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Math Help - Groups of Finite Order - IGNORE THIS - MOVED TO ALG.

  1. #1
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    Groups of Finite Order - IGNORE THIS - MOVED TO ALG.

    I have two questions, one of which I am pretty sure I have answered, but I would like your opinion on it.

    1) Let G be a finite group of order 12, is it possible that the center of this group has order 4?
    I am not even sure how to approach this. I did have to look up the definition of the center of a group and I know that the center of a group G is defined as:
    Z(G) = { z \in G | \forall g \in G, zg=gz}.

    2) Suppose that the order of some finite Abelian group G is divisible by 42. Prove that G has a cyclic subgroup of order 42.
    Let G be a finite abelian group of order 42. Remember that the Fundamental Theorem of Finite Abelian Groups states that any finite abelian group can be written as \mathbb{Z}_p_1 \oplus \mathbb{Z}_p_2 \oplus ... \oplus \mathbb{Z}_p_n where p_n are not necessarily distinct and where |G| = p_1 \cdot p_2 \cdot \cdot \cdot p_n. And a corollary follows stating that if m divides the order of a finite abelian group G, then G has a subgroup of order m.

    Now since 42 divides the order of G, and G is abelian then we know that G contains a subgroup of order 42. Let H be such a subgroup. Since G is abelian, H is abelian. Since H has order 42 = 2 \cdot 3 \cdot 7, we see that H \simeq \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_7 \simeq \mathbb{Z}_42. Since \mathbb{Z}_p is cyclic for any prime p, H is cyclic. \blacksquare

    Does that work?
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  2. #2
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    Re: Groups of Finite Order - IGNORE THIS - MOVED TO ALG.

    Posted this in the wrong section... oops... I'll move it to advanced algebra
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