User:J D Bowen/Math735 HW1

1.3.16) Suppose $n \geq m \ $, and consider the number of m-cycles in $S_n$.

An m-cycle can be represented by a selection of m elements from n without any repeats. The number of permutations of m elements from n possible elements is $n!/m! = n(n-1)...(n-m+1) \ $. However, each such string is merely a representation of an m-cycle; the cycle does not depend on the starting element in the string, and since there are m possible starting numbers, we must divide this number by m. Hence, the number of m cycles is

$\frac{n(n-1)(n-2)... (n-m+1)}{m}$

1.3.17) Consider a product of two disjoint 2-cycles (wx)(yz), where w,x,y,z are in {1,...,n}. The number of different such arrangements is then n(n-1)(n-2)(n-3). However, this does not account for the following relations:

(wx)(yz)=(xw)(yz)=(wx)(zy)=(xw)(zy)=(yz)(wx)=(yz)(xw)=(zy)(wx)=(zy)(xw)

Since there are 8 different arrangements in the set formed by the n choose 4 operation above, we must divide by 8. Hence, the number of element in $S_n$ which are the product of two disjoint 2-cycles is $n(n-1)(n-2)(n-3)/8$.

1.4.5) Let $F \ $ be a finite field of $m \ $ elements. Then the number of $n\times n \ $ matrices we can form from this field will be $mn^2 \ $, since in each entry of the matrix we have $m \ $ choices of element, and there are $n^2 \ $ such choices to be made.  This places an upper bound on the number of possible elements in $GL_n(F) \ $, since this group will cannot have any element not of this form.

1.4.7) Let $p \ $ be prime. Then the general linear group $GL_2(\mathbb{F}_p) \ $ will contain all matrices which are invertible, that is, have non-zero determinant.  In order for a 2 by 2 determinant to be zero, one row must be a multiple of another row, and of course there cannot be a row which is all zeroes.

Let us concern ourselves with the first row of an invertible matrix: For the first entry, $a_{11}$, we can have any of the p values in the field, and similarly for the second entry $a_{12}$. However, they cannot both be zero, or the determinant will be zero. So, there are $p^2 -1 $ possibilities for the first row - the $p^2$ from all possible options, minus the one option (0,0) which leads to a null determinant.

For the second row, we cannot have (0,0) either, or any multiple of the first row. Consider that (0,0) IS a multiple of the first row, if we multiply by 0. Hence, for the second row, there are $p^2-p$ possibilites - $p^2$ from the total possibilities, minus the p possibilities which are multiples of the first row.

Hence, the number of invertible matrices is $(p^2-1)(p^2-p)=p^4-p^3-p^2+p \ $.