The set of permutations which move a finite number of elements is normal

Let A be a nonempty set and let X \subseteq S_A. Let F(X) = \{ a \in A \ | \sigma(a) = a\ \mathrm{for\ all}\ \sigma \in X \} be the fixed set of X. Let M(X) = A \setminus F(X) be the elements which are moved by some element of X.

Define D = \{ \sigma \in S_A \ |\ |M(\sigma)| < \infty \}. Prove that D is a normal subgroup of S_A.

The identity permutation moves zero objects, and so 1 \in D. Now suppose \sigma,\tau \in D. Now M(\tau^{-1}) = M(\tau), and M(\sigma \circ \tau) \subseteq M(\sigma) \cup M(\tau), so that D is closed under identity, inversion, and multiplication, and hence is a subgroup.

Next we prove some lemmas.

Lemma 1: For all \sigma \in S_A and B \subseteq A, \sigma[A \setminus B] = A \setminus \sigma[B]. Proof: (\supseteq) Let x \in A \setminus \sigma[B]. Then x \notin \sigma[B], hence \sigma^{-1}(x) \notin B. Thus \sigma^{-1}(x) \in A \setminus B, and so x \in \sigma[A \setminus B]. (\subseteq) Let x \in \sigma[A \setminus B]. Then x = \sigma(y) for some y \in A \setminus B. So \sigma^{-1}(x) \notin B, hence x \notin \sigma[B], and thus x \in A \setminus \sigma[B]. \square

Lemma 2: If F(X) is defined as above, then for all \tau, \sigma \in S_A, \tau[F(\sigma)] = F(\tau \sigma \tau^{-1}). Proof: (\subseteq) If x \in \tau[F(\sigma)], then x = \tau(y) for some y \in F(\sigma). Thus \tau^{-1}(x) \in F(\sigma). Now (\tau\sigma\tau^{-1})(x) = \tau(\sigma(\tau^{-1}(x))) = (\tau\tau^{-1})(x) = x, so that x \in F(\tau\sigma\tau^{-1}). (\supseteq) If x \in F(\tau\sigma\tau^{-1}), then (\tau\sigma\tau^{-1})(x) = x. Thus \sigma(\tau^{-1}(x)) = \tau^{-1}(x), so that \tau^{-1}(x) \in F(\sigma). Hence x \in \tau[F(\sigma)]. \square

Lemma 3: If M(X) is defined as above, then for all \tau, \sigma \in S_A, M(\tau \sigma \tau^{-1}) = \tau[M(\sigma)]. Proof: We have M(\tau \sigma \tau^{-1}) = A \setminus F(\tau \sigma \tau^{-1}) = A \setminus \tau[F(\sigma)] = \tau[A \setminus F(\sigma)] = \tau[M(\sigma)]. \square

Now to the main result.

Let \sigma \in D and \tau \in S_A. Then M(\tau \sigma \tau^{-1}) = \tau[M(\sigma)] by Lemma 3. Since M(\sigma) is finite and \tau is a bijection, M(\tau \sigma \tau^{-1}) is also finite. Hence \tau\sigma\tau^{-1} \in D. Thus D \leq S_A is normal.

Post a comment or leave a trackback: Trackback URL.


  • Gobi Ree  On December 10, 2011 at 10:46 am

    It will be needed to check that D is a subgroup.

    • nbloomf  On December 12, 2011 at 11:43 am

      You’re right. Thanks!

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: