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Math Help - subgroups and generators

  1. #1
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    subgroups and generators

    Hi,
    Im really confused with subgroups and generators. May I know what is the difference?

    I was reading a book where they state that the subgroups of Z18 are <0>,<1>,<2>,<3>,<6>,<9> and the generators are 1,5,7,11,13 and 17. what does that mean?

    Thanks!
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  2. #2
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    Hi!

    Firstly let's revise the definition of subgroup:
    Let \langle G, \cdot, ^{-1}, e \rangle be a group ( \cdot is the binary operation, ^{-1} is unary operation of taking an inverse and e is the identity element). If a set H satisfies following conditions:
    1) H \subseteq G
    2) H contains the identity element e of G,
    3) H is closed under the operation ^{-1}, that is, for every a \in H also a^{-1} \in H
    4) H is closed under the operation \cdot, that is, for every a, b \in H also a \cdot b \in H,
    then we say that H is a subgroup of G.

    This means that the set H together with operation \cdot restricted to H and with the identity element e forms a group.

    In Z18, for example \{0, 6, 12\} is a subgroup, or whole Z18 is a subgroup, or \{0\} is a subgroup,... (check it!)

    Then, the definition of a generator of a group:
    Let G be a group and S \subseteq G be a set. We say that S is a generating set of G, if every element of G can be expressed as the product of finitely many elements of S and their inverses. We write this as <S> = G. When there is only one element in S, S = \{x\}, we write <x> = G and we say that the group G is generated by x, or that x is its generator.

    Now to your example. What are the generators of Z18? We see that 1 is surely a generator (every element of Z18 can be expressed as 1+1+..+1)
    Are there any other generators? Yes, exactly those stated in your book, let's see why. Firstly, can you see why the remaining elements are not generators?
    Spoiler:

    Every such element a has a common factor (>1) with 18. No matter how long the sequence a+a+...+a is, the result will always have this common factor with 18. So you can never get the element 1 (which does not have this common factor with 18)! Thus, a doesn't generate G.

    Next, let's verify that every of the elements 1,5,7,11,13 and 17 generates Z18. If you have heard of Bezout's identity, it is easy. It says that if a and b are nonzero integers with greatest common divisor d, then there exist integers x and y such that ax+by=d.
    In our case, take a= one of those 1,5,7,11,13 and 17, take b=18 and, of course, d=1. Can you finish from here?
    Last edited by Taluivren; August 30th 2009 at 07:45 AM.
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  3. #3
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    Quote Originally Posted by alexandrabel90 View Post
    Hi,
    Im really confused with subgroups and generators. May I know what is the difference?

    I was reading a book where they state that the subgroups of Z18 are <0>,<1>,<2>,<3>,<6>,<9> and the generators are 1,5,7,11,13 and 17. what does that mean?

    Thanks!
    <0>={0}
    <1>=<5>=<7>=<11>=<13>=<17>=Z_18 (the whole group)
    <2>={0,2,4,6,8,10,12,14,16}
    <3>={0,3,6,9,12,15}
    <6>={0,6,12}
    <9>={0,9}

    You can check that the above sets satisfy the group axioms under addition operation.

    You can generate a whole group using 1 only ( For instance, 5=1+1+1+1+1, 0 = 1+1+1+..... (18 times 1).

    There is a pattern for a generator 1, 5, 7, 11, 13, 17 of Z_18. They are all coprime with Z_18.

    The number of generators for Z_18 can be acquired by using an Euler totient function.

    \phi(18)=\phi(3)\phi(6)=2 \times 3 = 6. Those are 1, 5, 7, 11, 13, 17.
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  4. #4
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    thank you! i get it now..

    by the way, how do you check that it is closed under -1 and under the operation?

    I can never grasp the concept of something being closed.
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  5. #5
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    You check closure under some operation exactly as noted in 3) and 4) in my first post:

    "3) is closed under the operation , that is, for every also "

    This means if someone gives you an arbitrary element from the subset H, you have to prove that its inverse (which is known to exist and be unique because G is a group) also belongs to the subset H. If the inverse is not in H, then H can't be a subgroup!

    "4) is closed under the operation \cdot that is, for every also "

    This is similar, for arbitrary elements a,b from H you must verify that a \cdot b also belongs to H.

    So in your example we can check that <3>={0,3,6,9,12,15} is closed under taking inverses: they are all multiples of 3 from Z18, inverse of any element a of this set is 18 - a which is also a multiple of 3. And if a, b \in <3> then a+b is also a multiple of 3.
    So 3) and 4) are checked.
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  6. #6
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    thank you for the explanation!(:
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  7. #7
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    you're most welcome
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