# Thread: Group of order pq must have subgroups of order p and q.

1. ## Group of order pq must have subgroups of order p and q.

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.

2. If you want to be flash, you can say it follows directly from the first Sylow theorem:

First Sylow Theorem - ProofWiki

3. But I don't want to use such a powerful tool. There must be a simple proof which uses the fact that p <> q. Thank you for your reply.

4. Originally Posted by ENRIQUESTEFANINI
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.
suppose that every non-trivial proper subgroup of $G$ has order $p$ (respectively $q$). then, from the class equation, we'll get that $|Z(G)|$ is divisible by $q$ (respectively $p$). so $|Z(G)|=pq,$ i.e. $G$ is

abelian. let $H, K$ be two distinct subgroups of $G$ of order $p$ (respectively $q$). then $HK$ will be a subgroup of order $p^2$ (respectively $q^2$). contradiction! Q.E.D.

5. Originally Posted by NonCommAlg
suppose that every non-trivial proper subgroup of $G$ has order $p$ (respectively $q$). then, from the class equation, we'll get that $|Z(G)|$ is divisible by $q$ (respectively $p$). so $|Z(G)|=pq,$ i.e. $G$ is
Can you do this without using the class equation? When I was responding to this I started using the class equation but then figured that this is complete overkill.

6. Originally Posted by ThePerfectHacker
Can you do this without using the class equation? When I was responding to this I started using the class equation but then figured that this is complete overkill.
Assume the group is cyclic, then there is an element of order $pq$, call it $g$. Then $||=q$ and $||=p$ and we are done.

Else every element has order properly dividing $pq$. Assume they all have order $p$, then the group is a finite $p$-group and we know every finite $p$-group has order $p^n$ (as $p$ is prime), a contradiction. Assume they all have order $q$, then the group is a finite $q$-group and we know every finite $q$-group has order $q^n$ (as $q$ is prime), a contradiction.

Thus, there must exist elements of order $p$ AND elements of order $q$, and we are done.

Of course, I can't quite remember how we prove that every finite $p$-group has order $p^n$. This may use some big result...

7. I insist there must be a proof which uses neither Sylow's theorems, nor Cauchy's theorem that p / |G| implies G has an element of order p nor the class equation. Neither that a p-group is of order a power of p.

8. Originally Posted by Swlabr
This may use some big result...
This uses Cauchy's theorem which is a special case to prove the general first Sylow theorem.

9. 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.

10. Originally Posted by ENRIQUESTEFANINI
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.

11. But you are still using Cauchy's theorem. I thought maybe this is doable without any of that even though that is a simplified proof.

12. Originally Posted by ThePerfectHacker
This uses Cauchy's theorem which is a special case to prove the general first Sylow theorem.
Chauchy's Theorem is used to prove that finite groups with elements of order a power of $p$ are of order $p^n$. However, in this case the elements are of order $p$. This is a simpler case, and we can use the formula,

$|HK|=|H|.|K|/|H \cap K|$

That is to say, if we look at the problem in my proof each element has order $p$. As each element has order $p$ it will generate $p-2$ other elements of order $p$ and so if we take a minimum generating set, $P_1, P_2, \ldots, P_n$, the intersection of each pair is trivial, $G=P_1P_2 \ldots P_n$ and the formula can be applied. Thus,

$|G|=|P_1|.|P_2| \ldots |P_n| = p^n \neq pq$

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.

13. Originally Posted by Swlabr
Chauchy's Theorem is used to prove that finite groups with elements of order a power of $p$ are of order $p^n$. However, in this case the elements are of order $p$. This is a simpler case, and we can use the formula,

$|HK|=|H|.|K|/|H \cap K|$

That is to say, if we look at the problem in my proof each element has order $p$. As each element has order $p$ it will generate $p-2$ other elements of order $p$ and so if we take a minimum generating set, $P_1, P_2, \ldots, P_n$, the intersection of each pair is trivial, $G=P_1P_2 \ldots P_n$ and the formula can be applied. Thus,

$|G|=|P_1|.|P_2| \ldots |P_n| = p^n \neq pq$

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.
That is a nice approach. Yes, I know the formula and since
the formula can be proven combinatorically it is "elementary".

14. Originally Posted by ThePerfectHacker
That is a nice approach.
Thanks.

15. Thank you very much, swlabr.

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