# Show that G is abelian, if and only if (gh)^-1 = g^(-1) h^(-1)

1. Jun 17, 2012

### tonit

1. The problem statement, all variables and given/known data

Show that a group G is abelian, if and only if $(gh)^{-1} = g^{-1} h^{-1}$ for all $g,h\in G$

2. Relevant equations

3. The attempt at a solution

$gh = hg \\⇔ (gh)(gh)^{-1} = hg(gh)^{-1} \\⇔ 1 = hg(gh)^{-1} \\⇔h^{-1} = (h^{-1}h)g(gh)^{-1} \\ ⇔g^{-1}h^{-1} = (g^{-1}g)(gh)^{-1}\\ ⇔ (gh)^{-1} = g^{-1}h^{-1}$

The converse is clear from the above by taking the steps backwards

Thus the statement is proved. Correct?

Last edited: Jun 17, 2012
2. Jun 17, 2012

### algebrat

3. Jun 17, 2012

### tonit

where exactly do you see the mess? So that I can make it clearer?

4. Jun 17, 2012

### algebrat

if gh=hg, take the inverse of both sides, one step, done.

on the other hand, if (gh)^-1=g^-1h^-1, take the inverse of both sides, one step done.

5. Jun 17, 2012

### algebrat

clear? hmm. concise? no. as every good grader should tell you, this was harder to read than it should have been. it's like in an english class, you don't turn in your first draft. by the end of your first draft, you barely know what you're saying. the nice thing about math, you're not turning in 20 pages. (but sometimes your scratch work may feel like it.)

6. Jun 17, 2012

### algebrat

I'm assuming you know the rule (gh)^-1=h^-1g^-1

7. Jun 17, 2012

### tonit

If $gh = hg\\(gh)^{-1} = (hg)^{-1}\Rightarrow (gh)^{-1} = g^{-1}h^{-1}$

conversely, if $(gh)^{-1} = g^{-1}h^{-1}\\((gh)^{-1})^{-1} = (g^{-1}h^{-1})^{-1}\\\Rightarrow gh = (h^{-1})^{-1}(g^{-1})^{-1} = hg$

8. Jun 17, 2012

### algebrat

Yes, that looks a lot cleaner. Now I'm not sure if it was cleaner to prove both directions at the same time, play around with it, but it's right, I'll let you figure out how to pick a final format.

(The way you have it now you don't need the double arrow)

9. Jun 17, 2012

### tonit

yeah. Thanks for the help, I'm learning a lot at this forum really :D