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Homework Help: 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


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    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


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    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.
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