If you want to be flash, you can say it follows directly from the first Sylow theorem:
First Sylow Theorem - ProofWiki
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.
abelian. let be two distinct subgroups of of order (respectively ). then will be a subgroup of order (respectively ). contradiction! Q.E.D.
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.
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.