Subgroup of Solvable Group is Solvable/Proof 1

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Let $G$ be a solvable group.

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

Then $H$ is solvable.


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.