3

I need to prove that $G$ is abelian if and only if the function

$f : G \to G$

defined by $f(a)=a^{-1}$ is a homomorphism.

Assuming that $∀a ∈ G, (a^{-1})^{-1} = a.$

I don't quite understand how to do or approach the proof. The only thing I understand is it's called an abelian when $∀a,b ∈ G, a * b = b * a.$ Any help would be appreciated, thanks!

jasmine
  • 14,457
user138974
  • 33
  • You don't need to assume that $\forall a \in G. (a^{-1})^{-1} = a$: that's already true because $G$ is a group. (Well, you left that implicit...) To solve this problem: write down what it means for $f$ to be a homomorphism. – Magdiragdag Mar 30 '14 at 06:46
  • 2
    Write down what it means for $a\mapsto a^{-1}$ to be a group homomorphism... – anon Mar 30 '14 at 06:47

2 Answers2

0

Hint: In general $(ab)^{-1}=b^{-1}a^{-1}$ (and $\ne a^{-1}b^{-1}$), but here ...

Hagen von Eitzen
  • 374,180
0

If the map $f:G \to G$ defined by $f(x) = x^{-1}$ is a homomorphism, then $f(xy) = f(x)f(y) = x^{-1}y^{-1}$ for all $x, y \in G$. But we always have $(xy)^{-1} = y^{-1}x^{-1}$, whence $f(xy) = y^{-1}x^{-1}$ as well. So for all $x, y \in G$, $x^{-1}y^{-1} = y^{-1}x^{-1}$; taking the inverses of each side yields $yx = (x^{-1}y^{-1})^{-1} = (y^{-1}x^{-1})^{-1} = xy$, and so $G$ is abelian. Likewise, if $G$ is abelian then $x^{-1}y^{-1} = y^{-1}x^{-1}$, whence $f(xy) = (xy)^{-1} = y^{-1}x^{-1} = x^{-1}y^{-1} = f(x)f(y)$, showing $f$ is a homomorphism. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

Robert Lewis
  • 71,180