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Subgroup Question

  • 17-10-2010 7:15pm
    #1
    Closed Accounts Posts: 7


    What are the subgroups of the symmetric group s3?

    I know for a subgroup the following conditions hold :
    1) the subgroup can not be the null set
    2) for all elements a,b which are elements of the subset then a.b is also an element of the subgroup
    3) there exists an inverse in the subgroup
    4) the identity is the same in both the group and subgroup.

    What I can not understand is does the subgroup have to be of the same form as the group. ie several different permutations?????

    Really stuck here so any help would be appreciated....


Comments

  • Closed Accounts Posts: 6,081 ✭✭✭LeixlipRed


    Basically a subgroup is a subset of the set that your group operation operates on and that all the rules for your group also work for. Rule 3 there should be that each element in your subgroup has an inverse. But other than that you're good to go. So essentially your subgroups will be subsets of S3. There's not that many subsets of S3 that will work (for example, the order will have to divide the order of S3, which is 6, so a subset can only have 1,2,3 or 6(the group itself)).

    One example is {(1)}, the identity element is itself a subgroup of S6. It contains the identity, it's it's own inverse, etc,.


  • Registered Users, Registered Users 2 Posts: 966 ✭✭✭equivariant


    larry low wrote: »

    What I can not understand is does the subgroup have to be of the same form as the group. ie several different permutations?????

    Really stuck here so any help would be appreciated....

    Well, a subgroup is in particular a subset of the big group. So if the big group is a group of permutations then the subgroup is automatically a group of permutations.

    To find all the subgroups of S_3, you might want to think about Lagrange's theorem. What does that theorem tell you about the possible size of a subgroup of S_3? That will narrow down the possibilites that you will have to consider.


  • Closed Accounts Posts: 7 larry low


    Thats great thanks!!! :)


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