# Homework Help: Group Theory: Exponent Laws

1. Apr 21, 2016

### Bashyboy

1. The problem statement, all variables and given/known data
Let $x \in G$ and $a,b \in \mathbb{Z}^+$ Prove that $x^{a+b} = x^a x^b$.

2. Relevant equations

3. The attempt at a solution
If I am not mistaken, we would have to do multiple induction on $a$ and $b$ for the statement/proposition $P(a,b) : x^{a+b} = x^a x^b$. First we would have to show that $P(a,1)$ and $P(1,b)$ are true (base cases). But due to symmetry, it suffices to prove $P(a,1): x^{a+1} = x^a x$. This is where I had trouble; I thought I would have to do induction on the statement on the statement $P(a,1)$.

After trying many things, I decided to consult this https://crazyproject.wordpress.com/2010/01/04/laws-of-exponents-in-a-group/ This person appears to be doing the same as myself. But I found it unsettling that no justification was given for $x^{a+1} = x^a x$ (the base case); what is written is this: "For the base case, note that for all ."

The only way I could think of justifying it is as follows:

$x^{a+1} = \underbrace{x \cdot x \cdot ... \cdot x}_{(a+1)~terms}$

$= \underbrace{(x \cdot x \cdot ... \cdot x)}_{a ~terms} x$ (associative property)

$= x^a x$ (by definition of what $x^a$ means)

But if we can do that, then why not do similar splitting with $a^{a+b}$ , thereby avoiding all induction?

2. Apr 21, 2016

### Math_QED

Define G

3. Apr 21, 2016

### Bashyboy

Dummit and Foote doesn't specify, but I imagine they are regarding it as a group.

4. Apr 21, 2016

### SammyS

Staff Emeritus
I don't see the symmetry off hand, unless you have an "extended" associative property. (I suppose that's always implied.)
At any rate, it looks like induction on b is all that's required.
It looks like they just sketched the proof, leaving the reader to fill in some things they thought to be obvious, or at least fairly straight forward.
That shouldn't even require associativeity if normal order of operations is left to right.
$x^{a+1}= \underbrace{\left(\dots\left(\left(\left(x \cdot x \right)\cdot x\right)\cdot x\right) ... \cdot x\right)}_{a ~\text{factors}} \cdot x$​
.
(Surely you mean $\ x^{a+b}\$ in the above line.)

Induction isn't that difficult for this.

Assume that $\ x^{a+n} = x^a x^n\$ for some n ≥ 1. From that show that $\ x^{a+(n+1)} = x^a x^{n+1}\$ follows.

5. Apr 21, 2016

### Bashyboy

So, are you saying that this is a valid way of establishing $x^{a+1} = x^a x$? If so, then why can't I do the following:

$x^{a+b} = \underbrace{x \cdot x \cdot ... \cdot x}_{(a+b)-terms}$

$= \underbrace{x \cdot x \cdot ... \cdot x}_{a~terms} \cdot \underbrace{x \cdot x \cdot ... \cdot x}_{b~terms}$#

$=x^a x^b$,

thereby avoiding induction.

6. Apr 21, 2016

### SammyS

Staff Emeritus
I'm not saying you can or you can't do that.

It's partly a matter of what you have previously proved and what you have not proved.

Doing the inductive proof doesn't require you to assume much of anything.

Do you understand inductive proof ?

7. Apr 24, 2016

### Bashyboy

I am still wondering about the justification of the base case $x^{a+1} = x^a x$.

8. Apr 24, 2016

### SammyS

Staff Emeritus
I probably should have restated that it's not the associative law. It's basically just the definition of what is meant by $\ x^n\,,\$ perhaps along with the standard order of operations being left to right.

9. Apr 24, 2016

### Bashyboy

And we agree that the definition of $x^n$ is $x^n = \underbrace{x \cdot x \cdot ... \cdot x}_{n-terms}$? If so, then for $n=a+b$

$x^{a+b} = \underbrace{x \cdot x \cdot x ... \cdot x}_{(a+b)-terms}$

$= \underbrace{x \cdot x \cdot ... \cdot x}_{a} \cdot \underbrace{x \cdot x \cdot ... \cdot x}_{b}$.

$=x^a x^b$.

which is my problem. There is some subtle logic that is preventing me from seeing the need of any induction, as well as how we justify $x^{a+1} = x^a x$.

Last edited: Apr 24, 2016
10. Apr 24, 2016

### vela

Staff Emeritus
The subtlety may be that you have to prove the generalized associative property. You know (ab)c = a(bc), but what about when you have more than three elements?

11. Apr 24, 2016

### Bashyboy

Yes, we do have the generalized associative property; this is proved in the chapter before this problem.

12. Apr 24, 2016

### vela

Staff Emeritus
I just ran across a proof. In that class, $a^n = a^{n-1}a$ held by definition (along with $a^0 = e$ and $a^{-n} = (a^{-1})^n$) for $n \ge 1$.

13. Apr 24, 2016

### Bashyboy

Where did you find this? In my Dummit and Foote book, the only thing that holds by definition is $a^0 = e$.

14. Apr 24, 2016

### vela

Staff Emeritus
In notes for an abstract algebra class. You probably have some way to define what $x^n$ means other than the phrase "multiply $x$ together $n$ times."

15. Apr 24, 2016

### vela

Staff Emeritus
I found a copy of your book online, and it appears it is defined by that phrase. In that case, I think your non-inductive proof is fine.

On the other hand, if $x^n$ is defined as $x^{n-1}x$ with $x^0=e$, then I can see where you'd want to prove the law by induction.