 Let's talk a little bit more about cyclic groups and to that end we'll introduce a new idea So earlier we use the pigeonhole principle to prove that if we have a group any group at all and any element of that group then we know that a to some power p gives us the Identity element for some p among the natural numbers. So again, remember this is the result of performing the group operation Repeatedly with a and this says that we'll eventually get the identity element and we use the pigeonhole principle to prove that Well, since we know that at least one P exists then we can apply what's called the descent principle So P is a natural number and the descent principle says that anytime you have a set containing one or more natural numbers There has to be a least element So the set of natural numbers for which a to the p is the identity well that contains at least one element because we know that there is some Value p for which that's true that set contains at least one element So there's a least element of that set and we'll designate this p hat And we'll define the order of an element Let G be a group and take anything in that group the order of that element in G is the least value For which we get the identity All right, so let's see where this takes us So let p hat be the order of an element a in G Now if you want to think like a mathematician the question you want to always be asking yourself is well, what now? If p hat is the order of an element then I know a to p hat is the identity and a to the m is not the identity for anything less than p hat So let's think about that well I might consider the terms of the sequence a a squared a cubed and so on all the way up to a to the p hat which is the identity we know that's true and We claim that all elements of this sequence are distinct and the useful proof strategy if you want to claim something is Distinct if you want to claim that things are equal or not equal a common thing to do is to try and prove this by Contradiction we claim all these terms are distinct So let's assume that they're not and that I have two of these terms that are equal to each other So suppose that the terms are not distinct Which means that I have two powers of a that are equal for some I less than j less than p hat Here we've taken the liberty of assuming that I is the power of the first and j is the power of the second of these non-distinct terms well by the same argument as before I can prove that e the identity element is equal to a power j minus i make sure you can prove that and J minus i has to be less than p hat However p hat was the order of our element which means that nothing no power less than p hat can give us the identity element So a to the i can't be equal to a to the j for anything less than p And so it follows it all terms of our sequence have to be distinct Well, here's an even quicker result. Let p hat be the order of a in G Then the sequence forms an abelian subgroup of G and Here's a skeleton of a proof. You should be able to fill in the details We have closure. We want to check to make sure that the product of any two of these things is a term of the sequence And well, you might say oh well That's kind of obvious because this is just a multiplied by itself some number of times However, there's one worry that we have to consider what happens if m plus n is greater than p hat certainly if I multiply two things and the Exponent is less than p hat at someplace in here, but if I multiply two things and the exponent is greater than p hat It's not actually in this sequence. So that's a worry. We have to resolve We want to check to see if there's an inverse well actually That's pretty easy if I have a term in the sequence. What do I have to multiply it by to get a to p hat and Finally we want to establish that it's an abelian subgroup We want to determine whether we have commutativity We want to check to see if a to the m times a to the n does give us a to the n times a to the m And again the worry we have to be careful about here is that if we are Outside of this sequence of terms What are we going to do in that case? But all of these things are relatively easy to prove and we should be able to prove Relatively easily that this sequence will always be an abelian subgroup of any group that we start with even if the group itself is not abelian