# 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 $f:V\to V$ 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.

$$\begin{array}{cc}\hfill |SL(d,q)|={q}^{\left(\genfrac{}{}{0pt}{}{d}{2}\right)}\prod _{i=2}^{d}({q}^{i}-1),\hfill & \hfill |PSL(d,q)|=\frac{|SL(d,q)|}{(d,q-1)},\hfill \\ \hfill |GL(d,q)|={q}^{\left(\genfrac{}{}{0pt}{}{d}{2}\right)}\prod _{i=1}^{d}({q}^{i}-1),\hfill & \hfill |PGL(d,q)|=|SL(d,q)|,\hfill \\ \hfill |\mathrm{\Gamma}L(d,q)|=(q-1){q}^{\left(\genfrac{}{}{0pt}{}{d}{2}\right)}\prod _{i=1}^{d}({q}^{i}-1),\hfill & \hfill |P\mathrm{\Gamma}L(d,q)|=\frac{|\mathrm{\Gamma}L(d,q)|}{q-1}.\hfill \end{array}$$ |

###### Proof.

When $V=GF{(q)}^{d}$, the number of ordered bases can be counted. A basis
is a set $B=\{{b}_{1},\mathrm{\dots},{b}_{d}\}$ 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 $V-\u27e8{b}_{1}\u27e9$ 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:

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

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

$${q}^{\left(\genfrac{}{}{0pt}{}{d}{2}\right)}\prod _{i=1}^{d}({q}^{i}-1).$$ |

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^{} $det:GL(V)\to {k}^{\times}$. 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 $\mathrm{\Gamma}L(V)$ and $P\mathrm{\Gamma}L(V)$ simply notice $\mathrm{\Gamma}L(V)=GL(V)\u22ca{k}^{\times}$ 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 |
---|---|

Canonical name | ElementaryProofOfOrders |

Date of creation | 2013-03-22 15:56:52 |

Last modified on | 2013-03-22 15:56:52 |

Owner | Algeboy (12884) |

Last modified by | Algeboy (12884) |

Numerical id | 8 |

Author | Algeboy (12884) |

Entry type | Proof |

Classification | msc 11E57 |

Classification | msc 05E15 |