Let \(p\) be a positive prime number, \(\mathbb F_p\) the field with \(p\) elements, and let \(G=\text{GL}_2(\mathbb F_p)\).
Compute the order of \(G\), \(|G|\).
Write down an explicit isomorphism from \(\mathbb Z/p\mathbb Z\) to \[\begin{align*} U=\left\{ \begin{pmatrix} 1 & a\\ 0 & 1 \end{pmatrix} \bigg|a\in\mathbb F_p\right\}. \end{align*}\]
How many subgroups of order \(p\) does \(G\) have?
Hint: compute \(gu\inverseof{g}\) for \(g\in G\) and \(u\in U\); use this to find the size of the normalizer of \(U\) in \(G\).
Solution
1. The order of \(G=\operatorname{GL}_2(\mathbb F_p)\) is\[ |G|=(p^2-1)(p^2-p)=p(p-1)^2(p+1). \]
The first column of an invertible \(2\times2\) matrix may be any nonzero vector in \(\mathbb F_p^2\), giving \(p^2-1\) choices. Once it is chosen, the second column may be any vector not in its one-dimensional span, giving \(p^2-p\) choices.
2. The map\[ \varphi:\mathbb Z/p\mathbb Z\longrightarrow U, \qquad \bar a\longmapsto \begin{pmatrix}1&a\\0&1\end{pmatrix} \]is an isomorphism of groups.
Matrix multiplication gives \[ \begin{pmatrix}1&a\\0&1\end{pmatrix} \begin{pmatrix}1&b\\0&1\end{pmatrix} = \begin{pmatrix}1&a+b\\0&1\end{pmatrix}, \] so \(\varphi\) is a homomorphism from the additive group of \(\mathbb F_p\). It is visibly bijective.
3. The subgroup \(U\) is a Sylow \(p\)-subgroup of \(G\).
By 2, \(|U|=p\). From 1, \[ |G|=p(p-1)^2(p+1), \] and neither \(p-1\) nor \(p+1\) is divisible by \(p\). Hence the highest power of \(p\) dividing \(|G|\) is \(p\) itself.
4. The normalizer of \(U\) is the subgroup\[ N_G(U)= \left\{ \begin{pmatrix}a&b\\0&d\end{pmatrix}: a,d\in\mathbb F_p^\times,\ b\in\mathbb F_p \right\}. \]
Every nonidentity element of \(U\) has the form \[ u_t=\begin{pmatrix}1&t\\0&1\end{pmatrix},\qquad t\ne0, \] and its fixed-point space is exactly the line \[ L=\mathbb F_p e_1. \] If \(gUg^{-1}=U\), then for nontrivial \(u\in U\), the fixed line of \(gug^{-1}\) is \(gL\). But every nontrivial element of \(U\) has fixed line \(L\), so \(gL=L\). Thus \(g\) is upper triangular.
Conversely, if \[ g=\begin{pmatrix}a&b\\0&d\end{pmatrix} \] with \(a,d\ne0\), then a direct calculation gives \[ g\begin{pmatrix}1&t\\0&1\end{pmatrix}g^{-1} = \begin{pmatrix}1&(a/d)t\\0&1\end{pmatrix}\in U. \] Hence every invertible upper-triangular matrix normalizes \(U\).
5. There are exactly \(p+1\) subgroups of order \(p\) in \(G\).
By 3, the subgroups of order \(p\) are exactly the Sylow \(p\)-subgroups, and all Sylow \(p\)-subgroups are conjugate. Therefore their number is \[ [G:N_G(U)]. \] By 4, \[ |N_G(U)|=p(p-1)^2, \] so \[ [G:N_G(U)] =\frac{p(p-1)^2(p+1)}{p(p-1)^2} =p+1. \]