Subgroup of Solvable Group is Solvable/Proof 3

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Theorem

Let $G$ be a solvable group.

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


Then $H$ is solvable.


Proof

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.

$\blacksquare$


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