Proving an exponent law in group theory

[Mr Davis 97]

The textbook proves that $x^a x^b = x^{a+b}$ by an induction argument on b. However, is an induction argument really necessary here? Can't we just look at the LHS and note that there are a $a$ x's multiplied by $b$ x's, so there must be $a+b$ x's?

 

[ShayanJ]

That's the same as the induction argument, only without the details.

 

[Mr Davis 97]

That's the same as the induction argument, only without the details.

So the details are absolutely necessary?

 

[Stephen Tashi]

The textbook proves that $x^a x^b = x^{a+b}$

Can't we just look at the LHS and note that there are a $a$ x's multiplied by $b$ x's, so there must be $a+b$ x's?

If you look at the left hand side of that equation, you see two symbols "x" and on the right hand side you see one symbol "x". So the argument you are making involves more than inspecting the symbolic expressions on left and right hand sides. You are also making an interpretation of the symbolic notation in that equation as other symbolic notation and then you are implicitly making use of the associative property of the group's multiplication.

 

[fresh_42]

So the details are absolutely necessary?

No, not really, because it's "obvious" in this case. But if you want to be very picky, you have to avoid the "dots" and replace them by an induction argument. Additionally the associative property is required to justify the notation $x^a$ at all. So the answer is: it depends on how explicit you want to be.

 

[WWGD]

The textbook proves that $x^a x^b = x^{a+b}$ by an induction argument on b. However, is an induction argument really necessary here? Can't we just look at the LHS and note that there are a $a$ x's multiplied by $b$ x's, so there must be $a+b$ x's?

Are a,b positive integers?