Let $H\trianglelefteq G$ be a subgroup of some finite group $G$ and suppose that $N\trianglelefteq G/H$. Is it true that $\pi^{-1}(N)$ is normal in $G$? ($\pi$ here is just the quotient map from $G$ to $G/H$.) Trying to use this plausible lemma to prove the first part of the Jordan-Holder theorem in Dummit and Foote as well as address another problem in that section (3.4).

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    $\begingroup$ what is $N$ in $G$ for your $NH$ notation? That doesn't make sense, you should just stick with $\pi^{-1}(N)$. $\endgroup$ – Adam Hughes Nov 30 '16 at 16:36
  • $\begingroup$ For example, consider the classical $\Bbb Z/n$. The elements of this are just equivalence classes $[k]$ of integers. $\Bbb Z$ does not have equivalence classes as elements, it has individual integers, so if I take, eg $n=4$ so that $G =\Bbb Z/4$ and $N = \{[0],[2]\}$ then $N\not\subseteq\Bbb Z$. It's true that $\cup g_i N\subseteq G$, but $N$ itself is not, since it is a set of sets, not a set of elements. It's not that it has ruined your questions comprehensibility (the fourth isomorphism theorem is a well-known result, so we are able to read between the lines) but bad notation is what it is. $\endgroup$ – Adam Hughes Dec 3 '16 at 3:38
  • $\begingroup$ No, not by elements of $G$, but by subsets of G. Remember there is a difference between $\{x\}$ and $\{\{x\}\}$. Sets of subsets are not subsets, their union is. As you say, $\pi^{-1}(N)$ is a subset of $G$, but $g_iN\not\in G$, rather $g_iN\subseteq G$, so $\{g_iN\}$ is not a set of elements of $G$. $\endgroup$ – Adam Hughes Dec 3 '16 at 5:29
  • $\begingroup$ No, but you did say $N\le G$ which is patently false, that's all I'm commenting on, and that representation does not give you that, since the $g_iN$ represent how to do multiplication in $G/N$ not in $G$ itself. $\endgroup$ – Adam Hughes Dec 5 '16 at 5:08

Yes. There is a bijection between the subgroups of $G/H$ and the subgroups of $G$ which contain $H$. In this bijection, the normal subgroups of $G/H$ correspond exactly to the normal subgroups of $G$ which contain $H$.

  • $\begingroup$ Thanks. I imagine this isn't too tricky to prove, but... I think I can wait until after my qual. exam to do so haha...ooooh quals. The stress is heavy lately. $\endgroup$ – Tanner Strunk Dec 3 '16 at 3:26

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