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Order of 2x2 matrix group under mult. mod p

  1. Dec 14, 2007 #1
    This is the last problem on herstein's 2.3 problem set.

    So we want to count how many 2x2 matrices with entries being integers mod p have nonzero determinant mod p. p is prime.

    for p=3 there are 48 such matrices. I've broken the general case into three possibilities: matrices with no zeroes (the hardest to deal with), matrices with 1 zero and matrices with 2 zeros.

    For the last case there are two ways the pair of zeroes can be arranged, in either of the diagonals. And the remaning two elements can be ANY pair of nonzero integers mod p so that there are 2(p-1)^2 such matrices.

    For the matrices with one zero, the zero can occur in four different places and the remaining entries can be ANY nonzero integers mod p, so there are 4(p-1)^3 such matrices.

    For matrices with no zeroes there are (p-1)^4 possible matrices, but some of these will have zero determinant. This occurs if all entries are the same, and there are p-1 such matrices. It also occurs if the all entries in each row (or column) are the same. To calculate the number of these type of matrices is not so straightforward: first we know that a matrix with 2 pairs of the same entries (ie a and a and b and b) can be arranged in 6 ways for each pair a and b. But two of these are similar on the diagonals only and therefore don't necessarily have zero determinant. So we subtract another 4(p-1 choose 2) (this stands for binomial/combination notation) from (p-1)^4. The rest of the matrices that have zero determinant are as such because ad-bc may not be zero but is congruent to zero mod p. I.e is divisible by p. But this means ad is congruent to bc mod p.

    This last bit is confusing me because i think it overlaps with other possible ways to get zero determinant that I have already considered. I'll come back and add more after I think about it some more, for now though could someone give me feedback for what is already done.

  2. jcsd
  3. Dec 14, 2007 #2


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    You're overcomplicating things. Think linear independence of rows/columns.
  4. Dec 15, 2007 #3
    really? lol, i thought i was simplifying things. I AM thinking of linear independence of columns/rows, but I can't count all combinations of linearly independent columns/rows in a straightforward fashion.
  5. Dec 15, 2007 #4


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    You have two columns. If the matrix is going to be invertible, how many choices are possible for the first one? For each such column, how many choices can we have for the second column?
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