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Math Help - Group of order pq must have subgroups of order p and q.

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    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.
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  2. #2
    Super Member Matt Westwood's Avatar
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    If you want to be flash, you can say it follows directly from the first Sylow theorem:

    First Sylow Theorem - ProofWiki
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    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.
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    Quote Originally Posted by ENRIQUESTEFANINI View Post
    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.
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    Quote Originally Posted by NonCommAlg View Post
    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.
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    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 |<g^p>|=q and |<g^q>|=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...
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    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.
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    Quote Originally Posted by Swlabr View Post
    This may use some big result...
    This uses Cauchy's theorem which is a special case to prove the general first Sylow theorem.
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    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.
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    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by ENRIQUESTEFANINI View Post
    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.
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  11. #11
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    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.
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    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    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.
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    Quote Originally Posted by Swlabr View Post
    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".
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  14. #14
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    That is a nice approach.
    Thanks.
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  15. #15
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    Thank you very much, swlabr.
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