Define $G$ to be a solvable group if there exists a subnormal series $1=N_0\unlhd N_1\unlhd N_2\unlhd \dots\unlhd N_{i-1}\unlhd N_i=G$ such that the factor groups $N_{j+1}/N_j$ are all abelian.

I want to prove that the quotient groups of $G$ are solvable.

Suppose $H\unlhd G$. We can construct a subgroup series
$1=HN_0/H\le HN_1/H\le HN_2/H\le \dots\le HN_{i-1}/H\le HN_i/H=G/H$. I know that $H\unlhd G$ implies $H\unlhd HN_j$, so the quotients are well-defined. Since $HN_j\le HN_{j+1}$, by the lattice isomorphism thm., $HN_j/H\le HN_{j+1}/H$

but I cannot figure out why $HN_j/H\unlhd HN_{j+1}/H$, i.e., why $HN_j\unlhd HN_{j+1}$. It becomes messy when I write something like $(h'n_{j+1})(hn_j)(h'n_{j+1})^{-1}$ and showing that it is indeed an element of $HN_j$. Why is it so?

  • $\begingroup$ Instead of writing $HN_j$, write it as $\pi^{-1}(N_j)$, where $\pi: G \rightarrow G/H$ is the projection map. Would this make it look tidier for you? $\endgroup$ – MaudPieTheRocktorate Jun 3 '17 at 9:19
  • $\begingroup$ but $N_j$ is not an element of $G/H$. What do you mean? D: $\endgroup$ – user441558 Jun 3 '17 at 9:22
  • $\begingroup$ Sorry. I meant $\pi^{-1}(\pi(N_j)) = \{g \in G |\exists n\in N_j, \pi(g) = \pi(n)\}$ $\endgroup$ – MaudPieTheRocktorate Jun 3 '17 at 9:25

Hint: proving this will simplify matters $$HN_j/H = \pi(N_j)$$ where $\pi: G \rightarrow G/H$ is the projection map.

I'll do the easy direction. $$\forall n \in N_j, nH = 1\cdot nH \in HN_j/H$$ So $\pi(N_j)\subseteq HN_j/H$

For the other direction, $$\forall hnH \in HN_j/H$$

Can you find a $n'\in N_j$ such that $n'H = hnH$?

  • $\begingroup$ Hmm... yea, I understand that $HN_j/H = \pi(N_j)$. I was only struggling at proving the normality: $\pi^{-1}(\pi(N_j))\unlhd\pi^{-1}(\pi(N_{j+1}))$, which I think I can handle now. Thanks anyway! $\endgroup$ – user441558 Jun 3 '17 at 9:47
  • $\begingroup$ $\forall \pi(n)\in\pi(N_j), \pi(m)\in\pi(N_{j-1})$ we have $\pi(m)\pi(n)\pi(m)^{-1} = \pi(mnm^{-1})\in\pi(N_j)$ $\endgroup$ – MaudPieTheRocktorate Jun 3 '17 at 9:50

From what you wrote it follows that we only need to show $\;HN_j\lhd HN_{j+1}\;$ , as then normality is preserved in the quotient group. But for any $\;h,h'\in H\;,\;\;x_k\in N_k$:


$$\in x_{j+1}^{-1} HN_jx_{j+1}=\left[x_{j+1}^{-1} Hx_{j+1}\right]\left[x_{j+1}^{-1}N_jx_{j+1}\right]\in HN_{j}$$



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