M11

Let $\Delta = \{\delta_1, \delta_2, \delta_3, \delta_4, \delta_5\}$ and let $\Omega$ be the unordered pairs of these. We will use the names:

• 0: $\{\delta_1,\delta_2\}$,  1: $\{\delta_1,\delta_3\}$,  2: $\{\delta_1,\delta_4\}$
• 3: $\{\delta_1,\delta_5\}$,  4: $\{\delta_2,\delta_3\}$,  5: $\{\delta_2,\delta_4\}$
• 6: $\{\delta_2,\delta_5\}$,  7: $\{\delta_3,\delta_4\}$,  8: $\{\delta_3,\delta_5\}$
• 9: $\{\delta_4,\delta_5\}$

Let $G = A_\Delta$ (basically $A_5$) this acts transitively on $\Omega$!

• $a = (\delta_1\;\delta_2)(\delta_3\;\delta_5) = (1\;6)(2\;5)(3\;4)(7\;9)$
• $b = (\delta_1\;\delta_2)(\delta_4\;\delta_5) = (1\;4)(2\;6)(3\;5)(7\;8)$
• $g_2 = (\delta_2\;\delta_3)(\delta_4\;\delta_5) = (0\;1)(2\;3)(5\;8)(6\;7)$
• $g_3 = (\delta_1\;\delta_4)(\delta_2\;\delta_3) = (0\;7)(1\;5)(3\;9)(6\;8)$

you can clearly combine these to take 1 anywhere, which proves transitivity.

The stabilizer of 0 has size $|G_0| = |G|/|\Omega| = 60/10 = 6$ (orbit stabilizer) so $G_0 = \langle a, b \rangle = \{1,a,b,ab,ba,aba=bab\}$ and the $G_0$ orbits are:

• $\{0\}$
• $\{1,2,3,4,5,6\}$
• $\{7,8,9\}$

the double cosets corresponding to these suborbits are represented by $1,g_2$ and $g_3$.

gap> a:=(1,2)(3,5);;b:=(1,2)(4,5);;
gap> DoubleCosets(Group((1,2,3,4,5),(1,2,3)),Group(a,b),Group(a,b));
[ DoubleCoset(Group( [ (1,2)(3,5), (1,2)(4,5) ] ),(),Group( 
    [ (1,2)(3,5), (1,2)(4,5) ] )), 
  DoubleCoset(Group( [ (1,2)(3,5), (1,2)(4,5) ] ),(2,3)(4,5),Group( 
    [ (1,2)(3,5), (1,2)(4,5) ] )), 
  DoubleCoset(Group( [ (1,2)(3,5), (1,2)(4,5) ] ),(1,3)(2,4),Group( 
    [ (1,2)(3,5), (1,2)(4,5) ] )) ]

so take $\infty \not\in \Omega$ and set $\Omega^+ = \Omega \cup \{\infty\}$ then choose $$x = (\infty\;0)(2\;3)(4\;6)(8\;9)$$ then

• $x^2 = 1 \in G_0$
• $a^x = b$ and (since $x$ has order 2) $b^x = a$ so $G_0^x = G_0$.
• $g_2^x = (\infty\;1)(2\;3)(5\;9)(4\;7)=x^{g_2} \in G x G$.
• $g_3^x = (\infty\;7)(1\;5)(2\;8)(4\;9)=x^{g_3} a^b \in G x G$.

Thus $L = \langle G, x \rangle$ is a one point extension. It has order $660 = 11 \cdot 10 \cdot 6$ and is 2-transitive (hence a primitive permutation group). We saw that $G$ ($A_5$) - the stabilizer of $\infty$ in $L$ - is simple so if $L$ had a regular normal subgroup (with a transitive action on the 10+1 points, with all its stabilizers trivial so..) of order 11 which must (by Lagrange) be $C_{11}$. $|N_{S_{11}}(C_{11})| = 110$ (by counting cycles) which is strictly less than 660 so it can't be normal. Therefore by the proposition we have:

Theorem $L$ is simple.

$1$ and $x$ are $(G,G)$-double coset reps in $L$. Take $\omega \not \in \Omega^+$ and form $\Omega^\star = \Omega^+ \cup \{\omega\}$. $$y=(\omega\;\infty)(1\;4)(2\;5)(3\;6)$$ then

• $y^2 = 1 \in G = L_{\infty}$
• $G^y = G$ since $G = \langle a,b,g_2,g_3 \rangle$ and $a^y = a$, $b^y = b$, $g_2^y = g_2^b$, $g_3^y = g_3 a^b$.
• $x^y = y^x a^b \in L y L$

Therefore we have another one point extension $M_{11} = \langle L, y \rangle$, which is 3-transitive of order 7920 acting on 12 points: so it can't have any regular normal subgroup from the corollary as none of these groups have order 12) but since $L$ is simple by the proposition we have:

Theorem $M_{11}$ is simple.

From Jordan's theorem we see that we cannot perform another one point extension.

• Dodecahedral Faces of M_12