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Math Help - How to determine subgroups and generators

  1. #1
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    Exclamation How to determine subgroups and generators

    Hello folks!
    I'm new to this forum and I just want everyone to help me =)

    In our Abstract Algebra class, we are to find for the subgroup(s) of this group:

    G = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14} (this is mod 15)

    I already searched the net and found out that there are trivial and non-trivial subgroups for this group... But still, I don't know how to determine those subgroups.

    One more thing is how to get the generators of the group:
    G = {0,1,2,3,4,5,6,7,8,9} (this is mod 10)

    I don't know what generators are

    Please do help me...
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  2. #2
    Senior Member Dinkydoe's Avatar
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    So I assume we're looking at the G = \mathbb{Z}/15\mathbb{Z} with " +" as operation.

    I don't know what generators are
    Generators of a group are elements that "generate" the group, that is, with these elements you can make any other element of the group.

    Example: In G = (\mathbb{Z}/n\mathbb{Z}, +) the element "1" is allways a generator. Namely any element in G can be written as "1+1+1..+1" etc. Groups may have more then one generator. If a group can be generated by one element we call the group cyclic. Any additive group (\mathbb{Z}/n\mathbb{Z}, +) is cyclic.

    Even though such a group is cyclic, there may exist more then one element in G that can generate the whole group.

    Namely all a\in G with gcd (a,15) = 1 are generators! Proper subgroups are generated by elements a\in G with gcd (a, 15)\neq 1. (why?)

    Thus proper subgroups are: \left\langle 3\right\rangle, \left\langle 6\right\rangle, \left\langle 9 \right\rangle, \left\langle 12 \right\rangle,\left\langle 5 \right\rangle, \left\langle 10\right\rangle

    Now you try the other one: G = (\mathbb{Z}/10\mathbb{Z}, +)
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  3. #3
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    Thanks for a flash reply

    I think, there is no generator for the first group I posted...
    Because zero (0) is one of the elements... which cannot be generated using one (1)...
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  4. #4
    Senior Member Dinkydoe's Avatar
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    Yes it can be generated: 15 = 0

    "0" is the identity element of G

    Don't forget we're looking at residu-classes modulo 15. (with addition "+" as operator, so 14 + 1 = 0)
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  5. #5
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    Aw! Thanks again... very fast response

    so it goes like this:
    1 = 1 mod 15 = 1
    1 + 1 = 2 mod 15 = 2
    1 + 1 + 1 = 3 mod 15 = 3
    .
    .
    .
    1 + 1 + ... + 1 = 14 mod 15 = 14
    1 + 1 + ... + 1 + 1 = 15 mod 15 = 0

    Since all elements (0...14) are generated, then 1 is a generator? Thanks

    About subgroups, I don't know what do <3>, <6>, etc. represent?

    How could I write them as subgroup like this: subgroup1 = { numbers }
    is it like this
    Subgroup1 = { 3 }
    Subgroup2 = { 6 }?
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  6. #6
    Senior Member Dinkydoe's Avatar
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    (1)

    \left\langle 3\right\rangle = \left\{0,3,6,9,12\right\}
    \left\langle 6\right\rangle = \left\{0,6,12,3,9,\right\}
    \left\langle 9\right\rangle = \left\{0,9,3,12,6,\right\}
    \left\langle 12\right\rangle = \left\{0,12,9,6,3\right\}.


    Thus \left\langle 3\right\rangle = <br />
\left\langle 6\right\rangle = \left\langle 9\right\rangle = \left\langle 12\right\rangle

    " \left\langle a\right\rangle " means everything that can be generated by the element " a" under the operation "+" in the group.

    The same way we show that \left\langle 5\right\rangle = \left\langle 10\right\rangle. (a group with 3 elements)

    Thus we have 2 proper subgroups in \mathbb{Z}/15\mathbb{Z}

    And \mathbb{Z}/15\mathbb{Z} = \left\langle 1\right\rangle = \left\langle 2\right\rangle = \left\langle 4 \right\rangle = \left\langle 7 \right\rangle = \left\langle 8 \right\rangle = \left\langle 11\right\rangle = \left\langle 13 \right\rangle = \left\langle 14 \right\rangle
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