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.
- Subnormal series: A finite sequence of nested normal subgroups.
- Normal series: A subnormal series, but each subgroup is also a normal subgroup of the biggest group.
- Composition series: A subnormal series, and each factor group (quotients of adjacent groups in the series) are simple (contain no normal subgroups).
- Principal series: A normal series, and each factor group are simple (i.e. a composition series, but there’s got to be an underlying normal series).
- Solvable group: A group is solvable if it has a subnormal series such that all factor groups are abelian.
Flashcards
Basic definitions
@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 $G$.
Suppose:
- $\{H _ i\}$ is a subnormal series of a group $G$
- $\{K _ j\}$ is a subnormal series of a group $G$
@Define what it means for these two series to be isomorphic.
There is a bijection between the two groups appearing in each of the series such that each pair of factor groups $\{H _ {i+1} / H _ i\}$ and $\{K _ {j+1} / K _ j\}$ are isomorphic.
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).
@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.
Operations that preserve solvability
@Justify that if $G$ is solvable and $H$ is a subgroup of $G$, then $H$ is solvable.
Take the subnormal series with abelian quotients for $G$, and intersect each of the subgroups with $H$.
@Justify that if $G$ is solvable and $H$ is a normal subgroup of $G$, then $G/H$ is solvable.
Take the subnormal series with abelian quotients for $G$, and quotient each by $H$.
@Justify that if:
- $G$ is a group
- $H \trianglelefteq G$ is a normal subgroup
- $H$ is solvable
- $G/H$ is solvable
then:
- $G$ is solvable
Let $\pi : G \to G/H$ be the quotient homomorphism and consider
\[G \trianglerighteq \pi^{-1}(J_1) \trianglerighteq \cdots \trianglerighteq \pi^{-1}(J_n) = H \trianglerighteq H_1 \trianglerighteq \cdots \trianglerighteq \lbrace e \rbrace\]where
\[G/H \trianglerighteq J_1 \trianglerighteq \cdots \trianglerighteq J_n\]and
\[H \trianglerighteq H_1 \trianglerighteq \cdots \trianglerighteq H_n\]are subnormal series with abelian quotients for $H$ and $G/H$ respectively.
@Justify that if:
- $G$ is solvable
- $\varphi : G \to H$ is a surjective group homomorphism
Then:
- $\varphi(G)$ is solvable
Suppose
\[G = G_0 \triangleright G_1 \triangleright \cdots \triangleright G_n = \lbrace e \rbrace\]is a subnormal series with abelian quotients, we can apply $\varphi$ to obtain
\[\text{Gal}_F(m_{F, \alpha}) = \varphi(G_0) \triangleright \varphi(G_1) \triangleright \cdots \triangleright \varphi(G_n) = \lbrace e \rbrace\]and note that each $\varphi(G _ i)$ is normal in $\varphi(G _ {i-1})$ and the quotients remain abelian.
Jordan-Hölder theorem
@State the Jordan-Hölder theorem.
Any two composition series of a group $G$ are isomorphic.
Classes of solvable groups
What is special about a series of subgroups of any abelian group, and thus what can be said about the solvability of any abelian groups?
It is a normal series, and every abelian group is solvable.
$S _ n$ is solvable for $n < 5$, and not solvable for $n \ge 5$
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 $n \ge 5$.
Give a composition series for $S _ n$ where $n \ge 5$.
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).
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.
$G$ is solvable if it has a subnormal series $\{H _ i\}$ such that all the factor groups $H _ {i+1} / H _ i$ are abelian.
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.
$G$ is solvable if it has a subnormal series $\{H _ i\}$ such that all the factor groups $H _ {i+1} / H _ i$ are abelian.
is a composition series for $S _ 5$ (the only composition series), but $A _ 5$ is not abelian.