I am having troubles proving this:
Prove that any group on five elements is cyclic.
Lemma: If a set has no proper non-trivial subgroups, then it's cyclic.
Proof: Let meet the criterion. Let be arbitrary. We must then have that is a subgroup of that is non-trivial. Thus, by assumption we must have that is improper, thus equal to . The conclusion follows.
Now, using this lemma and Lagrange's theorem we see that any subgroup of where for some prime must be a divisor of . Thus, any subgroup must be of order . Therefore, the lemma implies that is cyclic for any prime-ordered group.
Your example is just a corollary.
I'm supposed to prove the statement by arguing that all the non-identity elements of G have order 5.
So far I have
Let such that .
if , then we are done, since .
Then and where .
Since they are all contradictory, x cannot have order 4. So no element in G can have order 4.
But I can't figure out the other cases
Not being a mathematician, I don't get it. Would you mind giving me the basic premise of the mathematical 5 elements concept?
In Taoism, any five cyclical system is represented by wood, fire, earth, metal, and water. How each supports each other is obvious - back down to water, the fifth element supporting wood, the first element all over again. This characterization is supposed to be a perfect analogy for anything set of five that are mutually supportive. I found a fuller description of the 5 elements