Exercise 4.6.6 (Infinite simple group)

Question

Let D be the subgroup of $S_\Omega$ consisting of permutations which move only a finite number of elements of $\Omega$ (described in Exercise 17 in Section 3) and let $A$ be the set of all elements $\sigma \in D$ such that $\sigma$ acts as an even permutation on the (finite) set of points it moves. Prove that $A$ is an infinite simple group. [Show that every pair of elements of D lie in a finite simple subgroup of D.]

Richard Ganaye · Accepted Answer

\begin{proof} As in Exercise 4.3.17, if $X \subseteq \Omega$, we define
$$F(X) = \{a \in A \mid \forall \sigma \in X,\ \sigma (a) = a\}$$
the fixed set of $X$, and 
$$M(X) = \Omega - F(X)$$
the set of elements of $\Omega$ which are moved by some element of $X$. Moreover
$$D = \{\sigma \in S_\Omega \mid |M(\sigma)| < \infty\}$$
is a (normal) subgroup of $S_\Omega$ by Exercise 4.3.17.

Here $A$ is the set of all elements $\sigma \in D$ such that $\sigma$ acts as an even permutation on the finite set of points it moves.

Since $\Omega$ is an infinite set, there are distinct points $a_1,a_2,\ldots, a_n, \ldots$ in $\Omega$. Then $(a_1\ a_2 \ a_3), (a_1\ a_2\ a_3\ a_4\ a_5), \ldots$, are distinct elements in $A$, so $A$ is infinite.

\bigskip
Suppose that $N \unlhd A$, where $N 
e \{1\}$. We want to prove $N = A$.

Let $	au$ be any element of $A$. There exists some $\sigma \in N, \ \sigma 
e 1$.

\bigskip
Since $\sigma, 	au$ are elements of $D$ then $|M(\sigma)| < \infty$ and $|M(	au)| < \infty$, thus
$$ M(\{\sigma, 	au\}) = M(\sigma) \cup M(	au),$$
the set of elements which are moved by $\sigma$ or $	au$, is finite. We take $C$ a finite set such that $M(\{\sigma, 	au\}) \subseteq C$ and $|C| \geq 5$, so that $\sigma$ and $	au$ fix every element which is not in $C$.

We define
\begin{align*}
H &= \{\sigma \in D \mid M(\sigma) \subseteq C\}\
&=  \{\sigma \in D \mid F(\sigma) \supseteq \Omega - C\}.
\end{align*}
the set of elements of $D$ which fix every element not in $C$.

(Then $\mathrm{id}_\Omega \in H$, and if $\lambda, \mu \in H$, then  $\lambda$ and $\mu$ fix every element of $\Omega - C$, thus $\lambda \mu \in H$, and $\lambda^{-1} \in H$, so $H$ is a subgroup of $D$.)

Note that
$$
\varphi
\left\{
\begin{array}{ccl}
H & 	o & S_C\
\sigma & \mapsto & \sigma |_C
\end{array}
ight.
$$
is an isomorphism, so $H$ is a finite subgroup of $D$. Then $H \cap A = \varphi^{-1}(A_C)$ is the subgroup of $H$ of even permutations in $H$, isomorphic to $A_C$, where $|C| \geq 5$, so $H \cap A$ is a simple group.

By definition of $H$, $\sigma \in H$ and $	au \in H$. Moreover, $\sigma \in N$, thus $\sigma \in A$, so $\sigma$ is even, i.e. $\sigma \in H \cap A$.

Consider the subgroup $H \cap N$ of $H$. Since $N \unlhd A$, then $H \cap N \unlhd H \cap A$, where $H \cap A$ is simple, thus $H \cap N = \{1\}$ or $H \cap N = H \cap A$. But $\sigma \in H \cap N$ and $\sigma 
e 1$, so $H \cap N 
e  \{1\}$. This shows that
$$H \cap N = H \cap A.$$
Since $	au \in H \cap A$, then $	au \in N$. Since this is true for any element $	au \in A$, this proves $A \leq N$, where $N \leq A$, so $N = A$. 

In conclusion, $A$ is an infinite simple group.
\end{proof}

Exercise 4.6.6 (Infinite simple group)

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Exercise 4.6.6 (Infinite simple group)

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