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

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• Sep 9th 2009, 10:36 PM
ENRIQUESTEFANINI
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.
• Sep 9th 2009, 10:58 PM
Matt Westwood
If you want to be flash, you can say it follows directly from the first Sylow theorem:

First Sylow Theorem - ProofWiki
• Sep 10th 2009, 08:14 AM
ENRIQUESTEFANINI
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.
• Sep 10th 2009, 04:10 PM
NonCommAlg
Quote:

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 \$\displaystyle G\$ has order \$\displaystyle p\$ (respectively \$\displaystyle q\$). then, from the class equation, we'll get that \$\displaystyle |Z(G)|\$ is divisible by \$\displaystyle q\$ (respectively \$\displaystyle p\$). so \$\displaystyle |Z(G)|=pq,\$ i.e. \$\displaystyle G\$ is

abelian. let \$\displaystyle H, K\$ be two distinct subgroups of \$\displaystyle G \$ of order \$\displaystyle p\$ (respectively \$\displaystyle q\$). then \$\displaystyle HK\$ will be a subgroup of order \$\displaystyle p^2\$ (respectively \$\displaystyle q^2\$). contradiction! Q.E.D.
• Sep 10th 2009, 06:56 PM
ThePerfectHacker
Quote:

Originally Posted by NonCommAlg
suppose that every non-trivial proper subgroup of \$\displaystyle G\$ has order \$\displaystyle p\$ (respectively \$\displaystyle q\$). then, from the class equation, we'll get that \$\displaystyle |Z(G)|\$ is divisible by \$\displaystyle q\$ (respectively \$\displaystyle p\$). so \$\displaystyle |Z(G)|=pq,\$ i.e. \$\displaystyle 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.
• Sep 11th 2009, 01:38 AM
Swlabr
Quote:

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 \$\displaystyle pq\$, call it \$\displaystyle g\$. Then \$\displaystyle |<g^p>|=q\$ and \$\displaystyle |<g^q>|=p\$ and we are done.

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

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

Of course, I can't quite remember how we prove that every finite \$\displaystyle p\$-group has order \$\displaystyle p^n\$. This may use some big result...
• Sep 11th 2009, 02:37 AM
ENRIQUESTEFANINI
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.
• Sep 11th 2009, 05:12 AM
ThePerfectHacker
Quote:

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.
• Sep 11th 2009, 10:54 AM
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.
• Sep 11th 2009, 11:44 AM
Matt Westwood
Quote:

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.
• Sep 11th 2009, 06:30 PM
ThePerfectHacker
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.
• Sep 12th 2009, 06:52 AM
Swlabr
Quote:

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 \$\displaystyle p\$ are of order \$\displaystyle p^n\$. However, in this case the elements are of order \$\displaystyle p\$. This is a simpler case, and we can use the formula,

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

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

\$\displaystyle |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.
• Sep 12th 2009, 07:22 AM
ThePerfectHacker
Quote:

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

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

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

\$\displaystyle |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".
• Sep 12th 2009, 08:10 AM
Swlabr
Quote:

Originally Posted by ThePerfectHacker
That is a nice approach.

Thanks.
• Sep 13th 2009, 11:17 PM
ENRIQUESTEFANINI
Thank you very much, swlabr.
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