Find the number of elements in the set: $A=\{\sigma\in S_4 |\thinspace \sigma\thinspace(3)=3\}$
I know that this would be $3!=6$. But are these the correct elements?
$$ \{e, (12), (24), (14), (142), (124)\} $$
I have written them in disjoint cyclic groups aside from $e$.
So if we have $\sigma\thinspace(a)=a$, in a group such as $S_4$, we would take a out $a$ from the group and be left with $\{b,c,d\}$. And thus we have $3!=6$ elements?
Yes, you are correct. There are several ways to think about this. Firstly, you could note that your subgroup just consists of the permutations of $1,2,4$, which is a copy of $S_3$. Secondly, you could argue that the orbit of $3$ under the action of $S_4$ on $\{1,2,3,4\}$ has $4$ elements (as we can send $3$ to anything with a suitable permutation). By the orbit-stabilizer theorem, the stabilizer of $3$ has $4!/4=6$ elements. Since you have listed $6$ distinct elements in the stabilizer, you are done.