# Thread: Group problem with Abelian

1. ## Group problem with Abelian

Prove that a group G is Abelian iff $(ab)^{-1} = a^{-1}b^{-1}$.

2. We have $(ab)^{-1}=b^{-1}a^{-1}$
Then the equality becomes $a^{-1}b^{-1}=b^{-1}a^{-1}$.
Multiply both sides at left by $a$:
$aa^{-1}b^{-1}a=ab^{-1}a^{-1}a\Rightarrow b^{-1}a=ab^{-1}$
Now, multiply both sides by $b$
$bb^{-1}ab=bab^{-1}b\Rightarrow ab=ba$

If G as abelian the identity is obvious:
$(ab)^{-1}=b^{-1}a^{-1}=a^{-1}b^{-1}$

3. I don't understand how you get $
(ab)^{-1}=b^{-1}a^{-1}
$
as I have $
(ab)^{-1} = a^{-1}b^{-1}
$
in the problem.

What allows you to switch them as we are proving the group is communtative, and not assuming it is. (From my understanding)

I don't understand how you get $
(ab)^{-1}=b^{-1}a^{-1}
$
as I have $
(ab)^{-1} = a^{-1}b^{-1}
$
in the problem.

What allows you to switch them as we are proving the group is communtative, and not assuming it is. (From my understanding)
$(ab)^{-1}$ is an element such that when multiplied by $ab$ produces $1$ (identity).

Since $(ab)(b^{-1}a^{-1})=1$ it means $(ab)^{-1}=b^{-1}a^{-1}$.

5. Thank you, sir, it is clear now.