1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help! Group Theory Question(s)

  1. Aug 2, 2007 #1
    1. The problem statement, all variables and given/known data

    Take G to be the cyclic group with 12 elements. Find an element g in G such that the equation x^2 = g has no solution.

    2. Relevant equations

    Notation: Z = set of integers.

    A group is said to be commutative or Abelian if the operation * satisfies the commutative law, that is, if for all g and h in G we have g*h=h*g.

    Some quotes from my textbook with relevant information:

    "Let G be a group and let g be an element of G. The set <g> = {g^n: n in Z} of all distinct powers of g is a subgroup, known as the subgroup generated by g. It has n elements if g has order n and it is infinite if g has infinite order.

    A group of the above type, that is, of the form <g> for some element g in it, is said to be cyclic, generated by g.

    Remark: It follows from the above theorem that a cyclic group is Abelian."

    3. The attempt at a solution

    I'm really confused with this entire chapter. It says G is cyclic with 12 elements, so does it look like this?

    G = {g^0, g^1, ... , g^10, g^11},
    such that g^12 = g^0, g^13 = g^1, etc. (?)

    But what is the question asking for? And how does it follow that a cyclic group is Abelian?
  2. jcsd
  3. Aug 2, 2007 #2


    User Avatar
    Homework Helper

    there's a bunch of difference g's that work. Start squaring all of the elements in G. Start with g^0=e, square it. what do you get? Next go to g. square it. What do you get? Next go g^2. Square it. what do you get? Do this for all 12 of the elements in G. What do you find?
  4. Aug 3, 2007 #3
    The most natural group to look at would be Z_12, as all cyclic groups with with order 12 are necessarily isomorphic to it. Hence, if you can show that Z_12 has such a property then all cyclic groups of order 12 also inherit this property.

    x^2 in Z_12 is really just adding twice and then reducing modulo 12. So if x^2=g then 2x=g+12k for some k. Can you find a g in {0,1,2,...11} in which no solutions exists? Why won't it work?

    In reality, we really don't need to work in Z_12. Since G has order 12 the order of the generator, say p, is 12. Hence, p^(12k+n)=p^12kp^n=p^n. Since x is in G, we have x=(p^m) for some m so that x^2=p^2m. Since g is in G we also have g=p^n, as g generates G. So, supposing x^2=g would mean p^(2m)=p^n. Therefore, 2m=n (mod 12). And it amounts to finding an n such that there is no solution. If you note, the two equations are precisely the same due to the isomorphism between G and Z_12.
    Last edited: Aug 3, 2007
  5. Aug 3, 2007 #4


    User Avatar
    Science Advisor
    Homework Helper

    Yes it does. The element g - whatever it is - is called the generator and g^0 is the unit element of the group. Think about g as being the natural number 1, and let the operation on the group be addition. Then if you calculate modulo 12, you get a cyclic group of order 12, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} and 11 + 1 = 0 again. Actually, you've done this for years already... when reading a clock! And if you know what an isomorphism is: any cyclic group of order 12 is isomorphic to this group (basically; it's the only cyclic group of order 12; if you don't know about isomorphisms yet, forget that).

    Can you find an element, y of the group such that there is no x in the group for which x^2 = y. Because the group is given explicitly, you have extra information you can use. For example, write x = g^n and y = g^m for some numbers n, m = 0, 1, 2, ..., 10, 11.

    Try proving this. Take any x, y in the group and try to show that xy = yx. Again, use the properties of this particular group.
  6. Aug 3, 2007 #5
    Find [tex]S=\{x^2(\bmod 12)\}[/tex] for [tex]x=0,1,...,11[/tex]. Now take the complement of [tex]S[/tex] under [tex]\mathbb{Z}_{12}[/tex] to get your answer.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Help! Group Theory Question(s)
  1. Group theory. Help (Replies: 1)

  2. Group theory question (Replies: 3)