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I have seen a couple of proofs online for problem #1.1.19 in Dummit and Foote's Abstract Algebra that use strong induction to prove the following claim;

Let $x\in G$ where $G$ is a group. Let $a,b\in \mathbb{Z}^+$. Prove $$x^{a+b}=x^ax^b$$

I have not seen any proofs of the following form:

$$x^{a+b}=\underbrace{x\cdot x\cdots x}_{a+b\text{ times}}=\underbrace{x\cdot x\cdots x}_{a\text{ times}}\cdot \underbrace{x\cdot x\cdots x}_{b\text{ times}}=x^ax^b$$

I would assume that with positive integer exponents, this would be the most obvious way to show this is true. So why the need for induction?

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  • $\begingroup$ Because, it's considered too informal. $\endgroup$ – quasi May 24 '17 at 13:19
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    $\begingroup$ Whenever you use dots ($\cdots$), you're likely using induction, just informally. $\endgroup$ – Eff May 24 '17 at 13:20
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    $\begingroup$ Here's an important technicality: but how do you prove that $$ \underbrace{x\cdot x\cdots x}_{a+b\text{ times}}=\underbrace{x\cdot x\cdots x}_{a\text{ times}}\cdot \underbrace{x\cdot x\cdots x}_{b\text{ times}}? $$ $\endgroup$ – Omnomnomnom May 24 '17 at 13:22
  • $\begingroup$ This $(⋯)$Does not make sense in math. $\endgroup$ – Motaka May 24 '17 at 13:23
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    $\begingroup$ @AveryJessup you're assuming that the only piece of information that matters is the number of elements $x$ which are being multiplied. Implicitly, you're moving around a lot of parentheses: $$ \underbrace{(((x\cdot x)\cdots )x)}_{a+b\text{ times}}=\underbrace{(((x\cdot x)\cdots )x)}_{a\text{ times}}\cdot \underbrace{(((x\cdot x)\cdots )x)}_{b\text{ times}} $$ $\endgroup$ – Omnomnomnom May 24 '17 at 13:26
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Note that multiplication is defined as a binary operator, not as something you do with a bunch of copies of $x$. As such, when you declare that $$ \underbrace{x\cdot x\cdots x}_{a+b\text{ times}}=\underbrace{x\cdot x\cdots x}_{a\text{ times}}\cdot \underbrace{x\cdot x\cdots x}_{b\text{ times}} $$ you're assuming that the only piece of information that matters is the number of elements x which are being multiplied. That is, you're implicitly moving around a lot of parentheses in the equation $$ \underbrace{(((x\cdot x)\cdots )x)}_{a+b\text{ times}}=\underbrace{(((x\cdot x)\cdots )x)}_{a\text{ times}}\cdot \underbrace{(((x\cdot x)\cdots )x)}_{b\text{ times}} $$ You'll notice that the inductive step in the proof only requires that we deal with one reordering at a time, as in $$ x^{a+1} \cdot x^b = (x \cdot x^a) \cdot x^b = x \cdot (x^a \cdot x^b) = x \cdot x^{a + b} = x^{a + b + 1} $$

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Because, it's considered too informal. To be formal, one needs to use the definition of $x^n$, and the definition is not the product of $n$ copies of $x$.

Since for nonnegative integers $n$, the definition of $x^n$ is an inductive definition, a formal proof of that identity is pretty much forced to use induction.

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  • $\begingroup$ If the definition of $x^n$ was repeated multiplication, would the proof in the OP still be considered too informal? $\endgroup$ – Ovi May 24 '17 at 13:26
  • $\begingroup$ Such a definition would itself lack formality. The whole point of the inductive definition was to make the meaning rigorous. $\endgroup$ – quasi May 24 '17 at 13:28
  • $\begingroup$ Okay. I think I understand $\endgroup$ – AveryJessup May 24 '17 at 13:29
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    $\begingroup$ Still, when we work with powers, we don't think inductively -- that's too primitive. We revert conceptually to the informal "repeated multiplication" version. And once the basic identities are proved, the inductive definition is essentially no longer needed. $\endgroup$ – quasi May 24 '17 at 13:31

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