Exercise 14.2.3

Proof. (a) Let $\phi :G\to S_3$ defined by $\tau =\phi (\sigma )$ iff $\sigma (R_i)=R_{\tau (i)}$. In other notations, this is the restriction to $G$ of the homomorphism of part (b) of Lemma 14.2.8, thus $\phi$ is an homomorphism.

$\text{∙}$ $\phi$ is surjective: Let $\tau$ be any permutation in $S_3$.

If $\tau$ is even, $\tau ={(123)}^k,k=0,1,2$. Let

$\sigma$ preserves the block structure defined by $R_1,R_2,R_3$, and $\sigma \in A_6$, so that $\sigma \in G=(S_3\wr S_2)\cap A_6$. Moreover $\sigma (R_1)=R_2,\sigma (R_2)=R_3,\sigma (R_3)=R_1$, thus $\phi (\sigma )=(123)$, and $\phi ({\sigma }^k)={(123)}^k=\tau$.

If $\tau$ is odd, then $\tau \in \{(12),(23),(13)\}$, and

Therefore $\phi$ is surjective.

$\text{∙}$ Let $\sigma \in S_6$. Then $\sigma \in \ker <!--\; FUNCTION\; APPLICATION\; -->\phi$ iff $\sigma \in A_6$ and $\sigma (R_1)=R_1,\sigma (R_2)=R_2,\sigma (R_3)=R_3$.

Morerover, for all $\sigma \in A_6$,

Verification: $6=|S_3|=|G∕\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )|=24∕4$. (b) Let $N$ be the number of $3$-Sylow subgroups of $G$. By the third Sylow Theorem,

Therefore $N=1$ or $N=4$. (c) Let $\tau \in S_6$ be a permutation of order 6. If $\tau ={\tau }_1\cdots {\tau }_k$ is the decomposition of $\tau$ in disjoint cycles, then the order of $\tau$ is the lcm of the order of ${\tau }_1,\dots ,{\tau }_k$. Therefore $\tau$ is a $6$-cycle or a product of a $2$-cycle by a $3$-cycle. In both cases $\tau$ is odd. Therefore $A_6$ has no element of order $6$. (d) Reasoning by contradiction, suppose that $G$ has only one $3$-Sylow subgroup $H$. Then, for all $g\in G$, $\mathit{gH}{g}^{-1}$ is a $3$-Sylow, thus $\mathit{gH}{g}^{-1}=H$, and $H$ is a normal subgroup of $G$.

Moreover $K=\ker <!--\; FUNCTION\; APPLICATION\; -->\phi =\{e,(12)(34),(12)(56),(34)(56)\}$ is normal in $G$, and has order 4. Therefore $H\cap K=\{e\}$.

The usual characterization of direct products (see Ex. 14.3.7) shows that, for all $h\in H$, all $k\in K$, $\mathit{hk}=\mathit{kh}$, and $\mathit{HK}$ is a normal subgroup of $G$ isomorphic to $H\times K$.

Take $h$ an element of order $3$ in $H,$ and $k$ and element of order 2 in $K$. Since $\mathit{kh}=\mathit{hk}$, the order of $\mathit{hk}\in A_6$ is 6, which is impossible by part (c).

Therefore $G$ has exactly four $3$-Sylow subgroups. (e) Write $X=\{H_1,H_2,H_3,H_4\}$ the set of $3$-Sylow subgroups of $G$, and $S(X)$ the set of permutations of $X$. Then $S(X)\simeq S_4$, and $g\cdot H=\mathit{gH}{g}^{-1}$ defines a left action of $G$ on $X$, so that

is a group homomorphism.

It is not obvious that $\psi$ is bijective. We prove first that $\psi$ is surjective (onto). We give explicitly the $3$-Sylow subgroups. Let

4 Then ${\lambda }_1,\dots ,{\lambda }_4\in G$ have order $3$, and $H_1=⟨{\lambda }_1⟩=\{e,{\lambda }_1,{\lambda }_1^2\},\dots ,H_4=⟨{\lambda }_4⟩=\{e,{\lambda }_4,{\lambda }_4^2\}$ are distinct, thus they are the four $3$-Sylow of $G$.

Now take

(We give a geometrical explanation of this choice in the final note.)

Then

thus $g{H}_1{g}^{-1}=H_2$, and since $g=g^{-1}$, $g{H}_2{g}^{-1}=H_1$. Moreover

thus $g{H}_3{g}^{-1}=H_3$, and since $\psi (g)$ is a permutation, $g{H}_4{g}^{-1}=H_4$.

Therefore $\psi (g)\in S(X)$ is the permutation $\left(\begin{array}{cccc}\hfill H_1\hfill & \hfill H_2\hfill & \hfill H_3\hfill & \hfill H_4\hfill \\ \hfill H_2\hfill & \hfill H_1\hfill & \hfill H_3\hfill & \hfill H_4\hfill \end{array}\right)$, which corresponds to the transposition $(12)\in S_4$. Similarly,

thus $h{H}_1{h}^{-1}=H_2,h{H}_2{h}^{-1}=H_3,h{H}_1{h}^{-1}=H_4$, and since $\psi (g)$ is a permutation, $h{H}_4{h}^{-1}=H_1$. Therefore $\psi (g)=\left(\begin{array}{cccc}\hfill H_1\hfill & \hfill H_2\hfill & \hfill H_3\hfill & \hfill H_4\hfill \\ \hfill H_2\hfill & \hfill H_3\hfill & \hfill H_4\hfill & \hfill H_1\hfill \end{array}\right)$corresponds to the $4$-cycle $(1234)$.

Since $\{(12),(1234)\}$ is a set of generators of $S_4$, $S(X)$ is generated by $\psi (g),\psi (h)$, so that $S(X)=\psi (G)$, and $\psi$ is surjective. Moreover, $|G|=|S(X)|=24$, thus $\psi$ is a bijection, and a group isomorphism:

(f) To conclude, using Exercise 2, we obtain

Note: We have proved in Exercise 7.5.10 that the symmetry group $G_0$ of the cube (or octahedron), is isomorphic to $S_4$. By composition with the indirect isometry $\sigma :v\mapsto -v$, which commutes with all elements in the group, we obtain the full symmetry group, isomorphic to $S_4\times S_2$.

We have a geometrical description of $G=(S_3\wr S_2)\cap A_6$ by regrouping the opposite faces of a cube in blocs: stick $1$ on a face of a dice, $2$ on the opposite face, and so on (I stuck labels on my Rubik’s cube). Then the $24$ rotations of the cube send opposite faces on opposite faces, so that the bloc structure $\{\{1,2\},\{3,4\},\{5,6\}\}$ is preserved by rotations.

We have proved in Exercise 7.5.10 that $G_0$ acts on the 4 long diagonals $D_1,D_2,D_3,D_4$ of the cube, so that $G_0\simeq S_4$. Each of the four $3$-Sylow of $G_0$ is generated by the rotation with angle $\frac{2\pi }{3}$ around such a long diagonal. They correspond to the $3$-Sylow $H_1,\dots ,H_4$ of $G$: this was useful for the above description of the $H_i$. Each $3$-Sylow corresponds to a long diagonal, so that $g{H}_i{g}^{-1}=H_j$ is equivalent to $\sigma (D_i)=D_j$, where $\sigma$ corresponds to $g$. It remains to find a rotation which acts on these diagonals as some given permutation in $S_4$, such that $(12)$ or $(1234)$. The corresponding permutations $g,h\in G$ are given in the text. □