Subgroup of Solvable Group is Solvable

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Theorem

Let $G$ be a solvable group.

Let $H$ be a subgroup of $G$.


Then $H$ is solvable.


Proof 1

Let $\set e = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G$ be a normal series for $G$ with abelian quotients.

For every $i = 1, 2, \ldots, n$ we have:

$\paren {H \cap G_i} \cap G_{i - 1} = H \cap G_{i - 1}$

From the Second Isomorphism Theorem for Groups:

$\dfrac {\paren {H \cap G_i} G_{i - 1} } {G_{i - 1} } \cong \dfrac {H \cap G_i} {\paren {H \cap G_i} \cap G_{i - 1} } = \dfrac {H \cap G_i} {H \cap G_{i - 1} }$

where $\cong$ denotes group isomorphism.

In particular, $H \cap G_{i - 1}$ is a normal subgroup of $H \cap G_i$.

We have that:

$\paren {H \cap G_i} G_{i - 1} \subseteq G_i$

and so from the Correspondence Theorem:

$\dfrac {\paren {H \cap G_i} G_{i - 1} } {G_{i - 1} } \le G_i / G_{i - 1}$


We have that $G_i / G_{i - 1}$ is abelian.

Thus from Subgroup of Abelian Group is Abelian:

$\dfrac {\paren {H \cap G_i } G_{i - 1} } {G_{i - 1} }$ is abelian.

Hence $\dfrac {H \cap G_i} {H \cap G_{i - 1} }$ is abelian.

Therefore, the series :

$\set e = H \cap G_0 \lhd H \cap G_1 \lhd \cdots \lhd H \cap G_n = H$

is a normal series with abelian factor groups for $H$.

Therefore $H$ is solvable.

$\blacksquare$


Proof 2

Firstly we, know that a group is solvable if and only if its derived series:

$\map D G = \sqbrk {G, G} \ , \ \map {D^i} G = \sqbrk {\map {D^{i - 1} } G, \map {D^{i - 1} } G}$

becomes trivial after finite iteration.

Meaning:


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$\map {D^j} G = \set 1$

for some finite $j$.

Now it is trivial that:


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$\map D H \le \map D G$

since $H$ is smaller than $G$.

Further since $\map {D^i} H$ is dominated by $\map {D^i} G$, it too has to become trivial after a finite amount of steps.


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$\blacksquare$


Proof 3

Let $H \leq G$ and $G$ be solvable with normal series:

$\set e = G_0 \lhd G_1 \lhd \dots \lhd G_m = G$

such that $G_{i + 1}/G_i$ is abelian for all $i$.

Define $N_i = G_i \cap H$.

We show that these $N_i$ will form a normal series with abelian factors.


Normality

Let $x \in N_i$ and $y \in N_{i + 1}$.

Then $y x y^{-1} \in N$ since $N$ as a group is closed.


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We also have $y x y^{-1} \in G_i$ since $G_i$ is normal in $G_{i + 1}$.

Hence $N_i$ is invariant under conjugation and therefore normal.


Abelian Factors

Note that:

\(\ds N_{i + 1} \cap G_i\) \(=\) \(\ds \paren {G_{i + 1} \cap H} \cap G_i\) Definition of $N_{i + 1}$
\(\ds \) \(=\) \(\ds H \cap \paren {G_{i + 1} \cap G_i}\) Intersection is Commutative, Intersection is Associative
\(\ds \) \(=\) \(\ds H \cap G_i\) Intersection with Subset is Subset: $G_i \subseteq G_{i + 1}$
\(\ds \) \(=\) \(\ds N_i\) Definition of $N_i$

Thus:

\(\ds \dfrac {N_{i + 1} } {N_i}\) \(=\) \(\ds \dfrac {N_{i + 1} } {N_{i + 1} \cap G_i}\)
\(\ds \) \(\cong\) \(\ds \dfrac {N_{i + 1} G_i} {G_i}\) Second Isomorphism Theorem for Groups
\(\ds \) \(\le\) \(\ds \dfrac {G_{i + 1} } {G_i}\)


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so the quotient $\dfrac {N_{i + 1}} {N_i}$ is isomorphic to a subgroup of the abelian group $\dfrac{G_{i + 1}} {G_i}$.

Hence it is abelian, proving the theorem.

$\blacksquare$


Sources

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