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Thread: Groups of order 27

  1. #1
    Feb 2011

    Groups of order 27


    I am given that G is a finite abelian group that contains
    8 elements of order three
    18 elements of order nine
    and an identity

    And am told to describe of possibilities for G, by giving explicit decomposition into cyclic groups up to an isomorphism

    So it seems like there are three possibilities for G

    $\displaystyle \mathbb{Z}_{27}$
    $\displaystyle \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_3$
    $\displaystyle \mathbb{Z}_9 \times \mathbb{Z}_3$ (is this isomorphic to $\displaystyle \mathbb{Z}_3 \times \mathbb{Z}_9$?)

    Now G cannot be $\displaystyle \mathbb{Z}_{27}$, for G has no element of order 27.

    I also think G cannot be $\displaystyle \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_3$, since no element of that group has order nine. I pretty sure every element of this group must have order less than or equal to 3, since if $\displaystyle (a,b,c) \in$ $\displaystyle \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_3$, then $\displaystyle 3*(a,b,c) = (0,0,0)$

    So I am left with $\displaystyle \mathbb{Z}_9 \times \mathbb{Z}_3$. Is there a quicker way to test the elements of this group besides considering the order of all 27 elements?

    i am pretty sure the element (1,a) \in $\displaystyle \mathbb{Z}_9 \times \mathbb{Z}_3$, would have order 9, but I dont think there's 18 of those.

    Thank you for your time.
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  2. #2
    Senior Member jakncoke's Avatar
    May 2010

    Re: Groups of order 27

    It is easy to see that G is $\displaystyle \cong$ to $\displaystyle Z_3 \oplus Z_9$ for the order of an element of this group if the lcm(a,b), where $\displaystyle a \in Z_3, b \in Z_9$ $\displaystyle Z_9$ has elements of order 9,3,1 (order of elements divides order of group, since cyclic these are the only 3 possible). So since $\displaystyle Z_9$ is cyclic, if has $\displaystyle \phi(9) = 6$ generators, or elements of order 9. So basically 6 * 3, elements of $\displaystyle Z_3 \oplus Z_9$ would be of the form (a,9), lcm(a,9) = 9. so 6*3 = 18 elements of order 9. 1 element of order 1, so 27-19, elements of order 3 (the only other possible order). or 8 elements of order 3.
    Last edited by jakncoke; Feb 21st 2013 at 03:20 PM.
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  3. #3
    MHF Contributor

    Mar 2011

    Re: Groups of order 27

    yes it has to be Z9xZ3 (which is isomorphic to Z3xZ9: it is easy to check that (a,b)-->(b,a) is an isomorphism).

    in general, for an abelian p-group (a finite group whose order is a power of a prime p), which p-group we have can be ascertained by examining how many elements of order p we have (this is particularly useful for p = 2).

    Z27, being cyclic, has just ONE subgroup of order 3, and therefore just TWO elements of order 3.

    Z3xZ3xZ3 has 26 elements of order 3.

    clearly, if (a,b) in Z9xZ3 has order 3, either a has order 3, or b has order 3, or both.

    ONLY a has order 3:


    ONLY b has order 3:


    a and b BOTH have order 3:


    there are your 8 elements of order 3. this is all of them.

    similarly reasoning applies to elements of order 9: a must have order 9, b might be of order 1, or 3 (in other words b can be anything). this gives us φ(9) = 6 choices for a, the 6 generators of Z9:

    1,2,4,5,7 and 8. so explicitly, the 18 elements of order 9 are:


    for example:

    (5,2) + (5,2) + (5,2) = (1,1) + (5,2) = (6,0), which is not the identity, so (5,2) is not of order 3. therefore (5,2) must be of order 9.

    since i have explicitly listed 8 elements of order 3, and 18 elements of order 9, the only element missing is (0,0), the identity. clearly, looking at the number of elements of order 3 was faster.
    Last edited by Deveno; Feb 24th 2013 at 12:30 PM.
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