14 Matrix Groups

#todo/school : Problems 1, 2, 4 📅 2025-01-14 ✅ 2025-01-14

1

Question

Prove that $| G L_{2} (F_{2}) | = 6$

Proof

Here (referring to Z over nZ - The Integers Modulo n) we have $F_{2} \equiv \frac{Z}{2 Z} \equiv {\overset{―}{0}, \overset{―}{1}}$ . So then all possible matrices of this group are $2 \times 2$ matrices where each entry is a binary number (0 or 1). Since we have 4 slots with 2 choices each, we have 32 total choices ( $2^{5} = 32$ ).

Now some of these choices are not allowed. For instance we cannot have the zero matrix (as it is non-invertible) and we cannot have the matrix:

(\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix})

since its determinant is 0 (so the matrix is non-invertible).

We can just check what solutions are valid. Consider the matrix:

(\begin{matrix} a & b \\ c & d \end{matrix})

where $a, b, c, d \in {0, 1}$ . For the purposes of this argument we'll represent $a b c d$ as a binary number, which is one-to-one with the matrix above. We need:

a d - b c \neq 0 \Rightarrow a d \neq b c

Going through it:

For the first column you have 3 choices $10, 01, 11$ (since then the result would be zeros)
Then for the second column you chose from the other 2 choices (ex: if $00$ was chosen then you either have $01$ or $11$ )
Thus then we have $3 \cdot 2 = 6$ .

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2

Question

Write out all the elements of $G L_{2} (F_{2})$ and compute the order of each element.

Proof

Similar to 14 Matrix Groups#1 we want to compute the order of each element. Let's look at each element we found in that part:

$0110$ is the same as:

(\begin{matrix} \overset{―}{0} & \overset{―}{1} \\ \overset{―}{1} & \overset{―}{0} \end{matrix})

which has order 2 it is the identity matrix (in this case):

{(\begin{matrix} \overset{―}{0} & \overset{―}{1} \\ \overset{―}{1} & \overset{―}{0} \end{matrix})}^{2} = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})

$0111$ has order 3 since:

{(\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix})}^{2} = (\begin{matrix} 1 & 1 \\ 1 & 2 \equiv \overset{―}{0} \end{matrix}) \Rightarrow {(\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix})}^{3} = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})

$1001$ is the identity matrix so it has order 1.

$1011$ has order 2 since:

{(\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix})}^{2} = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})

$1101$ has order 2 since:

{(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})}^{2} = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})

$1110$ has order 3 since:

{(\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix})}^{2} = (\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}) \Rightarrow {(\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix})}^{3} = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})

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4

Question

Show that if $n$ is not prime then $\frac{Z}{n Z}$ is not a field.

Proof

Suppose $n$ is not prime. Then there are two prime factors $m, k \in N$ where $m k = n$ (notice the trivial case where $n = 1$ , then we have $\frac{Z}{Z} \equiv Z$ is not a field, so we can ignore it).

Now notice that $\frac{Z}{n Z}$ fails to follow the property of a Field such such $(F - {0}, \cdot)$ is not an Abelian Group. Namely:

a b \neq b a

where $a, b \in \frac{Z}{n Z}$ . One such example is $m, k$ . Namely notice that $m, k ≢ \overset{―}{0}$ . Notice though that $m k \equiv \overset{―}{0}$ , so then $\cdot$ is not closed (so this cannot be a group anymore).

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