Modern Algebra: Proof of an exponent law

[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.

 

[jedishrfu]

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

 

[Mark44]

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.

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

 

[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.

 

[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?

 

[Mark44]

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?

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 (g^k)^-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?

 

[WWGD]

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.

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

Then you get :

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

 

[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.

 

[WWGD]

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

 

[Mark44]

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.

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

 

[Bashyboy]

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

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

 

[Bashyboy]

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

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.

 

[WWGD]

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

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 $$a^{-1}ab=b=a^{-1} $$

 

[Mark44]

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.

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

 

[Bashyboy]

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

 

[Mark44]

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

Are there any properties given for this operator?

 

[WWGD]

Are there any properties given for this operator?

II think this is just a general group operation.

 

[Bashyboy]

II think this is just a general group operation.

This is correct.

 

[Bashyboy]

But then you get that the identity 'e' on the left side equalsgng−n 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−1a^{-1} to get

a−1ab=b=a−1​

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

 

[Mark44]

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 ⋆ g^m = 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.

 

[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.

 

[Bashyboy]

No; we showed : gng−n=1:=id. 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_ gng−n=1:=id. g^n g^{-n}=1:=id. , so it follows that they are inverses of each other.

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?

 

[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.