# Modern Algebra: Proof of an exponent law

1. Sep 30, 2014

### Bashyboy

The problem is to verify $(g^n)^{-1} = g^{-n}$ is true $\forall n \in \mathbb{Z}$. Here is my proof:

$(g^n)^{-1} = (\underbrace{g \star g~ \star ~...~ \star g}_{n~ \mbox{copies}})^{-1} \iff$

$(g^n)^{-1} = [(g \star ~...~ \star g) \star g]^{-1}$

Using $(a \star b)^{-1} = b^{-1} \star a^{-1}$ (1), which I have already proven, gives

$(g^n)^{-1} = g^{-1} \star (\underbrace{g \star g~ \star ~...~ \star g}_{n-1~ \mbox{copies}})^{-1}$

If I apply (1) $n-1$ more times, I will arrive at

$(g^n)^{-1} = \underbrace{g^{-1} \star ~...~\star g^{-1}}_{n ~\mbox{copies}}$

The right hand side is by definition $g^{-n}$.

What is wrong with this proof? My professor said it is wrong.

Last edited: Sep 30, 2014
2. Sep 30, 2014

### Staff: Mentor

I think the preferred proof would be by induction where you do two steps:
- prove that when n=1 its true
- prove given its true for k then its true for k+1

3. Sep 30, 2014

### Staff: Mentor

Your prof is probably looking for a proof by induction. The "n copies" and "n - 1 copies" stuff is not rigorous.

4. Sep 30, 2014

### Bashyboy

Well, the definition includes the word "copy:"

"In general, if $n$ is a positive integer, then $g^n$ is the binary operation applied to $n$ copies of $g$."

I tried using induction; but I am having difficulty using induction to prove the procedure I am using.

Last edited: Sep 30, 2014
5. Sep 30, 2014

### Bashyboy

I think I need to use induction to show that my procedure is valid; the procedure being that I associate a certain amount of elements, and then apply (1). I could denote this procedure by P(n), and then show that P(n) is true for all natural numbers.

Does this appear correct?

6. Sep 30, 2014

### Staff: Mentor

Not quite. Along the lines of what jedishrfu said, do these things.
1. Show that the statement is true for n = 1.
2. Assume that the statement is true for n = k. IOW, assume that (gk)-1 = g-k.
3. Show (prove) that the statement is true for n = k + 1. You will need to use the assumption of step 2 to do this.

BTW, shouldn't the problem read "for all n $\in$ Z+; i.e., the positive integers?

7. Oct 1, 2014

### WWGD

Why not just left-multiply by $g^n$ in the first line?

Then you get :

$(g^n)(g^n)^{-1}=e = g^n g^{-n}$

8. Oct 1, 2014

### Bashyboy

But I am not certain at the moment that $g^{n}$ is the inverse of $g^{-n}$, as this would rely on the fact that $g^n g^m = g^{n+m}$, which I am trying to avoid using. However, I could write

$g^{n} \star g^{-n} = (g \star g \star ... \star g) \star (g^{-1} \star g^{-1} \star ... \star g^{-1} )$

From this I could the associative property a finite number of times.

9. Oct 1, 2014

### WWGD

No, I am not assuming that ;by def., $(g^n)^{-1}$ is the inverse of #g^n#

10. Oct 1, 2014

### Staff: Mentor

The equation above is precisely the type of thing you should use induction on to prove.

11. Oct 1, 2014

### Bashyboy

Yes, that is true. But it isn't immediately obvious that $g^{-n}$ is the inverse of $g^n$.

12. Oct 1, 2014

### Bashyboy

Yes, but I need to prove that the statement is true not only for the natural numbers, but also the negative integers. This is what I am having difficulty with.

13. Oct 1, 2014

### WWGD

But then you get that the identity 'e' on the left side equals$g^n g^{-n}$. This means precisely that #g^{-n}# is the inverse of #g^n#, since their product equals the identity. a is the inverse of b , by def., if ab=e:=id. then a is the inverse of b and viceversa; ab=id, then left- multiply both sides by $a^{-1}$ to get $$a^{-1}ab=b=a^{-1}$$

Last edited: Oct 1, 2014
14. Oct 1, 2014

### Staff: Mentor

Before we go any further, what is g? Is it an element of a group whose operation is multiplication?

15. Oct 1, 2014

### Bashyboy

$g$ is a group element with the arbitrary binary operator $\star$

16. Oct 1, 2014

### Staff: Mentor

Are there any properties given for this operator?

17. Oct 1, 2014

### WWGD

II think this is just a general group operation.

18. Oct 1, 2014

### Bashyboy

This is correct.

19. Oct 1, 2014

### Bashyboy

But would that not require assuming the equality $(g^{n})^{-1} = g^{-n}$ is true, which is precisely what we are trying to prove?

20. Oct 1, 2014

### Staff: Mentor

For n a positive integer, you're showing that gn ⋆ g-n = 1 (or whatever the identity is), using induction.
For m a negative integer, -m is a positive integer, show that g-m ⋆ gm = 1, also by induction. Sharper minds than mine might be able to give a reason to skip this second step.
If the group is known to be Abelian (commutive), then I don't think the second part above is needed, because a ⋆ b would be the same as b ⋆ a.