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Math Help - Need Permutation help...

  1. #1
    Junior Member ginafara's Avatar
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    Need Permutation help...

    Ok, so now matter how much I try and figure this out, I am lost...

    What is an example of a cyclic subgroup of order 4 in S4?

    is it just S4 itself??? My text is horrible when it comes to examples and explanation...

    Also, how would I be able to find out the max order of any element in an alternate group?

    TIA
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  2. #2
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    Quote Originally Posted by ginafara View Post
    Ok, so now matter how much I try and figure this out, I am lost...

    What is an example of a cyclic subgroup of order 4 in S4?

    is it just S4 itself??? My text is horrible when it comes to examples and explanation...

    Also, how would I be able to find out the max order of any element in an alternate group?

    TIA
    S4 is not cyclic because it's not generated by any single element. A good cue is that all cyclic groups are abelian -- S4 is not abelian, so can't be cyclic.
    S4 also has 24 elements, because Sn has n! elements for any n. But you want a subgroup with only 4 elements, so we can't be talking about S4.

    You need a group generated by an element that has order 4.

    Try the group generated by the element <(a b c d)>.
    I hope you are familiar with cycle notation.

    (a b c d)^1 = (a b c d) (obviously)
    (a b c d)^2 = (a b c d)(a b c d) = (a c)(b d)
    (a b c d)^3 = (a b c d)(a b c d)^2 = (a b c d)(a c)(b d) = (a d c b)
    (a b c d)^4 = (a b c d)(a b c d)^3 = (a b c d)(a d c b) = (I), the identity element.

    So this group has four elements in it, and it's cyclic. Of course its contained in S4, so it'll do for what you're looking for.
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  3. #3
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    The order of a generator is equal to the order of the cyclic group. So pick a cycle that has order 4. Now the order of a cycle is equal to its length. So try \left< (1,2,3,4) \right>
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  4. #4
    Junior Member ginafara's Avatar
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    Ok...that makes sense...now what about a non-cyclic subgroup of order 4...there are none right?

    I went through all of the 4 cycles I could think of they all give a similar result and then 3 cycles have order 3 and 2 order 2...etc.

    sorry for my ignorance but this material is hurting my very applied brain.

    TIA
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  5. #5
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    Quote Originally Posted by ginafara View Post
    Ok...that makes sense...now what about a non-cyclic subgroup of order 4...there are none right?

    I went through all of the 4 cycles I could think of they all give a similar result and then 3 cycles have order 3 and 2 order 2...etc.

    sorry for my ignorance but this material is hurting my very applied brain.

    TIA
    No, you can build a non-cyclic subgroup of order 4.
    For instance, look at the set H containing the cycles { (1) , (1 2)(3 4) , (1 3)(2 4) , (1 4)(2 3) }.

    It's a subgroup because it's finite and the product of the elements is still in H.
    (1 2)(3 4) * (1 3)(2 4) = (1 4)(2 3);
    (1 2)(3 4) * (1 4)(2 3) = (1 3)(2 4);
    ..and so on. I work out cycle composition from right to left (just so you know), but in this case it doesn't matter since all groups of order 4 are abelian.

    but it's not cyclic because every element (but the identity) has order 2, while the subgroup itself has order 4.

    And don't worry about ignorance. Constructing subgroups from scratch isn't easy until you learn about group structure itself, which typically comes a bit later.
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