The Sylow 2-subgroup of $SL(2,3)$.

Question

Why is the Sylow 2-subgroup of $SL(2,3)$ normal? I know that $n_2 \in \{1,3\} $, where $n_2$ is the number of Sylow 2-subgroups. But how do I show that $n_2 \neq 3$?

Answer 1

Suppose that there are $3$ Sylow $2$-subgroups, in this case there are at least $3\cdot(8-4)+4=16$ elements of order a power of $2$. Now observe that there are $4$ Sylow $3$-subgroups, so there are $4\cdot(3-1)=8$ elements of order $3$. Therefore every element of $SL(2,3)$ has order $3$ or a power of $2$, ( because $16+8=24$). But the element $$\begin{pmatrix} -1 & -1 \\ 0 & -1 \end{pmatrix}$$ has order 6.

Answer 2

Since I consider the ideas behind the counting of elements of the Sylow 2-subgroups to be non-trivial, I've decided to supplement mrprottolo's answer by elaborating on them for the sake of future readers.

---

Essentially, what the element count is stating is that if there are three Sylow 2-subgroups they must intersect as follows: there are 4 elements common to all three subgroups and each subgroup consists of those 4 common elements together with 4 elements unique to that subgroup. As we shall now demonstrate, this is the only possible structure when a group $G$ of order 24 has 3 Sylow 2-subgroups.

---

We shall denote the three Sylow 2-subgroups $P_1, P_2, P_3$. Firstly, we shall note that it sufficies to show that $|P_1\cap P_2\cap P_3|\geq4$ since, by Lagrange's theorem and the fact that the subgroups are distinct, this necessitates $|P_1\cap P_2\cap P_3|=|P_1\cap P_2|=|P_1\cap P_3|=|P_2\cap P_3|=4$, thus giving the required structure.

We proceed by considering the action of $G$ on the set of its Sylow 2-subgroups by conjugation. Since there are three such subgroups the action induces a homomorphism $\varphi:G\rightarrow S_3$

By the first isomorphism theorem we have: $$|G/ ker\varphi|=|Im\varphi|\leq|S_3|=6 \implies |Ker\varphi|\geq4$$ Second, we observe that, by definition, $x\in Ker\varphi$ if and only if $xPx^{-1}=P$ for every Sylow 2-subgroup $P$. That is, $Ker\varphi$ is exactly $N(P_1)\cap N(P_2)\cap N(P_3)$.

Lastly, by the Sylow thorems, for every Sylow 2-subgroup $P$ of $G$ the index of its normalizer $[G:N(P)]$ is equal to the number of Sylow 2-subgroups in $G$. In our case, this is 3. But that is also the index of the Sylow 2-subgroups themselves, and hence $N(P)=P$ for every Sylow 2-subgroup $P$.

Concluding all of the above results we have: $$Ker\varphi=P_1\cap P_2\cap P_3 \implies |P_1\cap P_2\cap P_3|= |ker\varphi| \geq 4$$ as desired.