# abstract algebra problem

• Sep 17th 2010, 11:00 AM
santiagos11
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.
• Sep 17th 2010, 11:16 AM
Bruno J.
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 $\displaystyle <c> \to <a>$...
• Sep 17th 2010, 04:30 PM
santiagos11
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>?