Hint : if are finite sets with the same cardinality, then an injection is also a surjection. Now consider the inclusion ...
Let G be a finite group. Let a and b be in G such that a != b.
Suppose |a|=|b|= d so |<a>|=|<b>|, and b is not in <a>.
...
Moreover suppose that c is in G such that |c|= d and c is in both <a> and <b>.
How does this implies that <a> = <c> = <b>
I clearly understand that <c> is contained in both <a> and <b>. But what argument shows that both <a> and <b> are contained in <c>
This is just part of a proof that will later yield the contradiction that b is in <a>. This proof is titled the number of elements of order d in a finite group.