5

My textbook doesn't explain this well at all.

I was thinking about how a group follows the axiom that $xx^{-1} = x^{-1}x = 1$, where $x$ is some element of the group, $1$ is the identity and $x^{-1}$ is $x$'s inverse. The book says that the powers of some $x$ work with the binary operation on itself.

For example I think for $(\mathbb{Z} , + )$, $1^5$ would be $1 + 1 + 1 + 1 + 1 = 5$. It then goes to say that you can manipulate exponents as usual... which makes me wonder. Since $xx^{-1} = x^{1-1} = 1 = x^0$, does that mean that an element of a group to the power of $0$ will always be the identity of that group?

Austin Mohr
  • 25,662
Bobby Lee
  • 1,030
  • 2
    @SanathDevalapurkar If your operation is $+$, and you use exponentiation to denote repeated operations, then, yes, $1^5=5$... – apnorton Jan 24 '14 at 01:06
  • If you want the law $a^{m+n}=a^ma^n$ to hold for every pair $(m,n)$ of integers, then you are forced to conclude that $a^{0}=e$ and $a^{-1}$ is the inverse of $a$. – André Nicolas Jan 24 '14 at 01:12
  • I'm sorry about that. –  Jan 24 '14 at 01:19

3 Answers3

5

You've basically given the proof, but maybe seeing it written this way is clearer.

Take any element $x$ from any group $G$. Since $G$ is a group, you can find $x^{-1} \in G$.

On the one hand, the definition of inversion implies $xx^{-1} = 1$ (the identity of $G$).

On the other hand, the definition of exponentiation implies $xx^{-1} = x^{1 - 1} = x^0$.

Comparing the previous two lines of reasoning, we see $x^0 = 1$, since both are equal to $xx^{-1}$.

Austin Mohr
  • 25,662
  • 3
    In other words, the assignment of $x^0 = 1$ plays nice with $x^ax^b = x^{ab}$ for all $a$ and $b$, a desirable property. – Eric Thoma Jan 24 '14 at 01:22
2

The answer is certainly morally correct. But it might be clearer to say that initially $x^0$ has no meaning, as per group axioms. We must define it. If we define it as $id$ then the original identity is a tautology. But the point is that this definition plays nicely with the definition of $x^n$, $n>0$, which I guess was in your book.

Yasha
  • 121
  • https://math.stackexchange.com/questions/4412683/proofs-request-proofs-that-five-exponention-rules-hold-for-positive-real-bases/4412691#4412691 – Ethan Bolker Sep 26 '23 at 19:01
0

It might be useful to give a sharper "answer" than in other answers and comments. Namely, first, naming/notating $x$-inverse as $x^{-1}$ does not actually allow an argument $e=x\cdot x^{-1}=x^{1-1}=x^0$, until we have established our notational conventions about exponents!

The main logical ordering that I know for this is, first, to inductively let $x^n=(x^{n-1})\cdot x$ for $1\le n\in \mathbb Z$. This part is uncontroversial, I think. Then define notation $x^0=e$ and $x^{-n}=$ $x^n$-inverse. Still, there is some checking to be done, that these definitions/notations do satisfy the (expected/desired) "rules of exponents". A few cases, etc.

But, so, then, the convention/definition $x^0=e$ is just a convention that is compatible with other stuff, as several other people noted. It is not innately "mathematically true".

paul garrett
  • 52,465