Proving simple rule by induction

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>>9019617
This is from Lang's book (Undergraduate Algebra).

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>>9019619
I first check that the *notation* [math]x_{1}\cdots x_{n}[/math] makes sense, since in the group axioms the product is only defined pairwise.

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>>9019624
Now I define [math]x^{n}:=x_{1}\cdots x_{n}[/math] where [math]x_{i}:=x[/math] for the various [math]i[/math]s, instead of [math]x^{n}:=x\cdots x[/math] with the subscript "n times".
This way you can prove the rule in OP for the various integers [math]m,n[/math] using induction instead of subscripts.
Here an example for positive integers.

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This is about the very basics of group theory so I encourage everyone with some experience on giving a read at it.

Here is an exercise. Can you give proof of the general case omitted here? Group axioms in the following post.

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>>9019641

>>9019617
Let n be fixed and do induction on m.
Base case m=1, you know [math] x^{n+1} = xx^n [/math].

Induction step: [math] x^{n+m+1} = x^{n+m}x = x^n x^m x = x^n x^{m+1} [/math] by applying the induction hypothesis in the third equality.

Let [math]m \,\in\, \mathbf Z[/math]. Let us show that [math]\forall n \,\in\, \mathbf N,\, x^{m \,+\, n} \,=\, x^m \, x^n[/math].

Initialization:
[eqn]x^{m \,+\, 0} \,=\, x^m \,=\, x^m \,1 \,=\, x^m \, x^0[/eqn]

Induction: Let [math]n \,\in\, \mathbf N[/math] such that [math]x^{m \,+\, n} \,=\, x^m \, x^n[/math].
[eqn]x^{m \,+\, \left( n \,+\, 1\right)} \,=\, x^{\left( m \,+\, 1\right) \,+\, n} \,=\, x^{m \,+\, 1} \, x^n \,=\, x^m \, x \, x^n \,=\, x^m \, x^{n \,+\, 1}[/eqn]

To extend the result to [math]n \,\in\, \mathbf Z[/math], note that for all [math]n \,\in\, \mathbf N[/math], using the previous result,
[eqn]x^{m \,-\, n}\, x^n \,=\, x^{m \,-\, n \,+\, n} \,=\, x^m \qquad\text{then}\qquad x^{m \,-\, n} \,=\, x^m \, x^{-n}.[/eqn]

>>9019650
Great.

>>9019650
>>9019672
I'd like to point out one thing in the inductive step that some brainlet like me might miss when reading the proof.

In the inductive step we use [math]x^{(n+m)+1}=x^{n+m}x^{1}[/math].
This is the case base. Since [math]x^{n+1}=x^{n}x^{1}[/math] is true for whatever integer [math]n[/math],
it remains true if we substitute [math]n[/math] with [math]n+m[/math].

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>>9019697
Cropping since latex fucked up.

>>9019697
>>9019699
If you say you wrote this because a brainlet like you might miss it, but a brainlet like you did not miss it, then why did you write it?

>>9019703
Because I overlooked it the fist time I read your proof.
I wrote it down for reference so I don't miss it again (I'm saving the thread) and so that anyone else willing to read the post can understand it fully.

>>9019641
Artin

>>9019668
Thank you.
When writing [math]xx^{n}=x^{n+1}[/math] in the inductive step,
you're using the induction hypothesis [math]x^{m+n}=x^{m}x^{n}[/math] with [math]m=1[/math], right?

>>9019728
Thanks for the reference.

>>9019733
You can either define integer powers inductively (in which case [math]x\, x^n \,=\, x^{n \,+\, 1}[/math] by definition) or add the [math]n \,=\, 1[/math] case in the initialization.

>>9019739
>You can either define integer powers inductively
Nice.
I have this feeling that sometimes Lang makes things more difficult to understand.