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Prove: $G$ is a solvable group and $H\trianglelefteq G$. Then $G/H$ is solvable.

Definition: A group $G$ is said to be solvable if there is a normal series of $G$ such that the factors are abelian, i.e. $G=G_0\trianglerighteq G_1\trianglerighteq \ldots \trianglerighteq G_n=\{1\},$ where $G_{i}/G_{i+1}$ are abelians.


What's wrong with the following proof? I searched on MSE and it seems that in order to quotient out $H$ many (such as here) consider the tower $G_iH$, but why can't we just take the image of canonical map as below?


Let $\phi: G \rightarrow G/H$ be the canonical map, then $\phi(G_i)$ forms a normal tower: $\phi(G_i)$ are groups as they are images of homomorphic maps, and given $\phi(a) \in \phi(G_i)$, $$ \phi(a)\phi(G_{i+1})\phi(a)^{-1} = \phi(aG_{i+1}a^{-1}) \subseteq \phi(G_{i+1}).$$ using normality of $G_{i+1}$ in $G_i$.

$\phi(G_i)/ \phi(G_{i+1})$ is abelian: We have $$ \phi(a)\phi(G_{i+1}) \phi(b) \phi(G_{i+1}) = \phi(ab) \phi(G_{i+1})$$ since $G_i/G_{i+1}$ is abelian, $abG_{i+1} = baG_{i+1}$ so, $ab= ba g'$, for some $g' \in G_{i+1}$. Hence, $$ \phi(ab) G_{i+1} = \phi(ba)\phi(g')\phi(G_{i+1})= \phi(ba) \phi(G_{i+1}). $$ so the group is abelian.

Thus, $G/H= \phi(G) \trianglerighteq \ldots \trianglerighteq \phi(1) = H$ is an abelian tower.

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    $\begingroup$ The terms are the same by the second isomorphism theorem, no? $\endgroup$ – tehjh Mar 26 '17 at 3:42
  • $\begingroup$ Sorry, may you elaborate? $\endgroup$ – Bryan Shih Mar 26 '17 at 8:01
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They are the same thing because for each term in the tower quotiented by $H$ i.e. $G_iH/H$ we have $$G_iH/H\cong G_i/G_i\cap H$$ by the second isomorphism theorem and the group on the right is just the image you get when you subject $G_i$ to the map $\phi$ since the kernel is just $G_i\cap H$ and the first isomorphism theorem applies.

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