Let $A$ be a non-empty set and let $X$ be a subset of $S_A$
Now let $F(X) = \{a \in A : \sigma(a) = a, \forall \sigma \in X\}$, $M(X) = A\backslash F(X)$, and $D = \{ \sigma \in S_A : \mid M(\sigma)\mid < \infty \}$
Prove $D$ is a normal subgroup. The straight forward one-step or two-step subgroup test with added normal subgroup test seems 'easy' enough but I am trying to do it via showing $D$ is the kernal of some group homomorphism
I am a little unsure on what mapping to take, something like
$\varphi : S_A \times A \to S_A$ defined by $(\sigma, a) \mapsto \sigma a \sigma^{-1}$
HINT: Just prove it directly. Suppose that $\sigma\in D$ and $\tau\in S_A$, and consider which elements of $A$ are moved by $\tau\sigma\tau^{-1}$. Specifically, what does $\tau\sigma\tau^{-1}$ do to the set $\tau[F(\sigma)]$?