I'm finally attempting to conquer D&F (3rd ed) and I want to build good proof habits and fix mistakes early on. Here is the exercise (Ch. 1 ex. 32) and the following is my proof.
Prove that for $x \in G$, where $G$ is a group and $x$ has finite order $n$, all of $1, x, x^2, \ldots, x^{n-1}$ are distinct and deduce that $|x| \leq |G|$.
Proof. Let $a,b \in [0,n-1]\cap \mathbb{N}$ such that $x^a=x^b$. Then $1 = x^ax^{-a}=x^bx^{-a} = x^{b-a}$. Since $1 \leq a \lt n$ and $1 \leq b \lt n$, $b-a\lt n$. So $x^{b-a}=1 \implies b-a=0$ and $a=b$. Since $a$ and $b$ were arbitrarily chosen, $x^c$ is distinct for all $c \in [0,n-1] \cap \mathbb{N}$. We know $x^c\in G$ for all $c \in [0, n-1]\cap \mathbb{N}$ by definition of group closure, and $|[0,n-1]\cap \mathbb{N}|=n$. So $G$ contains at least $n$ elements, and $|x|\leq |G|$.
Is this a convincing proof? I feel like there might be some circular logic, and perhaps not enough detail. I was thinking of establishing that the elements form a cyclic group, and show this group is a subgroup of $G$. Any tips or comments would be greatly appreciated.
Best Answer
Almost.
You assert $b-a < n$. I think you want to assume $b \ge a$ (without loss of generality) from the start, to make $b-a$ nonnegative as well. Otherwise you need to handle the case when it's not.