# Homework Help: 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.

21. Oct 1, 2014

### WWGD

No; we showed : $g^n g^{-n}=1:=id.$. This means, by def. that a,b are inverses of each other; in a group G, two elements are said to be inverses of each other if ab=1. We _showed_ $g^n g^{-n}=1:=id.$, so it follows that they are inverses of each other.

22. Oct 1, 2014

### Bashyboy

But how did you show this? All I see is that you wrote $g^n \star g^{-n} = e$. How do I know that the product of $g^n$ and $g^{-n}$ is the identity element?

23. Oct 1, 2014

### TeethWhitener

You're almost there. You just have to combine what everyone's told you. WWGD is right, but you have to use it in conjuction with an inductive step. So you assume that $(g^n)^{-1}=g^{-n}$ and you prove that $(g^{n+1})^{-1}=g^{-(n+1)}$, which you can do by left-multiplying with $g^{n+1}$ and using associativity and the inductive hypothesis.