Number of Sylow $5$-subgroups

Question

Suppose $G=SL(2,4)$. Could you please suggest, is there is any simple argument to show that $G$ cannot have the unique Sylow $5$-subgroup, i.e. $|Syl_5(G)|>1$ without quoting the isomorphism $A_5 \cong G$? My final goal is to prove the existence of a subgroup of order $12$. I've dealt with everything except the fact that $|Syl_5(G)|=1$ is not the case (then $|Syl_5(G)|=6$). Also, can we use this result to show that $G$ is simple (and then it is clearly isomorphic to $A_5$?

Answer 1

$$\text{SL(2,4)$\cong A_5$}$$

Notes: We can represent $F_4=\{0,1,a,a^2\}$ s.t. $1+a+a^2=0$ as $F_4\cong Z_2[x]/x^2+x+1$.

And $n_p$ denotes number of the Sylow-$p$ subgroup and $H_p$ represent a sylow-$p$ subgroup.

Fact$0:$ You can easily check that these two matrices $\begin{bmatrix} a & a \\a & 1 \end{bmatrix}\quad$ and $\begin{bmatrix} a^2 & a^2 \\a^2 & 1 \end{bmatrix}\quad$ has order $5$ in $SL(2,4)$. So, $n_5>1$.

Fact$1$: You can easily check that these two matrices $\begin{bmatrix} a^2 & 1 \\0 & a \end{bmatrix}\quad$ and $\begin{bmatrix} a^2 & 0 \\0 & a \end{bmatrix}\quad$ has order $3$ in $SL(2,4)$. So $n_3>1$.

Fact$2$: Any group of order $30$ has subgroup of order $15$.

proof: After showing that it has either normal sylow$3$ subgroup or sylow-$5$ by counting the elements, you can conclude that it has subgroup of order $15$.

Starting: Let $N$ be a proper normal subgroup of $G=SL(2,4)$.

Case$1$: If $5$ divides $|N|$;

Then $H_5\leq N$ and since $N$ is normal it includes all conjugates of $H_5$ so it includes all sylow-$5$ subgroup of $G$. Since $n_5>1 \implies n_5=6 \implies$ there are $6(5-1)=24$ elements with oreder $5$ in $N$ . Thus, we must have $|N|=30$.

Now, Since $3$ divides $|N|$, it also includes all sylow-$3$subgroup.Since $n_3>1$ by fact$1$ then $4\leq n_3$ then $4(3-1)+24=32$ we have more then $30$ contradition.

Case2: If $3$ divides $|N|$;

By case $1$; we can say that $|N|\leq 12$ and since $3$ divides $|N|$ and $N$ is normal, it must includes all sylow $3$ subgroup of $G$ and since $4\leq n_3$ it has atleast $4(3-1)=8$ elements. So we have $8\leq |N| \leq 12$ then we must have $|N|=12$ and $n_3=4$.

Now, Since $n_3=4$ then $|G:N_G(H_3)|=4$ then $N_G(H_3)$ is a group with oreder $15$ and so it is cyclic and it includes a $H_5$ which is normal in $N_G(H_3)$ which means $N_G(H_3)\leq N_G(H_5)$ but since $n_5=6$ then $N_G(H_5)$ has order $10$ contradiction.

Case3: If $2$ divides $|N|$;

By case$1$ and case$2$; we have $|N|\leq 4$

subcase$1$: If $|N|=2$;

Then, $G/N$ is a group of order $30$ and has subgroup $H/N$ of order $15$ by fact$2$. Then, $H$ is a subgroup of $G$ with order $30$. Since $|G:H|=2$ , $H$ is normal in $G$. Since $5$ divides $|H|$, we have contradiction by case$1$.

subcase$2$: If $|N|=4$;

$G/N$ is a group of order $15$ and cyclic and has normal subgroup $H/N$ of order $5$. Then by coresspondence theorem; $H$ is normal subgroup of $G$ with order $20$ . Then $H$ must includes all of elements with order $5$ but there are $6(5-1)=24$ elements of $G$ with order $5$ which is a contradiction.

Conclusion: Now, we can say that it has only trivial normal subgroups,thus $G$ is simple and since $A_5$ is uniqe simple subgroup with order $60$ up to isomorphism, we must have $SL(2,4)\cong A_5$ .