If you want to be flash, you can say it follows directly from the first Sylow theorem:
First Sylow Theorem - ProofWiki
Hi:
Notation:
o(G) order of group G.
(a) subgroup generated by a.
I have a problem which is stated as follows: let o(G) be pq, where p,q are primes, p > q. Then G has a subgroup of order p and a subgroup of order q.
I have been able to prove G has a subgroup of order q, say (b). Then I have taken a not belonging to (b) and I have seen (a) intersection (b) = (e). What I say is this: if o(a) = p the statement is proved. If o(a) = q,
I say o(ab) = p. But I cannot prove it. Any hint would be welcome. Thanks for reading.
If you want to be flash, you can say it follows directly from the first Sylow theorem:
First Sylow Theorem - ProofWiki
suppose that every non-trivial proper subgroup of has order (respectively ). then, from the class equation, we'll get that is divisible by (respectively ). so i.e. is
abelian. let be two distinct subgroups of of order (respectively ). then will be a subgroup of order (respectively ). contradiction! Q.E.D.
Assume the group is cyclic, then there is an element of order , call it . Then and and we are done.
Else every element has order properly dividing . Assume they all have order , then the group is a finite -group and we know every finite -group has order (as is prime), a contradiction. Assume they all have order , then the group is a finite -group and we know every finite -group has order (as is prime), a contradiction.
Thus, there must exist elements of order AND elements of order , and we are done.
Of course, I can't quite remember how we prove that every finite -group has order . This may use some big result...
In I.N.Herstein, Topics in Algebra I read: "Were Cauchy's theorem our ultimate and only goal, we could prove it, using the barest essentials of group theory, in a few lines [The reader should look up the charming, one-paragraph proof of Cauchy's theorem found by McKay and published in the American Mathematical Monthly, Vol. 66 (1959), page 119.]"
Herstein uses the class equation to prove Cauchy's theorem.
If I had McKay's proof, I could use it to construct a proof of my statement (which is a problem in Herstein's book). Unfortunately, I have no access to the American Mathematical Monthly. But perhaps somebody has.
This proof is given here:
Cauchy's Group Theorem - ProofWiki
I think it can be found in Allan Clark's "Abstract Algebra" but don't quote me cos I'm not sure of my ground on that last statement.
Chauchy's Theorem is used to prove that finite groups with elements of order a power of are of order . However, in this case the elements are of order . This is a simpler case, and we can use the formula,
That is to say, if we look at the problem in my proof each element has order . As each element has order it will generate other elements of order and so if we take a minimum generating set, , the intersection of each pair is trivial, and the formula can be applied. Thus,
and we have the contradiction needed to salvage my earlier proof.
I hope that makes sense...
It should be noted that that formula has an elementary proof, and so I believe that my proof will count as "elementary". I hope.