Exhibit all of the Sylow 3-subgroups of a special linear group

Exhibit all Sylow 3-subgroups of SL_2(\mathbb{F}_3).


By this previous exercise, [GL_2(\mathbb{F}_3) : SL_2(\mathbb{F}_3)] = 2. We also know that |GL_2(\mathbb{F}_3)| = (3^2-1)(3^2-3) = 2^4 \cdot 3. Thus |SL_2(\mathbb{F}_3)| = 2^3 \cdot 3. By Sylow’s Theorem, the number n_3 of Sylow 3-subgroups is congruent to 1 mod 3 and divides 8. Thus n_3 \in \{1,4\}.

Let A \in SL_2(\mathbb{F}_3), with A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}.

Note that A^2 = \begin{bmatrix} a^2+bc & ab+bd \\ ac+dc & bc+d^2 \end{bmatrix} and A^3 = \begin{bmatrix} a^3+2abc+bcd & a^2b+b^2c+d^2b+abd \\ a^2c+c^2b+d^2c+adc & d^3+2bcd+abc \end{bmatrix}. If A^3 = I_3, then the entries of A must satisfy the following four equations. Note that \alpha^3 = \alpha for all \alpha \in \mathbb{F}_3.

  1. a+2abc+bcd = 1
  2. a^2b+b^2c+d^2b+abd = 0
  3. a^2c+c^2b+d^2c+adc = 0
  4. d+2bcd+abc = 1

Suppose c = 0. Then equations (1) and (4) give a = d = 1. Now b \in \mathbb{F}_3, but b \neq 0 since otherwise A = I_3. Thus b \in \{1,2\}. In either case, we generate the Sylow 3-subgroup H_1 = \left\langle \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \right\rangle = \left\{ I, \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \right\}.

Now suppose b = 0. Then equations (1) and (4) give a = d = 1. Now c \in \mathbb{F}_3, but c \neq 0 since otherwise A = I_3. Thus c \in \{1,2\}. In either case, we generate the Sylow 3-subgroup H_2 = \left\langle \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \right\rangle = \left\{ I_3, \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \right\}.

Henceforth we may assume that c and b are not zero.

Suppose c = 1.

  • If a = 0, then equation (1) gives us bd = 1. Clearly then b = d, as otherwise bd = 2. Moreover, equation (2) gives b^2(b+1) = 0. Since b \neq 0, we have b = 2 = d. Thus A generates the Sylow 3-subgroup H_3 = \left\langle \begin{bmatrix} 0 & 2 \\ 1 & 2 \end{bmatrix} \right\rangle = \left\{ I_3, \begin{bmatrix} 0 & 2 \\ 1 & 2 \end{bmatrix}, \begin{bmatrix} 2 & 1 \\ 2 & 0 \end{bmatrix} \right\}.
  • If a = 1, then equation (1) gives us b(2+d) = 0. Since b \neq 0, we have d = 1. Then equation (3) gives b = 0, a contradiction.
  • If a = 2, then equation (1) gives us b(1+d) = 2. Then b \neq 1+d, as otherwise b(1+d) = 1. If b = 1, then 1+d=2, so that d = 1. Then equation (2) gives us 2 = 0, a contradiction. Thus b = 2, so that 1+d = 1 and we have d = 0. Thus A generates the Sylow 3-subgroup H_4 = \left\langle \begin{bmatrix} 2 & 2 \\ 1 & 0 \end{bmatrix} \right\rangle = \left\{ I_3, \begin{bmatrix} 2 & 2 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 2 & 2 \end{bmatrix} \right\}.

Suppose c = 2.

  • If a = 0, then equation (1) gives us 2bd = 1, so that bd = 2. If b = 2, then d = 1 and equation (2) gives us 1 = 0, a contradiction. Thus b = 1 and d = 2; in this case A generates H_4.
  • If a = 1, then equation (1) gives us b(1+d) = 0. Since b \neq 0, we have 1+d = 0, so that d = 2. But then equation (1) gives b = 0, a contradiction.
  • If a = 2, then equation (1) gives us b(1+d) = 1. Then b = 1+d, as otherwise b(1+d) = 2. If b = 2, then 1+d = 2, so that d = 1. Then equation (3) gives 1 = 0, a contradiction. Thus b = 1, so that d = 0, and A generates H_3.

Thus n_3 = 4, and we have exhausted the Sylow 3-subgroups of SL_2(\mathbb{F}_3).

Post a comment or leave a trackback: Trackback URL.

Comments

  • 1AND2IS3  On June 23, 2011 at 9:03 pm

    At the top it should say the number of Sylow 3 subgroups is congruent to 1 mod 3 and divides 8, so it’s 1 or 4.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: