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.

  • 2
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.