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Math Help - Group: order 35

  1. #1
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    Group: order 35



    Having troubles with this, i know that 35 = 7x5, which are both prime and subsequently 5 doesn't divide 6, so G is cyclic.

    But I don't understand how to show there are elements of order 5 and 7 exactly, and that it's commutative, i.e abelian. Thanks in advance.
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  2. #2
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    Hi

    I can't see how you deduce that G is cyclic from what you've written before.

    Lagrange's theorem (or one of his corollary): in a finite group, the order of an element divides the order of the group.
    (For what follows, I assume you don't use Sylow's theorems.)

    So the elements of G have order 1, 5, 7 or 35. The only of order 1 is the identity element e. Can the others all be of order 5? We want to show that no.

    Let x be an element of order G. The subgroup generated by x,\ <x>, has five elements, and four different from e. Using Lagrange's theorem, what are their order?
    Now, do you agree that in a finite group of order n, an element of order n is a generator of the group?
    So let x,y be two elements of order 5 in G, what if <x>\cap <y>\neq \{e\} (i.e. they have a common element)?

    Therefore, if all elements in G-\{e\} have order 5, all cyclic subgroups <x> have their intersections equal to \{e\}, and the number of elements in G-\{e\} is a multiple of 4, let's say 4k where k is a positive integer. Is 4k+1=35 possible? No? Then contradiction, and all elements can't have order 5.
    Same thing with 7.
    Finally, if some element a\in G has order 35 (i.e. is a generator of G), what about a^5 and a^7 ?

    I guess that we've proved that G contains at least an element g of order 5 and an element h of order 7.


    Now, what we want to show is that given a G of order 35, if G is abelian, then it is cyclic.

    So if G is commutative, what about gh order ? Conlude
    Last edited by clic-clac; March 13th 2009 at 01:20 AM. Reason: error in a letter
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  3. #3
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    This is a proof that uses a Sylow theorem.

    Lemma 1. If there is only one Sylow p-subgroup P in G, then P is normal in G.

    By Sylow's theorem, G has subgroups of order 5 and order 7.

    If the order of G is pq, then the number of Sylow q-subgroup is kq+1 and divides pq.
    We see that there exists only one Sylow 5-subgroup and one Sylow 7-subgroup in G. Thus, both are normal in G by lemma 1.

    Since both <5> and <7> are normal in G and intersects only at e in G, G = <5> \times <7> \cong Z_5 \oplus Z_7 . Thus G is abelian. Since Z_5 \oplus Z_7 \cong Z_{35} (see finitely generated abelian group), G is cyclic.
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