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Math Help - Groups of order p^{e}m contains subgroups of order p^r for every integer r<=e.

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    Groups of order p^{e}m contains subgroups of order p^r for every integer r<=e.

    Is this problem stated correctly? Doesn't p have to be a prime for this to be true?



    Similarly, doesn't p have to be prime?

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    Quote Originally Posted by CropDuster View Post
    Is this problem stated correctly? Doesn't p have to be a prime for this to be true?


    Yes p should be prime. That said, this isn't a trivial trivial theorem per se. That said, unless you know the first Sylow theorem and the fact that p-groups have a subgroup of every order dividing the group.


    Similarly, doesn't p have to be prime?

    Yes, similarly you should assume that p is prime here too. Also, not a trivial exercise.
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    Quote Originally Posted by CropDuster View Post
    Is this problem stated correctly? Doesn't p have to be a prime for this to be true?



    Similarly, doesn't p have to be prime?


    6.4 - Yes, it must be p a prime in the first case:

    A_4\,,\,\,|A_4|=6\cdot 2 , is a counterexample (p=6)


    6.5 - The second question is false, too:

    A_5\,,\,\,|A_5|=30\cdot 2 , is a counterexample (p=30) .

    Even the condition |G|>p is very weird, given the data of the question and,

    of course, it should say "a proper non-trivial subgroup, otherwise {1}

    makes the question itself trivial.

    Or, of course, I'm missing something.

    Tonio
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    Quote Originally Posted by Drexel28 View Post
    Yes p should be prime. That said, this isn't a trivial trivial theorem per se. That said, unless you know the first Sylow theorem and the fact that p-groups have a subgroup of every order dividing the group.



    Yes, similarly you should assume that p is prime here too. Also, not a trivial exercise.
    Ok, so I've been banging my head against the wall here for a while. I can't figure out how to proceed.

    6.4
    I know that there is an element in G that has order p. And I know that there is a Sylow p-subgroup in G, say H\subseteq G such that |H|=p^e. And I also know that the center of H is not trivial. I also know that if x is an element in G, that is of order p then x is a positive power of p. That is, if x^k has order p then k=p^{e-1}, since Ord (x^k)=\frac{|H|}{\gcd\bigl[k,|H|\bigr]}=p. But I just don't know how to proceed from here. Can anyone give me some hints?
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    Quote Originally Posted by CropDuster View Post
    Ok, so I've been banging my head against the wall here for a while. I can't figure out how to proceed.

    6.4
    I know that there is an element in G that has order p. And I know that there is a Sylow p-subgroup in G, say H\subseteq G such that |H|=p^e. And I also know that the center of H is not trivial. I also know that if x is an element in G, that is of order p then x is a positive power of p. That is, if x^k has order p then k=p^{e-1}, since Ord (x^k)=\frac{|H|}{\gcd\bigl[k,|H|\bigr]}=p. But I just don't know how to proceed from here. Can anyone give me some hints?
    Let p^k\| p^e m then by the first Sylow theorem there is a Sylow p-subgroup, say H. But, since |H|=p^k and e\leqslant k one has that (by the fact I said) that H and thus G has a subgroup of every \ell\leqslant k and thus every \ell\leqslant e.


    Are you asking why a p-group has a subgroup of every order dividing it? Try using the fact that the converse of Lagrange's theorem is true for abelian groups (or prove the result is true for abelian p-groups...this is easy), the center of a p-group is non-trivial and inducting on the power of p...this is one of many ways. Ask if you get stuck.
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    I'm sorry, I think I left some ambiguity as to what assumption I can make. According to my text (Artin), a Sylow p-subgroup of a group G, where |G|=p^{e}m is a subgroup of G that has order p^e.

    That said, the way the first Sylow theorem is stated in my book (and the way my class is using it) is as follows:

    A finite group whose order is divisible by a prime p contains a Sylow p-subgroup.

    So I feel like what you are suggesting depends on my knowing that there is a subgroup of order p^k where k\leq e. Based on the way we (my class/professor) are defining the first Sylow theorem, I don't think I can assume this. I realize that some books state the first Sylow theorem differently, and if I could apply that statement, then this problem would be proven just as you have suggested.
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    Quote Originally Posted by CropDuster View Post
    I'm sorry, I think I left some ambiguity as to what assumption I can make. According to my text (Artin), a Sylow p-subgroup of a group G, where |G|=p^{e}m is a subgroup of G that has order p^e.

    That said, the way the first Sylow theorem is stated in my book (and the way my class is using it) is as follows:

    A finite group whose order is divisible by a prime p contains a Sylow p-subgroup.

    So I feel like what you are suggesting depends on my knowing that there is a subgroup of order p^k where k\leq e. Based on the way we (my class/professor) are defining the first Sylow theorem, I don't think I can assume this. I realize that some books state the first Sylow theorem differently, and if I could apply that statement, then this problem would be proven just as you have suggested.
    If you look at my suggestion it's still applicable. What I suggest is using the fact (Sylow's first theorem mine and yours) that a finite group has a subgroup of the maximal power of any prime dividing the order.
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