What does $(aba^{-1})^n$ mean? What does $a^n b^n a^{-n}$ mean? What would have to be true in order for both of the expressions to be equal? Does the order matter? (I think you can guess what I am hinting the answer to your question is.)

I guess, according to Thm 2.8, those expressions are equal only if the group is abelian, which I can't assume.

On the other hand, according to this theorem, $(g^m)^n = g^{mn}$. So, what's the deal? I have to assume $(g \cdot h)^n \neq g^nh^n$, right? The exponent laws as I know them don't work anymore?

I feel like Prop. 2.4 could be useful here, $(ab)^{-1} = b^{-1}a^{-1}$.

Cathy Fulkerson Ajamie

On the other hand, according to this theorem, $(g^m)^n = g^{mn}$. So, what's the deal?

The deal is that an element always commutes with itself. There's only $g$ here!

I have to assume $(g \cdot h)^n \neq g^nh^n$, right?

I think it would be more accurate to say that you cannot assume anything one way or the other.

The exponent laws as I know them don't work anymore?

I was trying to jump up and down in class when I was trying to make this point. Some of the exponent laws that you are familiar with for real numbers are the way they are *precisely* because $\mathbb{R}$ is abelian. By definition, we have

And on the other hand, we have

(2)The only way these two expression can be equal is if we can commute $g$ and $h$.

Have you tried writing down what $(aba^{-1})^n$ actually means?

How do you do box text like you did in your answer?

And

(1)So, if I multiplied equation #2 by $n$ copies on its inverse, $(aba^{-1})^{-1}$ it would equal 1.

(2)I think if this is true, $(h^{-1}g^{-1})^{-n}= (gh)^n$ then $(aba^{-1})^{-n}= (a b^{-1}a^{-1})^n,$ but does it help?

Cathy Fulkerson Ajamie

To get

How do you do box text like you did in your answer?

you type

`> How do you do box text like you did in your answer?`

Make sure you have a space to the right of `>`.

Cathy wrote:

$(aba^{-1})=\underbrace{(aba^{-1})(aba^{-1})\cdots (aba^{-1})}_{n\text{ copies}}$

Yes! Stare at this until you see what to do. Keep staring. See it yet? If not, keep staring. It's all right before your eyes.

Is this it? Since $n=n$, the same copies of $b^n$ on the left will equal the number of copies of $b$ on the right. The rest of the algebra is just right or left multiplying by the approriate inverses. All I can say is duh…

BTW, when you copied my text, the $n$ was omitted from the $(aba^{-1})$. Here's what was meant to be copied:

$(aba^{-1})^n = \underbrace{(aba^{-1})(aba^{-1}) \cdots (aba^{-1})}_{n\text{ copies}.$

*Proof*. Let $a,b \in G$ and$n\in \Bbb{R}$. Assume that $G$ is a group. We need to show that

We see that

(2)So, since $1 = 1$, both sides of the equation are equal.

$\square$

Cathy Fulkerson Ajamie

Awesome, you just proved that $1=1$:) OK, go back to the line I told you to stare at. Staring at it? Since groups are associative, the parentheses don't matter, right? Hint: CANCEL. How many $b$'s do you have?

LOL I can't stand it! How do you?

OK,

(1)We see that

(2)So, I suppose that I have proved this for $b^2$… So, I continue to struggle with how to get the $n$ in the right-hand side of the equation to be an exponent for $b$. Like I said, I understand that the number of copies of $(a b a^{-1})$ that I have on the right will yield the same number of $b$s that will be present on the left because $n=n$.

Wait, do I get to present this in class?

Cathy Fulkerson Ajamie

We don't "see" $ab^na^{-1}=(aba^{-1})^n$ until we prove it. You can't start with this since this is what you are trying to prove. You can, among other things, start with one side and do allowable stuff to it to get to the other. There are other ways, too. Take a step back; I think you're trying to hard. It's like Chinese finger cuffs.

Perhaps someone else has something constructive to say.

Which is what we want.

\underbrace{gg \cdots g}_{n\text{ copies}}

Cathy Fulkerson Ajamie

So, just to be explicit about this $n$ thing… I'm allowed to write $b^n$ in the end because I have $bbb...b$ - writing those ellipses makes it "legal" to write $b^n$. My whole hang-up for most of this proof is how to "legally" get $n$ from outside those parentheses to the $b$ on the inside…

Cathy Fulkerson Ajamie

It's not just legal, that's exactly what it means! BTW, you should use `\cdots` instead of `\dots` (it's the multiplication that's continuing). Also, in your second to last post, you should probably use an `\underbrace` in the appropriate places to denote how many copies of stuff you have. In addition, in the third line in your string of equations, you should write $ebe$ (or something like that) in the middle instead of just $e$ to make the pattern more explicit.