# Find the number of $2$-Sylow subgroups in $GL_2(\mathbb F_5)$

First, we can calculate the order of $$GL_2(\mathbb F_5)$$ to find the order of the $$2$$-Sylow subgroups.

$$|GL_2(\mathbb F_5)|=(5^2-1)(5^2-5)=24\cdot 20=2^5\cdot 3\cdot 5$$

Thus we have $$|P|=2^5$$.

Now by using Sylow theorems, we can try to find $$m$$, the number of $$2$$-Sylow subgroups.

$$m | \frac{|GL_2(\mathbb F_5)|}{|P|}, m \equiv 1 \mod p$$

$$m | 3\cdot 5,m \equiv 1 \mod 2$$

The solution given in my textbook involves the third Sylow theorem, $$m = \left[ G : N _ { G } ( P ) \right]$$ and $$m=6$$. While this solution would be simpler since $$N _ { G } ( P )$$ is the group of upper triangular matrices, I was wondering where was the error in my reasoning. $$6$$ does not divide $$3\cdot 5$$ so there must be an error.

EDIT: Here is the correction given by my teacher:

Taking as P the subgroup, we claim that $$N _ { G } ( P ) = \{ T | T \in \mathrm { GL } _ { 2 } \left( \mathbb { F } _ { 5 } \right)$$ is upper triangular $$\}$$.

$$A U A ^ { - 1 } = ( a d - b c ) ^ { - 1 } \left[ \begin{array} { l l } { a } & { b } \\ { c } & { d } \end{array} \right] \left[ \begin{array} { l l } { 1 } & { x } \\ { 0 } & { 1 } \end{array} \right] \left[ \begin{array} { c c } { d } & { - b } \\ { - c } & { a } \end{array} \right]$$ $$= ( a d - b c ) ^ { - 1 } \left[ \begin{array} { c c } { a d - b c - a c x } & { a ^ { 2 } x } \\ { - c ^ { 2 } x } & { a d - b c + a c x } \end{array} \right]$$ $$= \left[ \begin{array} { c c } { 1 - acx( a d - b c ) ^ { - 1 } } & { a ^ { 2 } x ( a d - b c ) ^ { - 1 } } \\ { - c ^ { 2 } x ( a d - b c ) ^ { - 1 } } & { 1 + a c x ( a d - b c ) ^ { - 1 } } \end{array} \right]$$ $$\text { Therefore, } A U A ^ { - 1 } \in P \text { is equivalent to: }$$ $$\left\{ \begin{array} { l l } { a c x = 0 } & { \forall x \in \mathbb { F } _ { 5 } } \\ { c ^ { 2 } x = 0 } & { \forall x \in \mathbb { F } _ { 5 } } \end{array} \right.$$ which, in turn, is equivalent to $$c = 0$$. $$\text { Therefore, } N _ { G } ( P ) = \{ T | T \in \mathrm { G } \mathrm { L } _ { 2 } \left( \mathbb { F } _ { 5 } \right) \text { is upper triangular } \}$$

$$m = \left[ G : N _ { G } ( P ) \right] = \frac { | G | } { \left| N _ { G } ( P ) \right| } = \frac { 480 } { 80 } = 6$$

• If they describe the subgroup of upper triangular matrices as the normalizer of a Sylow subgroup my educated guess is that there is a typo and they wanted you to figure out the number of Sylow $5$-subgroups. Anyway, you are right, the number of Sylow $2$-subgroups is always odd. – Jyrki Lahtonen Jan 7 at 18:15
• One Sylow $2$-subgroup of this group is the group of monomial matrices $M$. That is, invertible matrices with a single non-zero entry on both rows and columns. There will be $4\cdot4=16$ diagonal matrices in $M$, and $16$ matrices with zeros on the diagonal. – Jyrki Lahtonen Jan 7 at 18:18
• @JyrkiLahtonen I added the correction given by the teacher. Your comment makes more sense, with $32$ we would have $480/32=15=3 \cdot 5$ which is precisely the result required by both Sylow theorems – NotAbelianGroup Jan 7 at 18:27
• Ok, the edit clinches it. The matrices of the form $$\pmatrix{1&x\cr0&1\cr}$$ form a Sylow $5$-subgroup. After all, there are five choices for $x$. Your teacher is looking for the normalizer of that group. – Jyrki Lahtonen Jan 7 at 18:27
• But do check with your teacher. Is the question about Sylow-2 or Sylow-5 subgroups? – Jyrki Lahtonen Jan 7 at 18:31