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Abstract Algebra mod 6

  1. Feb 16, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm working with a mod 6 addition table.
    I want to compute the subgroups <0>,<1>,<2>,<3>,<4>,<5>
    I also want to find what elements are generators of the group mod 6.
    Then I wnat to use do a subgroup diagram.


    2. Relevant equations



    3. The attempt at a solution
    I am not sure about how to compute sungroups. Doe sthis incolve something like:
    1+1+1 and so on?
    I know an element is a generator if <a>=G. My problem is figuring out which elelments are. Is it when 2 subgroups are equal?
     
  2. jcsd
  3. Feb 16, 2009 #2
    Ok, is there a general way to determine a subgroup from a table. I think if I can figure out one, I'll be fine on the rest.
     
  4. Feb 16, 2009 #3

    Tom Mattson

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    Yes. And when you start getting repeat elements, you're done.

    Just do it! Work out all 6 cyclic subgroups then it will be obvious when two of them are equal.
     
  5. Feb 16, 2009 #4
    What do you mean by repeat elements. So I for
    <0> =0+0=0+0=0
    <1>=1+1=2+1=3+1=4+1=5+1=0+1=1+1=2=<2,3,4,5,0>
    <2>=2+2=4+2=0+2=2+2=4=<4,0,2>
    <3>=3+3=0+3=0
    <4>=4+4=2+4=0+4=4+4=2=<2,0,4>
    <5>=5+5=4+5=3+5=2+5=1+5=0+5=5+5=4=<4,3,2,1,0>
    So,<<0> and <3> are generators. Is what I did correct?
     
  6. Feb 16, 2009 #5

    Tom Mattson

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    I mean that when the elements start to repeat, then stop.

    So for this one, [itex]<0>=\{0\}[/itex]. There's nothing more you can get out of it.

    Your notation is very sloppy. You can't just set all of those things equal to each other like that. You've got 1+1=2+1, which is obviously not true. Plus you didn't take it far enough. 1 should be in that subgroup too. Also you shouldn't list the elements of a subgroup in angled brackets. You should use curly braces instead. So in this case, [itex]<1>=\{0,1,2,3,4,5\}[/itex].

    Bad notation aside, you got the subgroups [itex]<2>[/itex] and [itex]<4>[/itex] correct. You didn't take [itex]<5>[/itex] far enough, and I can't tell if you know what you're doing on [itex]<3>[/itex] because you didn't write your answer down.

    No, neither of those are generators. I suggest you look up the definition of "generator of a group".
     
  7. Feb 16, 2009 #6
    Ok, so I want to make sure, I have every possible number in the subgroup.
    <5>=5+5=4+5=3+5=265=1+5=0+5=5+5=4
    <5>=<4,3,2,1,0,5>

    <3>=3+3=0+3=3+3=0
    <3>=<0,3>

    An element a of a group G generates G and is a generator for G if <a>=G.
    Something about the definition doesn't make sense.

    My thought is that <1>, <5>, <2>, and <4> are generators because they have the same elements.
     
  8. Feb 16, 2009 #7

    Tom Mattson

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    You really must stop doing this. "5+5=4+5" is simply not true.

    Yes, but you should use the curly braces as I said. What you've written here means something else. The notation [itex]<a,b>[/itex] means "the group generated by a and b", not "the set whose elements are a and b."

    Bad notation aside, this is correct.

    It makes perfect sense. Either [itex]<a>=G[/itex] or it doesn't. There's no ambiguity, so it's a fine definition.

    Now that doesn't make sense. [itex]<1>[/itex] isn't a generator, it's a subgroup. The fact that [itex]<1>=G[/itex] means that 1 is a generator. 5 is also a generator, but 2 and 4 are not. Do you see why, in light of the definition of "generator"?
     
  9. Feb 16, 2009 #8
    Ok, I think I know where my confusion is I understand <a> is a subgroup, but I don't understand what the group G is.
    Oh, so G consists of the elements 0,1,2,3,4,5 and <1> and <5> have all these elements, so they are generators?
     
  10. Feb 16, 2009 #9

    Tom Mattson

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    Yes, that's right.

    More precisely, <1> and <5> have all these elements, so 1 and 5 are generators. A generator is an element of a group, not a group itself.
     
  11. Feb 16, 2009 #10
    Ok, that makes sense. I have to make an addition table of these subgroubs.

    Ok, so <1>+<2>= ?
    I'm not sure what to do when the sizes of the groups aren't the same?
    Would <2> +<4>=<0>
     
  12. Feb 16, 2009 #11

    Tom Mattson

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    <1>+<2> doesn't make any sense. You don't add subgroups, you add elements. What exactly were you asked to do?
     
  13. Feb 16, 2009 #12
    Give the subgroup diagram for the subgroups of mod 6.

    I thought this menat creating an addition table.
     
  14. Feb 16, 2009 #13

    Tom Mattson

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    No, it doesn't mean that. Isn't there an example of such a thing in your book?
     
  15. Feb 16, 2009 #14
    Not really:
    My book doesn't explain very well.
    i think I start with 1 and create a diagram connecting with 2, 3, 4, 5, 0?
     
  16. Feb 16, 2009 #15

    Tom Mattson

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    What book are you using?
     
  17. Feb 16, 2009 #16
  18. Feb 16, 2009 #17

    Tom Mattson

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    That's the book I learned from. It explains subgroup diagrams with examples.
     
  19. Feb 16, 2009 #18
    Ok, so I think I have an idea the subgroup <0> stems out from <3>
     
  20. Feb 16, 2009 #19
    Ok, I think I'll get it eventually. It's just a different way of thinking.
     
  21. Feb 16, 2009 #20
    Ok, I figured it out. I was looking at it all wrong.
     
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