# Abstract Algebra Question: order, stabilizer, and general linear groups

1. Dec 7, 2011

### sleventh

3) Let X be the set of 2-dimensional subspaces of F_{p}^{n}, where n >= 2.
(a) Compute the order of X.
(b) Compute the stabilizer S in GL_{n}(F_{p}) of the 2-dimensional subspace U = {(x1, x2, 0, . . . , 0) ε F_{p}^{n} | x1, x2 ε F_{p}}.
(3) Compute the order of S.
(4) Show that GL_{n}(F_{p}) acts transitively on X.

for 3.a I still need clarity on why this is the case, but the order of F_{p}^{n} is (p^{n}-1)(p^{n}-p) My reason for thinking this is because there are p^{n} possible vectors we then subtract one to eliminate the null case, we also must consider all multiples of the p^{n}-1 vectors, since it is a finite field there are (p^{n}-p) possible scalar combinations to multiply by, For the 2-D subspaces we must first consider the fact the order of F_{p}^{n} includes all multiples of 2-D spaces, so we must mod redundant multiples, the multiples are linear combinations of the 2-D space which is GL_{2}(F_{p}) so we must divide the order of F_{p}^{n} by the order of GL_{2}(F_{p}) i.e. |F_{p}^{n}| / |GL_{2}(F_{p})| = (p^{n}-1)(p^{n}-p) / (p^{2}-1)(p^{2}-p)
I am not sure if my reasoning for dividing by the order of the 2-d special linear group is correct

3.b I am having a bit of difficulty with this one. GL_{n}(F_{p}) will be a nXn matrix with elements from the prime ordered field F with a non-zero determinant. The stabilizer will be the nXn matrices that whose product with U (defined above) produce U. so G*U = U . To do this we get a block matrix defined below:
top left 2x2 matrix is the identity. after the first two rows are zero. after second two columns any element from F_{p} may be used.

| I [ any elements] |
| 0 0 [any elements] |
| . . [ ] |
| . . [ ] |
| 0 0[ ] |

the order of this matrix is the order of the (n-2)Xn matrix minus the cases that produce determinant zero
I will wait to get too involved to see if what i've done till now is on the right path