
abstract algebra problem
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.

Hint : if $\displaystyle S,T$ are finite sets with the same cardinality, then an injection $\displaystyle S \to T$ is also a surjection. Now consider the inclusion $\displaystyle <c> \to <a>$...

Sure, and clearly <c>=<a>= d since c=a=d. But how does this shows that everything that is in <a> is also in <c>?