For any prime $p$, consider the group $G=\mathrm{GL}_{2}(\mathbb{F}_p) $. Then which of the following are true?

Question

For any prime $p$ , consider the group $ G=\operatorname{GL}_2 (\mathbb{F}_p) $. Then which of the following are true?

1. $G$ has an element of order $p$

2. $G$ has exactly one element of order $p$

3. $G$ has no $p-$Sylow subgroups

4. Every element of order $p$ is conjugate to a matrix $ \begin{bmatrix} 1 & a \\ 0 & 1 \\ \end{bmatrix} $ , where $ a\in (\mathbb{F}_p)^* $

Since the order of the group $G$ is $p (p-1)(p^2-1) $ , clearly $G$ has $p-$Sylow subgroups of order $p $. So option 3 is false.

Also, the $p-$Sylow subgroups are cyclic so option 1 is true.

For option 2 , I chose $p=2$ and I saw there are $3$ elements of order $2$ namely $\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} $ , $\begin{bmatrix} 1 & 0 \\ 1 & 1 \\ \end{bmatrix} $ and $\begin{bmatrix} 1 & 1 \\ 0 & 1 \\ \end{bmatrix}$ . So option 2 is false .

For option 4 I chose $p=2$ and consider the matrix $\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} $ which is of order 2 but not conjugate to $\begin{bmatrix} 1 & a \\ 0 & 1 \\ \end{bmatrix} $ , here $a=1$ since $ (\mathbb{F}_2)^* $ contains only 1

But option 4 is correct in the answer key. Is my explanation for option 4 wrong? I don't understand how.

Any help for option 4 would be great. THANKS.

Answer 1

Hint: Use Sylow's theorem to conclude that 4. is true. For $p=2$, you can also check directly that the conjugate of $\begin{bmatrix} 1 & 1 \\ 0 & 1 \\ \end{bmatrix}$ by either of $\begin{bmatrix} 1 & 1 \\ 1 & 0 \\ \end{bmatrix}$ and $\begin{bmatrix} 1 & 0 \\ 1 & 1 \\ \end{bmatrix}$ is $\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}$.

Moreover, for 2., you should show that this is false also for $p\neq 2$. The case of $p=2$ is the only nontrivial one, of course, but you want to show that this has to be false, regardless of $p$, not that it fails for some $p$.

Answer 2

The fourth claim:

Working modulo two in $\operatorname{GL}(2,\Bbb F_2)$ we have the relation $$ \begin{aligned} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} ^{-1} &= \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \\ &= \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \ . \end{aligned} $$ So there is no contradiction in the 4.th claim for the value $p=2$, for the special chosen matrix.

This can be seen also in a more abstract way. The group $G= GL(2,\Bbb F_2)$ is isomorphic to $S_3$, the symmetric group of all bijections of the set $\{1,2,3\}$ with the composition. Then all three transpositions (group elements of order two) are conjugated. It is enough to show how to conjugate $(1,2)$ using $\sigma$ in $(2,3)$, which should be equal to $(\sigma(2),\sigma(3))$. We have two chances to do this, either map the list $1,2$ into the list $2,3$ by $\sigma$, so $\sigma$ is the $3$-cycle $(1\to 2\to 3\to 1)$, or map the list $1,2$ into the list $3,2$ by $\sigma$, so $\sigma$ is the transposition $(1\leftrightarrow 3)$.

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One can also use computer support for this. I am always showing the code, when this was my search, and this was the case, sage code was used:

sage: G = GL(2, GF(2))
sage: G.list()
(
[0 1]  [0 1]  [1 0]  [1 0]  [1 1]  [1 1]
[1 0], [1 1], [0 1], [1 1], [0 1], [1 0]
)
sage: a,b,e,c,d,f = G.list()
sage: for s in G.list():
....:     if s * a * s.inverse() == d:
....:         print "Possible conjugation via\n%s\n" % s
....:         
Possible conjugation via
[0 1]
[1 1]

Possible conjugation via
[1 0]
[1 1]

sage: 

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Let us show this fourth claim in general.

Let $F=\Bbb F_p=\Bbb Z/p$ be the field with $p$ (prime) elements, realized as integers modulo $p$. Let $A\in G:=\operatorname{GL}(2,F)$ be a matrix of order $p$ in the group $G$ of order $(p^2-1)(p^2-p)$.

(The case of the (even, and thus) oddest prime $p=2$ is special, and was considered explicitly. The following does not assume $p$ odd, but if it is really needed, please tell me.)

Let $p$ be (an odd) prime. The matrix $A$ satisfies an equation $A^2-tA+d=0$, $s,d$ being the trace, and the determinant.

• If the polynomial $X^2-sX+d$ has (at least) one, thus both roots in $\Bbb F_p$, then there exist an eigenvector to the (first) eigenvalue, and by base change we can arrange to move this eigenvector to the column vector with entries $1,0$, so we may and do assume $A$ to be of the shape $$ A= \begin{bmatrix} d & a\\ 0 & d' \end{bmatrix} \ . $$ Because of $A^p=1$, we have $d=d^p=1$, and with the same argument $d'=1$. And we are done.

• Else, if there is any else, we proceed similarly. After a base change to the quadratic extension $L$ of $F=\Bbb F_p$ we have both roots in the field, now a field $L$ with $p^2$ elements. Here we find an eigenvalue, take the corresponding eigenvector, a base change over $L$ brings $A$ in a triangular form with diagonal elements $d,d'\ne 0$ having multiplicative order $p$, which is relatively prime to the order $p^2-1$ of the group of units in $L$, so again, $d=d'=1$. The characteristic polynomial of $A$ is thus $X^2-(d+d')X+dd'=X^2-2X+1=(X-1)^2$. So at least one root lives over $F$. Contradiction.

This proves the fourth claim.

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Let us now review the other claims.

The second claim is false, any matrix of the shape $$ U(a)= \begin{bmatrix} 1 & a\\ 0 & 1 \end{bmatrix} \ ,\ a\ne 0\ , $$ having order $p$. In characteristic $2$ we have a special argument.

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The first claim follows using $U(1)$, and the relation $U(1)^a=U(a)$, so $U(1)^p=U(p)=U(0)=1$, and $a=p$ is the minimal integer $a>0$ with "this property".

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The third claim is false, since the order of $G$ is $p^2-p)(p^2-1)$, and $p$ divides exactly (so the terminology "Sylow" is properly used) this order.