Exercise 4.3.17 ($D = \{\sigma \in S_A \mid |M(\sigma)| < \infty\}$ is a normal subgroup of $S_A$)

Proof. If $σ \in D$ , then $σ$ fixes all the elements of $A$ , except for a finite number of them, i.e., the elements of $M (σ)$ .

Let $σ, τ$ be any elements of $S_{A}$ . Put $σ^{'} = 𝜏𝜎 τ^{- 1}$ .

If $a \in F (σ)$ , then
$σ^{'} (τ (a)) = (𝜏𝜎 τ^{- 1}) (τ (a)) = τ (σ (a)) = τ (a),$
so $τ (a) \in F (σ^{'})$ .
If $a \in M (σ)$ , then $σ (a) \neq a$ , and $τ$ is injective, thus $τ (σ (a)) \neq τ (a)$ . Therefore
$σ^{'} (τ (a)) = τ (σ (a)) \neq τ (a),$
so $τ (a) \in M (σ^{'})$ .

Since $M (σ)$ and $F (σ)$ are complementary sets, this shows that for all $a \in A$ ,

a \in M (σ) ⟺ τ (a) \in M (σ^{'}) :

if $τ (a) \in M (σ^{'})$ , then $τ (a) \notin F (σ^{'})$ , thus $a \notin F (σ)$ , so $a \in M (σ)$ .

This proves $M (σ) = τ^{- 1} (M (σ^{'}))$ , or equivalently

\begin{align} τ (M (σ)) = M (𝜏𝜎 τ^{- 1}) (σ, τ \in S_{A}) . & (1) \end{align}

Since $τ$ is bijective, $| M (σ) | < \infty ⟺ | τ (M (σ)) | < \infty$ Therefore, for all $σ \in S_{A}$ ,

\begin{array}{l} σ \in D & ⟺ | M (σ) | < \infty \\ ⟺ | τ (M (σ)) | < \infty \\ ⟺ | M (𝜏𝜎 τ^{- 1}) | < \infty & (by (1)) \\ ⟺ 𝜏𝜎 τ^{- 1} \in D . \end{array}

Consequently, $D$ is a normal subgroup of $S_{A}$ . □

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\begin{proof} 
If $\sigma \in D$, then $\sigma$ fixes all the elements of $A$, except for a finite number of them, i.e., the elements of $M(\sigma)$.

Let $\sigma, \tau$ be any elements of $S_A$. Put $\sigma' = \tau  \sigma \tau^{-1}$.
\begin{itemize}
\item If $a \in F(\sigma)$, then
$$\sigma'(\tau(a)) = (\tau \sigma \tau^{-1})(\tau(a)) = \tau(\sigma(a)) = \tau(a),$$
so $\tau(a) \in F(\sigma')$.

\item If $a \in M(\sigma)$, then $\sigma(a) \ne a$, and $\tau$ is injective, thus $\tau(\sigma(a)) \ne \tau(a)$. Therefore
$$\sigma'(\tau(a)) = \tau(\sigma(a)) \ne \tau(a),$$
so $\tau(a) \in M(\sigma')$.
\end{itemize}
Since $M(\sigma)$ and $F(\sigma)$ are complementary sets, this shows that for all  $a \in A$,
$$a \in M(\sigma) \iff \tau(a) \in M(\sigma'):$$
if $\tau(a) \in M(\sigma')$, then $\tau(a) \not \in F(\sigma')$, thus $a \not \in F(\sigma)$, so $a \in M(\sigma)$.

This proves $M(\sigma) = \tau^{-1}(M(\sigma'))$, or equivalently
\begin{align}
\tau(M(\sigma)) = M(\tau \sigma \tau^{-1})\qquad  (\sigma, \tau \in S_A).
\end{align}
Since $\tau$ is bijective, $|M(\sigma)| < \infty \iff |\tau(M(\sigma))| < \infty$ Therefore, for all $\sigma \in S_A$,
\begin{align*}
\sigma \in D &\iff  |M(\sigma)| < \infty\\
&\iff |\tau(M(\sigma))| < \infty\\
&\iff |M(\tau \sigma \tau^{-1})| < \infty&(\text{by (1)})\\
&\iff \tau \sigma \tau^{-1} \in D.
\end{align*}

Consequently, $D$ is a normal subgroup of $S_A$.

\end{proof}

Exercise 4.3.17 ($D = \{\sigma \in S_A \mid |M(\sigma)| < \infty\}$ is a normal subgroup of $S_A$)

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