Normal Subgroups

Normal Subgroup

A subgroup

H

G

is called a normal subgroup of

G

a H = H a

for every

a \in G

, and we write

H ⊴ G

Normal Subgroup Test

A subgroup

H

G

is normal iff

x H x^{- 1} \subseteq H

for every

x \in G

Quotient Group

Suppose that

G

is a group and that

H ⊴ G

. Then we define

G / H : = {a H : a \in G}

Some may call a quotient group a factor group

Solvable

Show that

S_{3}

is solvable

We'll consider $G_{0} = {e}, G_{1} = {e, (123), (132)}, G_{2} = S_{3}$ . Recall that from group theory that if $H$ is a subgroup of $G$ and that the index of $H$ in $G$ is 2, then $H$ is a normal subgroup of $G$ , therefore in our context we can observe that $G_{1}$ is a normal subgroup of $S_{3}$ as it has index 2.

It's noted that trivially ${e}$ is a normal subgroup of $G_{1}$ . We'll now move to showing that $G_{0} / G_{1}$ and $G_{1} / G_{2}$ are abelian, for the first we just have to realize that $G_{1}$ is abelian itself because $(123) (321) = (321) (123) = e$ and the rest of the comparisons are trivial, so also $G_{1} / G_{0}$ is abelian. For $S_{3} / G_{1}$ , we can actually find that this is isomorphic to $D_{6}$ which we already know is abelian, therefore by definition we've shown that $S_{3}$ is solvable

The Symmetric Group for n Greater Than or Equal to 5 Is not Solvable

S_{n}

for

n \geq 5

is not solvable

Suppose for the sake of contradiction that

S_{n}

is solvable, therefore there is a chain of subgroups of the form

{e} ⩽ G_{1} \dots G_{n - 1} ⩽ G_{n} = S_{n}

wherein consecutive quotients are abelian, if we consider

G_{n} / G_{n - 1}

, then if

G_{n}

contains all the three cycles, then by what we proved proviously so does

G_{n - 1}

. Then if

G_{n - 1}

contains all the three cycles so does

G_{n - 2}

continuing on, we'll eventually arrive at a contradiction, therefore

S_{n}

is not solvable.