Exercise 1.1.32 ($|x| \leq G$)

Proof. Assume for the sake of contradiction that there are integers $i,j$ such that

Then $j-i>0$ and $x^{j-i}=1$. By definition of the order, $n\le j-i$, but $j-i<j\le n$, so $j-i<n$. This is a contradiction, which proves that the elements $1,x,x^2,\dots ,x^{n-1}$ are all distinct.

This shows that the set $\{1,x,x^2,\dots ,x^{n-1}\}$ has $n$ elements, and $\{1,x,x^2,\dots ,x^{n-1}\}\subseteq G$, therefore

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