# elementary proof of orders

When possible, our proofs avoid matrices so that the proofs retain some value to infinite dimensional settings. When we use $k$ we mean any field, and $GF(q)$ indicates the special case of a finite field of order $q$. $V$ is always our vector space  .

The basic starting point in computing orders of classical groups is an application of elementary linear algebra rephrased in group theory terms.

###### Proposition 2.

$GL(V)$ acts regularly on the set of ordered bases of vector space $V$ over a field $k$.

###### Proof.

Given any two bases $B=\{b_{i}:i\in I\}$ and $C=\{c_{i}:i\in I\}$ of a vector space $V$, define the map $f:V\rightarrow V$ by

 $\left(\sum_{i\in I}l_{i}b_{i}\right)f=\sum_{i\in I}l_{i}c_{i}.$ (1)

[Note the above sum has only finitely many non-zero $l_{i}\in k$.] It follows $f$ is an invertible linear transformation so $f\in GL(V)$. Furthermore, any linear transformation $g\in GL(V)$ with $(b_{i})g=c_{i}$ must satisfy (1) to be linear so indeed $g=f$. Therefore $GL(V)$ acts regularly on ordered bases of $V$. ∎

###### Theorem 3.
 $\begin{array}[]{cc}|SL(d,q)|=q^{\binom{d}{2}}\prod_{i=2}^{d}(q^{i}-1),&|PSL(d,% q)|=\frac{|SL(d,q)|}{(d,q-1)},\\ |GL(d,q)|=q^{\binom{d}{2}}\prod_{i=1}^{d}(q^{i}-1),&|PGL(d,q)|=|SL(d,q)|,\\ |\Gamma L(d,q)|=(q-1)q^{\binom{d}{2}}\prod_{i=1}^{d}(q^{i}-1),&|P\Gamma L(d,q)% |=\frac{|\Gamma L(d,q)|}{q-1}.\end{array}$
###### Proof.

When $V=GF(q)^{d}$, the number of ordered bases can be counted. A basis is a set $B=\{b_{1},\dots,b_{d}\}$ of linearly independent  vectors. So $b_{1}$ may be chosen freely from $V-\{0\}$, providing $q^{d}-1$ possible choices. Next $b_{2}$ must be chosen independent form $b_{1}$ so $b_{2}$ can be freely chosen from $V-\langle b_{1}\rangle$ leaving $q^{d}-q$ choices. In a similar  fashion $b_{3}$ has $q^{d}-q^{2}$ possiblities and so continuing by induction  we find the total number of ordered bases to be:

 $(q^{d}-1)(q^{d}-q)\cdots(q^{d}-q^{d-1}).$

Now we treat $q$ as the variable of a polynomial and factor this number into:

 $q^{\binom{d}{2}}\prod_{i=1}^{d}(q^{i}-1).$

As $GL(V)$ acts regularly on ordered bases of $V$, this is the order of $GL(V)$.

 $[GL(d,q):SL(d,q)]=|GF(q)^{\times}|=q-1$

so that $|SL(d,q)|=|GL(d,q)|/(q-1)$.

In a similar process, $PGL(V)=GL(V)/Z(GL(V))$ so if we derive the order of the center of $GL(V)$ we derive the order of $PGL(V)$. The central transforms are scalar (they must preserve every eigenspace  of every linear transform) so $Z(GL(V))$ is isomorphic  to $k^{\times}$. Thus the order of $PGL(V)$ is the same as the order of $SL(V)$.

For $\Gamma L(V)$ and $P\Gamma L(V)$ simply notice $\Gamma L(V)=GL(V)\rtimes k^{\times}$ so when $k=GF(q)$ we get an additional $q-1$ term.

Finally, we consider $PSL(V)=SL(V)/(SL(V)\cap Z(GL(V)))$. The order of $SL(V)\cap Z(GL(V))$ must be computed. So we require scalar transforms with determinant 1. As the such, if $r$ is the eigen value of the scalar transform we need $r^{d}=1$ in $GF(q)$. From finite field theory we know $GF(q)^{\times}\cong\mathbb{Z}_{q-1}$. As this group is cyclic we know that every element $r\in GF(q)^{\times}$ satisfying $r^{d}=1$ also satisfies $r^{q-1}=1$ and $r$ lies in the unique subgroup of order $(d,q-1)$ of $GF(q)^{\times}$. Thus $|SL(V)\cap Z(GL(V))|=(d,q-1)$. ∎

Title elementary proof of orders ElementaryProofOfOrders 2013-03-22 15:56:52 2013-03-22 15:56:52 Algeboy (12884) Algeboy (12884) 8 Algeboy (12884) Proof msc 11E57 msc 05E15