Why is $SL_2(\mathbb F_4)\cong \operatorname{Alt}(5)$? What's wrong with my work?

Question

Why is $SL_2(\mathbb F_4)\cong \operatorname{Alt}(5)$? They both have order $60$ and if a group of order $60$ has more than one Sylow $5$-subgroup, it must be $\operatorname{Alt}(5)$. But I just can't find an element in $SL_2(\mathbb F_4)$ which is of order $5$, let alone two of them! What's more, there's no element in $\operatorname{Alt}(5)$ of order $4$, $\begin{pmatrix}1&1\\0&1\\\end{pmatrix}\in SL_2(\mathbb F_4)$ is an element of order $4$. There has to be something wrong here, which I can't figure out. What's wrong here though?

Answer 1

The problem seems to be not knowing what the field $\Bbb{F}_4$ looks like. Most notably, you should not confuse it with the residue class ring $\Bbb{Z}/4\Bbb{Z}$. That is not a field, because the class $\overline{2}$ has no multiplicative inverse.

The field $\Bbb{F}_4$ is an extension field of $\Bbb{F}_2$, in other words $\Bbb{F}_2\subset \Bbb{F}_4$ in such a way that both addition and multiplication are extended. This extension field can simply be described as

$$\Bbb{F}_4=\{0,1,\omega,\omega+1\},$$ where the arithmetic is determined by the obvious properties of $0$ and $1$, and the relations $1+1=0$ and $\omega^2=\omega+1$ (consequently $\omega^3=1$). If you are familiar with the method of constructing extension fields as quotient rings of the ring of polynomials over the base field, then you can use $$\Bbb{F}_4\simeq\Bbb{F}_2[x]/\langle x^2+x+1\rangle,$$ where the isomorphism sends $\omega\mapsto x+\langle x^2+x+1\rangle$. Here it is crucial that $x^2+x+1$ is an irreducible polynomial in $\Bbb{F}_2[x]$.

With these misunderstanding out of the way you can quickly verify that the set of matrices of the form $$ \left(\begin{array}{cc}1&x\\0&1\end{array}\right) $$ with $x$ ranging over $\Bbb{F}_4$ gives you a Sylow 2-subgroup isomorphic to the Klein 4-group. All the matrices of this form commute with each other, and have squares equal to $I_2$.

On the other hand, the matrix $$ B=\left(\begin{array}{cc}0&1\\1&\omega\end{array}\right) $$ can be shown to be of order five (check it to gain familiarity with the arithmetic of $\Bbb{F}_4$!). I constructed it as a companion matrix of the polynomial $$ p(x)=x^2+\omega x+1\in\Bbb{F}_4[x]. $$ If $\alpha$ is a zero of $p(x)$, we can check, repeatedly using the fact $p(\alpha)=0$, that $$ \begin{aligned} \alpha^2&=\omega\alpha+1,\\ \alpha^3&=\alpha\alpha^2=\omega\alpha^2+\alpha=(\omega^2+1)\alpha+\omega=\omega\alpha+\omega,\\ \alpha^4&=\alpha\alpha^3=\omega\alpha^2+\omega\alpha=(\omega+\omega^2)\alpha+\omega=\alpha+\omega,\\ \alpha^5&=\alpha\alpha^4=\alpha^2+\omega\alpha=1. \end{aligned} $$ So the zeros of $p(x)$ are two elements of order five in an extension field of $\Bbb{F}_4$.