Subgroup of Solvable Group is Solvable/Proof 1
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Theorem
Let $G$ be a solvable group.
Let $H$ be a subgroup of $G$.
Then $H$ is solvable.
Proof
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.
$\blacksquare$