# Why do you need induction to show that $x^{a+b}=x^ax^b$ if $x\in G$

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?

• Because, it's considered too informal. – quasi May 24 '17 at 13:19
• Whenever you use dots ($\cdots$), you're likely using induction, just informally. – Eff May 24 '17 at 13:20
• Here's an important technicality: but how do you prove that $$\underbrace{x\cdot x\cdots x}_{a+b\text{ times}}=\underbrace{x\cdot x\cdots x}_{a\text{ times}}\cdot \underbrace{x\cdot x\cdots x}_{b\text{ times}}?$$ – Omnomnomnom May 24 '17 at 13:22
• This $(⋯)$Does not make sense in math. – Motaka May 24 '17 at 13:23
• @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: $$\underbrace{(((x\cdot x)\cdots )x)}_{a+b\text{ times}}=\underbrace{(((x\cdot x)\cdots )x)}_{a\text{ times}}\cdot \underbrace{(((x\cdot x)\cdots )x)}_{b\text{ times}}$$ – Omnomnomnom May 24 '17 at 13:26

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 $$x^{a+1} \cdot x^b = (x \cdot x^a) \cdot x^b = x \cdot (x^a \cdot x^b) = x \cdot x^{a + b} = x^{a + b + 1}$$
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.
• If the definition of $x^n$ was repeated multiplication, would the proof in the OP still be considered too informal? – Ovi May 24 '17 at 13:26