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Ring: Proof of Group

  1. Aug 5, 2014 #1
    1. The problem statement, all variables and given/known data
    Let R be the ring of all 2*2 matrices, over Zp, p a prime. Let G be the set of elements x in the ring R such that det x ≠ 0. Prove that G is a group.


    2. Relevant equations
    Matrix is invertible in ring R.


    3. The attempt at a solution
    Group properties and ring properties are similar I think.
    Group and Ring - closure, associativity, identity (zero in Rings?)
    Is this how I am supposed to approach the problem? By proving the common properties of a group and ring?
     
  2. jcsd
  3. Aug 5, 2014 #2

    jbunniii

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    ##G## will be a group under multiplication, not addition. So you can ignore the addition and address the following questions:

    1. What is the multiplicative identity? Is it contained in ##G##?
    2. Is matrix multiplication associative?
    3. Is the set of matrices with nonzero determinant closed under multiplication and inverses?

    Hint for part 3: can you prove and apply the identities ##\det(AB) = \det(A)\det(B)## and ##\det(I) = 1##?
     
  4. Aug 5, 2014 #3

    Mark44

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    The elements of G already belong to a ring, so these elements already satisfy all of the properties of a ring (which must include an addition operation and a multiplication operation). If you list the properties that a group must satisfy, this should be pretty easy to prove.

    Groups are much more basic than rings.
     
  5. Aug 5, 2014 #4
    1. Multiplicative identity in a ring is 1, right? But 1 is not in G, because G consists of the set of matrices for which ad - bc ≠ 0, but (1) is not a 2*2 matrix.
    2. Yes; [(ab)c]ij = (ab)ik ckj = (ailblk)ckj = ail(blkckj) = ail(bc)lj = [a(bc)]ij
    3. Yes, for A= (a b c d), and B= (e f g h), it was proven that det(A)det(B)=det(AB), and det(I)= 1 by 1/ad-bc (ad-bc 0 0 ad-bc) since 1/ad-bc where ad-bc≠0 is the inverse of matrix M. The set is closed.
     
  6. Aug 5, 2014 #5

    jbunniii

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    We often use the notation "1" for the multiplicative identity in a ring, but it doesn't have to literally mean the number 1. In this case, the ring consists of 2x2 matrices over ##Z_p##, so if there is an identity, it must be a 2x2 matrix over ##Z_p##. What is the definition of "identity"?

    Seems like there are some ##\sum## symbols missing?
    Where did you use the fact that the matrix elements are from ##Z_p## (where ##p## is prime), as opposed to ##Z_n## for some nonprime integer? Also, can you explain in more detail how you concluded that the set is closed? How do the determinant formulas prove this?
     
  7. Aug 5, 2014 #6
    It means 1*n = n. The identity should be (1 0 0 1).

    Sorry, I completely forgot to insert them.

    I failed to mention that the elements are from Zp, p being a prime. Zp is a field if and only if p is prime, and the properties hold true because of this.
    Since the determinant is nonzero, and the inverse has been solved for: Det(M)* I = 1, the set is closed under multiplication and inverses. A group must contain an element and its inverse, which this set of matrices does.
     
  8. Aug 5, 2014 #7

    jbunniii

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    This equation doesn't make any sense. If I is the identity matrix, then the left hand side is a matrix. But the right hand side is (apparently) the scalar 1.
    Why? Suppose that ##A \in G##. You need to explain in detail why ##A^{-1}## exists, and why ##A^{-1} \in G##.

    Similarly, suppose that ##A,B \in G##. Why is ##AB \in G##?
     
  9. Aug 5, 2014 #8
    M * M-1 = I. The original matrix times its inverse is equal to the Identity Matrix.

    If an element exists, then its inverse must exist because the identity element is always in a group (definition of group).
    Groups are always closed under products, as well.
     
  10. Aug 5, 2014 #9

    jbunniii

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    1. How do you know that ##M## has an inverse?
    2. How do you know that ##\det(M^{-1}) \neq 0##?
    But your goal is to prove that ##G## is a group. You are effectively arguing that "it's a group because it's a group."
     
  11. Aug 5, 2014 #10
    M has an inverse because M= (a b c d) has another matrix such that M * (other matrix) = Identity Matrix. We know that the det(M-1) ≠ 0 because det(M) ≠0, and 1/det(M)≠0, ever. This ring is specifically an integral domain, I believe.

    The property of a ring is closure under multiplication and identity exists.
    So since the ring is a subgroup of the M(2*2) then it is a group itself.
     
  12. Aug 5, 2014 #11

    jbunniii

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    How do you know this is true? It would not be true if the matrix elements were from ##Z_n## where ##n## is not prime.
    OK, so you're using ##\det(M)\det(M^{-1}) = \det(M M^{-1}) = \det(I) = 1##. You need to state this more explicitly.
    An integral domain is commutative by definition. Matrix multiplication is not commutative.
    What ring is a subgroup of ##M_{2\times 2}##? I'm not following your logic. You need to explain why ##G## is closed under multiplication. If ##A## and ##B## are two matrices with nonzero determinant, what can you say about ##\det(AB)##?
     
  13. Aug 5, 2014 #12
    If Det(A) ≠0 and Det(B)≠0, then Det(AB)≠0 too.
     
  14. Aug 5, 2014 #13

    jbunniii

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    ...because ##\det(AB) = \det(A)\det(B)##. And how do you know that if ##\det(A)## and ##\det(B)## are nonzero, then so is ##\det(A)\det(B)##? Keep in mind that we are working in ##Z_p##, not the real numbers, so you have to justify what may seem like standard facts. Again, this would not be true in ##Z_n## for nonprime ##n##.
     
  15. Aug 5, 2014 #14
    For A, B in Zp, where p is prime, Det(A)Det(B) = Det(AB).
     
  16. Aug 6, 2014 #15

    pasmith

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    No, it's true for every matrix ring that [itex]\det(AB) = \det(A)\det(B)[/itex]. The question here is: if [itex]\det(A)[/itex] and [itex]\det(B)[/itex] are both non-zero, is it necessarily the case that [itex]\det(A)\det(B)[/itex] is non-zero?

    This is equivalent to showing that, for [itex]a \in \mathbb{Z}_p[/itex] and [itex]b \in \mathbb{Z}_p[/itex], if [itex]a \neq 0[/itex] and [itex]b \neq 0[/itex] then [itex]ab \neq 0[/itex].

    This isn't true for general [itex]\mathbb{Z}_n[/itex]: for example in [itex]\mathbb{Z}_{10}[/itex] we have [itex]2 \times 5 = 0[/itex]. So what's special about [itex]\mathbb{Z}_p[/itex] for prime [itex]p[/itex]?

    EDIT: You may also at some point want to make use of the fact that in any ring of 2x2 matrices, [tex]
    \begin{pmatrix} a & b \\ c & d \end{pmatrix}
    \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}
    = \begin{pmatrix} ad - bc & 0 \\ 0 & ad - bc \end{pmatrix}
    [/tex] and [tex]
    \det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc.
    [/tex]
     
    Last edited: Aug 6, 2014
  17. Aug 8, 2014 #16
    Is G, technically the non-abelian group GL(2, R)? I approached this problem in that manner.
     
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