For any prime $p$, consider the group $G=\mathrm{GL}_{2}(\mathbb{F}_p) $. Then which of the following are true? For any prime $p$ , consider the group $ G=\operatorname{GL}_2 (\mathbb{F}_p) $. Then which of the following are true?


*

*$G$ has an element of order  $p$

*$G$ has exactly one element of order  $p$

*$G$ has no $p-$Sylow subgroups 

*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. 
 A: 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 $p$.
A: 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 $(1\leftrightarrow 3)$.

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: 


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.

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.

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".

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.
