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.

Remark 1.

There are many alternative methods for computing orders of classical groups, for instance observing special subgroups (http://planetmath.org/TheoryFromOrdersOfClassicalGroups2) or from Lie theory and the study of Chevalley groups. The method explored here is intended to be elementary linear algebra.

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 $V$. ∎

In the world of group theory, a regular action is a typical substitute for knowing the order of a group. In particular, any two groups, even infinite, have the same order if they have a regular action on the same set. However, we are presently after specific order of finite groups so we return to the case of $V$ a finite dimension vector space over a finite field $k=GF(q)$. We do however attempt to establish the orders through bijections with other sets and groups so that the results apply in more general contexts as well.

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 $GL(V)$.

For the order of $SL(V)$ recall the $SL(V)$ is the kernel of the determinant homomorphism $\det:GL(V)\rightarrow k^{\times}$. Furthermore, $\det$ is surjective as a diagonal matrix can be used to exhibit any determinant we seek. We conclude

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

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