The Homomorphic Image of a Solvable Group is Solvable

# The Homomorphic Image of a Solvable Group is Solvable

 Proposition 1: Let $G$ and $H$ be groups. If $G$ is a solvable group and $\phi : G \to H$ is a group homomorphism from $G$ to $H$ then $\phi(G)$ is a solvable group.
• Proof: Let $G$ be a solvable group and let $\{ e \} = G_0 \leq G_1 \leq ... \leq G_n = G$ be a finite chain of successive subgroups of $G$ such that $G_i$ is a normal subgroup of $G_{i+1}$ for all $0 \leq i \leq n - 1$ and $G_{i+1}/G_i$ is abelian for all $0 \leq i \leq n - 1$.
• For each $0 \leq k \leq n$ let $H_k = \phi(G_k)$. Note that the homomorphic image of each $G_k$, $\phi(G_k)$ is a subgroup of $\phi(G)$ and that:
(1)
\begin{align} \quad \{ e \} = H_0 \leq H_1 \leq ... \leq H_n = \phi(G_n) = \phi(G) \end{align}
• Let $0 \leq i \leq n - 1$. Since $G_i$ is a normal subgroup of $G_{i+1}$ we have that $\phi(G_i)$ is a normal subgroup of $\phi(G_{i+1})$. To see this, let $x \in \phi(G_{i+1})$ and let $z \in \phi(G_i)$. Since $x \in \phi(G_{i+1})$ there exists $g_x \in G_{i+1}$ such that $\phi(g_x) = x$. Since $z \in \phi(G_i)$ there exists $g_z \in G_i$ such that $\phi(g_z) = z$. Since $G_i$ is a normal subgroup of $G_{i+1}$ we have that:
(2)
\begin{align} \quad g_xg_zg_x^{-1} \in G_i \end{align}
• Therefore:
(3)
• Since this holds for all $x \in \phi(G_{i+1})$ and for all $x \in \phi(G_i)$ we see that $\phi(G_i)$ is a normal subgroup of $\phi(G_{i+1})$. So the quotient group $\phi(G_{i+1})/\phi(G_i)$ is well-defined for each $0 \leq i \leq n - 1$.
• Let $f : G_{i+1}/G_i \to \phi(G_{i+1})/\phi(G_i)$ be defined for all $G_i \in G_{i+1}/G$ by:
• Then $f$ is a homomorphism from $G_{i+1}/G_i$ to $\phi(G_{i+1})/\phi(G_i)$ since for all $gG_i, hG_i \in G_{i+1}/G_i$ we have that:
• Moreover, $f$ is surjective since for all $x \phi(G_i) \in \phi(G_{i+1})/\phi(G_i)$ since $x \in \phi(G_{i+1})$ there exists a $g_x \in G_{i+1}$ such that $\phi(g_x) = x$, and so $g_xG_i \in G_{i+1}/G_i$ is such that $f(g_xG_i) = \phi(g_x) \phi(G_i) = x \phi(G_i)$.
• Since $G_{i+1}/G_i$ is abelian, so is $f(G_{i+1}/G_i) = \phi(G_{i+1})/\phi(G_i) = H_i$. A similar argument applies for each $0 \leq i \leq n - 1$.
• So $\{ 0 \} = H_0 \leq H_1 \leq ... \leq H_n = \phi(G)$ is such that $H_i$ is a normal subgroup of $H_{i+1}$ for all $0 \leq i \leq n - 1$ and $H_{i+1}/H_i$ is abelian for all $0 \leq i \leq n - 1$. Therefore $\phi(G)$ is solvable. $\blacksquare$