Subgroup of Solvable Group is Solvable/Proof 3

Theorem

Let $G$ be a solvable group.

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

Then $H$ is solvable.

Proof

Let $H \le G$ 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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