Notes - Galois Theory HT25, Solvable groups

This content isn’t in the course notes but instead in the [[Part A]]^U short-option Group Theory (which I didn’t take). Most of the definitions here come from [[Abstract Algebra, Judson]]^N.

Explain the steps that allow you to conclude that the alternating group $A _ n$ is simple for all $n \ge 5$.

If $N \trianglelefteq A _ n$ for $n \ge 4$ and $N$ contains a 3-cycle, then $N = A _ n$.
For $n \ge 5$, every nontrivial normal subgroup $N$ of $A _ n$ contains a 3-cycle.
Therefore $A _ n$ is simple for $n \ge 5$.

@Define a subnormal series of a group $G$.

A finite sequence of subgroups

\[G = H_n \trianglerighteq H_{n-1} \trianglerighteq \cdots \trianglerighteq H_1 \trianglerighteq H_0 = \{e\}\]

where each $H _ i$ is a normal subgroup of $H _ {i+1}$.

@Define a normal series of a group $G$.

A finite sequence of subgroups

\[G = H_n \trianglerighteq H_{n-1} \trianglerighteq \cdots \trianglerighteq H_1 \trianglerighteq H_0 = \{e\}\]

where each $H _ i$ is a normal subgroup of $H _ {i+1}$ (i.e. a subnormal series), and each $H _ i$ is a normal subgroup of $G$.

What is special about a series of subgroups of any abelian group?

It is a normal series.

Suppose:

@Define what it means for these two series to be isomorphic.

There is a one-to-one correspondence between $\{H _ {i+1} / H _ i\}$ and $\{K _ {j+1} / K _ j\}$.

Suppose $\{H _ i\}$ is a subnormal series of a group $G$. @Define what it means for this series to be a composition series.

All the factor groups are simple (equivalently, none of the factor groups contain a normal subgroup).

Suppose $\{H _ i\}$ is a normal series of a group $G$. @Define what it means for this series to be a principal series.

All the factor groups are simple (equivalently, none of the factor groups contain a normal subgroup).

Give a composition series for $S _ n$ where $n \ge 5$.

\[S_n \trianglerighteq A_n \trianglerighteq \{(1)\}\]

since

\[S_n / A_n \cong \mathbb Z_2\]

and both $\mathbb Z _ 2$ and $A _ n$ are simple (this is where $n \ge 5$ is used).

@State the Jordan-Hölder theorem.

Any two composition series of a group $G$ are isomorphic.

@Define what it means for a group $G$ to be solvable.

It has a subnormal series $\{H _ i\}$ such that all the factor groups $H _ {i+1} / H _ i$ are abelian.

We have the definition that:

$G$ is solvable if it has a subnormal series $\{H _ i\}$ such that all the factor groups $H _ {i+1} / H _ i$ are abelian.

@Justify why $S _ 4$ is solvable.

\[S_4 \trianglerighteq A_4 \trianglerighteq \{(1), (12)(34), (13)(24), (14)(23)\} \trianglerighteq \{(1)\}\]

and each intermediate factor group is abelian.

We have the definition that:

$G$ is solvable if it has a subnormal series $\{H _ i\}$ such that all the factor groups $H _ {i+1} / H _ i$ are abelian.

@Justify why $S _ 5$ is not solvable.

\[S_n \trianglerighteq A_n \trianglerighteq \{(1)\}\]

is a composition series for $S _ 5$ (the only composition series), but $A _ n$ is not abelian.