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Math Help - Subgroups involving disjoint cycle notation.

  1. #1
    Super Member Showcase_22's Avatar
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    Subgroups involving disjoint cycle notation.

    This is a question from a practice paper which I have an exam on the first day back. I just wanted to run it by someone else to see if what I was doing was correct.

    Which, if any, of the following subsets of S_4 are subgroups? Give brief reasons for your answers.

    i). {i, (1 2)}
    ii). {(1 2 3), (1 3 2)}
    iii). {i, (1 2)(3 4), (1 4)(2 3)}
    iv). {i, (1 2), (1 4), (2 4), (1 4 2), (1 2 4)}.
    I structured my entire answer on the properties of a group:
    1). Closed
    2). Associativity
    3). The existence of a neutral element
    4). The presence of inverses.

    i). Is not associative since starting at 1, there is no path to 3 or 4 (The definition of associativity states that " \forall g_1, g_2 \in G \exists g_3 \in G s.t g_1g_2=g_3". Here the " \forall g_1, g_2" statement fails.).
    All the other properties hold.
    Therefore i) is not a subgroup of S_4.

    ii). For this case, all the properties hold except for the presence of a neutral element. Starting at any of 1,2,3 or 4 there is no path that takes either of these numbers back to themselves without stopping at another number first.
    Hence ii). is not a subgroup of S_4.

    iii). All the properties hold for this set.
    Therefore iii). is a subgroup of S_4.

    iv). All the properties once again hold except for associativity. Starting from 1,2 or 4 it is impossible to get to 3.
    Therefore iv). is not a subgroup.

    I think I have done this correctly but my answers to i) and iv) are the same. There is also only one property for a subgroup failing each time.

    I would appreciate it if someone could check these answers and see if it is what they would put.
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  2. #2
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    i) Why would you check a subgroup os associativity ? and why are you complicating the definition of associativity ? I learnt it as a \circ (b \circ c ) =( a \circ b ) \circ c  and a subgroup will trivially inherit the associatively property.

    also (12) is short hand for (12)(3)(4).


    for ii) there is no identity, so move on, don't bother checking for everything when one property clear fails.

    iii) did you prove it was closed ?

    iv) refer to i)


    Bobak
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  3. #3
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    Quote Originally Posted by bobak View Post
    i) Why would you check a subgroup os associativity ? and why are you complicating the definition of associativity ? I learnt it as a \circ (b \circ c ) =( a \circ b ) \circ c  and a subgroup will trivially inherit the associatively property.
    I didn't know that. We were taught to check each part separately.

    for ii) there is no identity, so move on, don't bother checking for everything when one property clear fails.
    I wanted to see if it had more than property failing. I'm not sure how "brief" the reasons would have to be.

    iii) did you prove it was closed ?
    I drew the cycle out and saw that all the points were connected and there were no incomplete cycles. I would have put my diagram on here as well but these library computers don't seem to have paint on them. =S
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    you don't need to check associativity as it is inherited from the group. ASSOCIATIVITY IS GUARATEED, bcause the subset is under the same operation as the group.

    i) is a subgroup. I don't understand your reason. As I said associativity does not need to be checked.

    ii) Bobak is right, if there is no identity, it cannot be a subgroup. It certainly has no closure if there is no identity because for any element g, g^n where n is the cardinality of the group is e, by Lagrange's theorem.

    iii) is not closed... multiply 2

    iv) again, wrong reason, associativity is always present. It is not closed if you check say (1,2) and (1,4). I don't know what you are on about as not being able to get to 3. 3 remain unchaged under any of these permutations, which is not a problem at all...
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  5. #5
    Super Member Showcase_22's Avatar
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    Quote Originally Posted by lordcrusade9 View Post
    you don't need to check associativity as it is inherited from the group. ASSOCIATIVITY IS GUARATEED, bcause the subset is under the same operation as the group.
    hmmm, that's a very good point!

    i) is a subgroup. I don't understand your reason. As I said associativity does not need to be checked.
    Okay, I understand this now.

    ii) Bobak is right, if there is no identity, it cannot be a subgroup. It certainly has no closure if there is no identity because for any element g, g^n where n is the cardinality of the group is e, by Lagrange's theorem.
    I understand this bit too! Since there is no e then g^n is not equal to anything so n cannot be calculated. Is that about right?

    iii) is not closed... multiply 2
    What do you mean by this? I thought all these groups would be closed....

    iv) again, wrong reason, associativity is always present. It is not closed if you check say (1,2) and (1,4). I don't know what you are on about as not being able to get to 3. 3 remain unchaged under any of these permutations, which is not a problem at all...
    Why isn't it closed for (1,2) and (1,4)?

    I do understand what you mean about the number 3 though.
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  6. #6
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    sorry for 3, I didn't finish the message. If you multiply the non identity elements.

    You start with

    1,2,3,4

    applying (1 2)(3 4)
    applying (1 4)(3 2)

    1-> 2 -> 3
    2-> 1 -> 4
    3-> 4 -> 1
    4-> 3 -> 2

    (1 3) (2 4) not in the group


    for ii, it is not g^n is not equal to anything, you mean g^n is not in the subgroup. I am not sure what you mean n cannot be calculated. It is always a factor of the cardinality of the group, so g^|G| is always identity, hence it must always be in the subgroup.
    Last edited by lordcrusade9; January 1st 2009 at 08:35 AM.
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  7. #7
    Super Member Showcase_22's Avatar
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    I was just reviewing this today but (2 4 1) is in the group, it's just written (1 2 4) instead.
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  8. #8
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    Wtf are you on...
    so (1,2,4) (2,4,1) (4,1,2) are the same thing?
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  9. #9
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    sorry my bad

    I just realised I made an epic error.
    The last one is actually a subgroup. I think it is actually S3
    Last edited by lordcrusade9; January 1st 2009 at 08:36 AM.
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