Subgroup of a group of bijections

Question

Let $G$ be the group of bijections of the set $X$, which is assumed to have more than two elements. Let $x_0$ ∈ $X$ and $H$ = { $f$ ∈ $G$: $f$( $x_0$)= $x_0$}. Is $H$ normal or not?

Answer 1

There is a stronger statement. Note that if $|X|=n$, then the set of bijections from X to X form a group, which will be isomorphic to the symmetric group of order $n$. Further, the symmetric group contains only one normal subgroup called the alternating subgroup, which has exactly half the size ( and hence is normal). The characterization of the subgroup you have given would mean that it has $(n-1)!$ order, which isn't half of $n!$, the total order. Further, this alternating subgroup exists only when $n \geq 5$, so if $|X|<=4$, then anyway you will be wrong, since the group then won't have any normal subgroup (called simple), except if $|X| = 4$, but here the normal subgroup is of order $4$, while we would like one of order $6$.

If $|X|$ is infinite, then the finitary symmetric group properties will assert the same, i.e. only the alternating group is a normal group, and it won't be equal to the set you have specified above.