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Math Help - tricky group proof

  1. #1
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    tricky group proof

    Show that any group of order 10 is either cyclic (so isomorphic to
    Z10),
    or is isomorphic to the dihedral group
    D5 of order 10 (symmetries of a regular

    pentagon).

    I have no idea how to even get started on this
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  2. #2
    Super Member TheChaz's Avatar
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    Just a thought from the logical side of things. You have:

    |G| = 10 \Rightarrow (G \cong \mathbb{Z}_{10} \vee  G \cong D_5)

    So we can start by assuming that G is NOT isomorphic to Z10. This is a common approach to proofs of the form P -> (Q v R), where "v" is the logical operator "or".

    In other words, not being Z10 means no elements have order ten. But the order of an element must divide the order of the group, so look for elements of orders 5 and 2.
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  3. #3
    Senior Member Tinyboss's Avatar
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    Have you studied Sylow's theorems, semidirect products, and the fundamental theorem of finitely-generated abelian groups yet? Those are the usual way to attack this kind of problem, although when the order is this small, you can usually make more basic arguments.

    For instance, if the group is abelian it's cyclic, since 2 and 5 are coprime. You also know (in any case) that there's a normal subgroup of order 5 (Cauchy's theorem gives you the existence of an order-5 element, and the cyclic subgroup generated by it has index 2 in G, so it's normal).
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  4. #4
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    Let G be a group with 10 elements. The number of Sylow 5-subgroups must divide 10 (the possibilities are 1, 2, 5, 10) and be congruent to 1 modulo 5. Thus there is only one Sylow 5-subgroup, P, which in thus a normal subgroup of G. Since the order of P is 5, P is cyclic, say P = <x>. G also has at least one Sylow 2-subgroup, so let y be an element of order 2. Thus the elements of G are \{e, x,x^2,x^3,x^4,y,yx,yx^2,yx^3,yx^4\}. Since P is normal, yxy^{-1} is in P, and so is of the form x^i for some i.

    If i = -1, then yx = x^{-1}y, and G is isomorphic to

    D(5) = <x, y | x^5 = e = y^2, yx = x^{-1}y>.

    Assume i \ne -1. Since y^2 = e, (yx)^2 = yxy^{-1}x = x^{i+1}. So the even powers of yx are powers of x, whereas the odd powers of yx are of the form yx^j for some j. The order of yx divides 10, and so is 1, 2, 5, or 10. We see that yx is not of order 1 (otherwise x = y), itís not of order 2 (since (yx)^2 = x^{i+1}), and itís not of order 5 (since its 5th power is yx^j for some j). Thus yx has order 10, and G is cyclic (thus, G is abelian, and yx = xy).
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by poirot View Post
    Show that any group of order 10 is either cyclic (so isomorphic to
    Z10),
    or is isomorphic to the dihedral group
    D5 of order 10 (symmetries of a regular

    pentagon).

    I have no idea how to even get started on this
    People have given you a general idea (one of a few) ways to prove this. As a point of interest it's true that every group of order 2p where p is prime is either cyclic or dihedral.
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  6. #6
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    Quote Originally Posted by Tinyboss View Post
    Have you studied Sylow's theorems, semidirect products, and the fundamental theorem of finitely-generated abelian groups yet? Those are the usual way to attack this kind of problem, although when the order is this small, you can usually make more basic arguments.

    For instance, if the group is abelian it's cyclic, since 2 and 5 are coprime. You also know (in any case) that there's a normal subgroup of order 5 (Cauchy's theorem gives you the existence of an order-5 element, and the cyclic subgroup generated by it has index 2 in G, so it's normal).
    No we haven't studied any of those theorems. Whats that result about abelian and co prime? Its irrelevant but just interested.
    Anyway, I was thinking of proof along these lines.
    By lagrange's theorem, elements (except identity) must have orders 2,5 or 10. If there is an element order 10, it is cyclic. If not such and such must be true. But I am not given any information on the elements.
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