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  "body": "\n## Proof\n\nWe push a series of $G$ forward along $\\varphi$.\n\n1. Since $G$ is [[solvable|0P3KZK0FTV]], it has a [[subnormal series|096HN851YZ]] $\\{e\\} = G_0 \\trianglelefteq G_1 \\trianglelefteq \\cdots \\trianglelefteq G_n = G$ in which each factor $G_i/G_{i-1}$ is abelian.\n2. For $0 \\leq i \\leq n$, let $K_i = \\varphi(G_i)$. The [[image|0112DTS464]] of a subgroup under a [[homomorphism|0HV9Y7WG5R]] is a subgroup, since $\\varphi(x)\\varphi(y)^{-1} = \\varphi(xy^{-1})$.\n3. We have $K_0 = \\{e\\}$, and $K_n = \\varphi(G) = K$ because $\\varphi$ is surjective.\n4. Fix $i$ with $1 \\leq i \\leq n$. Let $a = \\varphi(x) \\in K_i$ and $b = \\varphi(y) \\in K_{i-1}$, with $x \\in G_i$ and $y \\in G_{i-1}$. Then $aba^{-1} = \\varphi(xyx^{-1})$.\n5. Since $G_{i-1}$ is [[normal|0W77CVBCHQ]] in $G_i$, we have $xyx^{-1} \\in G_{i-1}$. So $aba^{-1} \\in K_{i-1}$, and $K_{i-1} \\trianglelefteq K_i$.\n6. So $\\{e\\} = K_0 \\trianglelefteq \\cdots \\trianglelefteq K_n = K$ is a subnormal series of $K$.\n7. For a group $A$ with normal subgroup $B$, the [[quotient|0YYWCNTPWR]] $A/B$ is abelian if and only if $u^{-1}v^{-1}uv \\in B$ for all $u, v \\in A$. Indeed, $uvB = vuB$ holds if and only if $(vu)^{-1}uv \\in B$.\n8. Let $a = \\varphi(x)$ and $b = \\varphi(y)$ be in $K_i$, with $x, y \\in G_i$. Since $G_i/G_{i-1}$ is abelian, step 7 gives $x^{-1}y^{-1}xy \\in G_{i-1}$.\n9. Then $a^{-1}b^{-1}ab = \\varphi(x^{-1}y^{-1}xy) \\in K_{i-1}$. By step 7, $K_i/K_{i-1}$ is abelian.\n10. So $K$ has a subnormal series with abelian factors, and $K$ is solvable.\n11. For a normal subgroup $N$ of $G$, the projection $\\pi \\colon G \\to G/N$, $\\pi(g) = gN$, is a surjective homomorphism. By step 10, $G/N$ is solvable.\n\n## Idea\n\nApply $\\varphi$ to each term of a series of $G$. Normality and commutativity modulo the next term are statements about products, and a homomorphism preserves products. So the image series has abelian factors.\n",
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