I'm starting my group theory course and we arrived to the following demonstration:
Let $G$ be a finite group, then, for every $a\in G$, $O(a) \leq |G|$ where $O(a)$ is the order of the element $a$ and $|G|$ is the order of the group
In the demonstration we used the following fact (without any previous treatment or demonstration): because $G$ is finite, there exists $i,j \in \mathbb{N}$ such that $a^i = a^j$; this, naively, comes quick: because the group is finite, operating an element with itself enough times will arrive to the result repeating, because there's only so many elements in $G$.
My question then is: Is there a way of proving that, in a finite group, there exists $i, j \in \mathbb{N}$ such that $a^i = a^j$, I know there must be a way, but I only seem to find convincing arguments. Any tips?
You consider the list $a,a^2,a^3,\ldots$, which are elements of the group. If at no point you have $a^i = a^j$, then all the elements in the list are distinct, so there are infinitely many elements in the group, a contradiction.