Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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

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

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.