for any Abelian group? If it is true I would like to know why?

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Yes, $(ab)^{-1} = b^{-1}a^{-1}$ in general but since abelian this is the same as $a^{-1}b^{-1}$. –  Deven Ware Oct 11 '12 at 19:04
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up vote 8 down vote accepted

When using $+$ for the group operation, it is more traditional to write inversion as negation: i.e.

$$ (-a) + (-b) = -(a+b) $$

Anyways, that equation is true. A similar statement is true for any group:

$$ a^{-1} b^{-1} = (b a)^{-1} $$

The proof is straightforward: $(ba)^{-1}$ is the unique group element satisfying the equation

$$ x (ba) = 1 $$

and $a^{-1} b^{-1}$ is a solution for $x$.

In additive terms, the equation would be that $-(b + a)$ is the unique solution to

$$ x + (b + a) = 0 $$

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It is true, but normally if the group is Abelian and you're using "$+$" as the symbol for the group operation, you'd write the inverse of $a$ as $-a$ rather than as $a^{-1}$.

To see that this is true, recall that $(ab)^{-1}$ (with $a$ to the left of $b$) is equal to $b^{-1}a^{-1}$ (with $b$ to the left of $a$), and then recall that since the group is Abelian, it doesn't matter which is on the left.

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