Let $G$ be some group and $H$ be a normal group. Prove that $G$ is solvable if and only if $H$ is solvable and $G / H$ is solvable.
This is not quite trivial. The manual answer for solvability of $H$ is straightforward:
Suppose $G$ be solvable. That is, there exists a tower of $G$ - a sequence of normal sub-groups of $G$ that contains each other consecutively. Namely, there exists $G_1, ..., G_m$ such that
$$G = G_0 \supseteq G_1 \supseteq ... \supseteq G_m$$
For each $i = 0, ..., m - 1$, $G_i / G_{i + 1}$ is abelia(of course, $G_{i + 1} \triangleleft G_{i}$)
Then, consider $H_i = H \cap G_i$. It is obvious that $$H = H_0 \supseteq H_1 \supseteq ... \supseteq H_{m}$$ Furthermore, $$H \triangleleft H \Rightarrow H \triangleleft (H \cap G_i)$$ $$G_{i + 1} \triangleleft G_i \Rightarrow G_{i + 1} \triangleleft (H \cap G_i)$$
So, it must hold that $$(H \cap G_{i + 1}) \triangleleft (H \cap G_i) \Leftrightarrow H_{i + 1} \triangleleft H_i$$ This is the key argument: Also, there exists an injective map(inclusion map, specifically) from $$(H \cap G_{i}) / (H \cap G_{i + 1}) \rightarrow G_{i} / G_{i + 1}$$ Since $G_{i} / G_{i + 1}$ is abelian, so is $H_{i} / H_{i + 1}$. Thus, $H$ is indeed solvable.
I would like to repeat this argument for $G / H$ and also the reverse direction $$(H \text{ solvable and } G / H \text{ solvable}) \Rightarrow G \text{ solvable}$$ I only manage to prove the case $G / H$ but stuck at finding some injective map for the reverse direction.
My attempt: Proof for solvability of $G / H$ is easy. Let us define the sequence $G_1 / H, ..., G_n / H$. This is clearly valid due to $$H \triangleleft G \Rightarrow H \triangleleft G_k$$ Now, we want to show that $G_{k + 1}/ H \triangleleft G_{k} / H$. Indeed, let $g_kH \in G_{k} / H$ and $g_{k + 1}H \in G_{k + 1} / H$. Clearly, $$(g_kH)g_{k + 1}H(g_{k}H)^{-1} = (g_kg_{k + 1}g_k^{-1})H$$ Due to $G_{k + 1} \triangleleft G_{k}$, $g_kg_{k + 1}g_k^{-1} \in G_{k + 1}$, which shows $(g_kH)(g_{k + 1}H)(g_{k}H)^{-1} \in G_{k + 1} / H$. Also, consider the map $$\Phi: (G_{k} / H) / (G_{k + 1} / H) \rightarrow G_{k} / G_{k + 1}$$ $$\Phi: (g_kH)(G_{k + 1} / H) \rightarrow g_k(G_{k + 1})$$ Now, let $\Phi(g_kH(G_{k + 1} / H)) = \Phi(g_k'H(G_{k + 1} / H))$. Then, $$g_{k}G_{k + 1} = g_k'G_{k + 1} \Rightarrow (g_kg_k'{-1} \in G_{k + 1})$$ So, $$(g_kg_k'^{-1})H \in G_{k + 1} / H$$ , which proves $$(g_k)H(G_{k + 1} / H) = g_{k}'H(G_{k + 1} / H)$$ The map is indeed injective. The reverse is a bit more tricky. What I'm trying to do is: Suppose $H$ solvable and $G / H$ solvable. Then, there exists abelian towers of $G / H$ such that $$G / H \supseteq M_1 \supseteq ... \supseteq M_m$$ Now, define $$G_k = \{x \in G: xH \in M_k\}$$ We can easily check that such choice forms groups with $$G_0 \supseteq ... \supseteq G_m$$ To prove $G_{k + 1}$ normal in $G_{k}$: Again, let $g_{k + 1} \in G_{k + 1}$ and $g_k \in G_k$. Then, $$(g_{k}g_{k + 1}g_{k}^{-1})H = (g_{k}H)(g_{k + 1}H)(g_{k}H)^{-1}$$ Here, $g_{k}H \in M_{k}$. Since $M_{k + 1}$ normal in $M_{k}$, we must have $$(g_{k}g_{k + 1}g_{k}^{-1})H \in M_{k + 1} \Rightarrow g_{k}g_{k + 1}g_{k}^{-1} \in G_{k + 1}$$ as desired.
My question is:
- One is for proof verification.
- I can't somehow find the injective map of the reverse case. How can I proceed?
