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Let $G$ be a group and $N\trianglelefteq G$ with $[G:N]=4$. Show that there is an $N'\trianglelefteq G$ with $[G:N']=2$.

Proof. We have $|G/N|=4$, so there must exist a $M\trianglelefteq G/N$ of order 2. This corresponds to a $N'\trianglelefteq G$ with $N\subseteq N'$. We now have $4=[G:N]=[G:N']\cdot[N':N]$. How can I obtain $[N':N]=2$ from $|M|=2$?

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  • $\begingroup$ Perhaps you should highlight more precisely where the gap is present. You could appeal (for example) to the Third Isomorphism Thm. for some of the last step. $\endgroup$
    – hardmath
    Nov 8, 2017 at 19:06
  • $\begingroup$ See also this question. Don Antonio also answered your question. $\endgroup$ Nov 8, 2017 at 19:09
  • $\begingroup$ I don't understand what you mean by highlighting where the gap is present. My gap is that I cannot conclude $[N':N]=2$. I already used the isomoprhism theorem $G/N\cong (G/N')/(N'/N)$ to get the formula for the indices. $\endgroup$
    – Buh
    Nov 8, 2017 at 19:10
  • $\begingroup$ @DietrichBurde I don't understand the immediate equivalence he doesn't bother to prove. $\endgroup$
    – Buh
    Nov 8, 2017 at 19:11
  • $\begingroup$ He does. He writes $M$ as quotient modulo $N$, and then uses the correspondence theorem. $\endgroup$ Nov 8, 2017 at 19:15

1 Answer 1

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Since $|M|$ =2, $M$ has $2$ cosets $N$ and $gN$ where $g\in G - N$. Thus $N'= N\cup gN$ so $[N':N]=2$

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  • $\begingroup$ I don't understand how we get to $N'=N\cup gN$. You are using the correspondence theorem, aren't you? In which way does the bijective map obtain order? I don't know of any proof or theorem that makes a statement about this. $\endgroup$
    – Buh
    Nov 9, 2017 at 21:45
  • $\begingroup$ What do you think $G/N$ is? It is the set of 4 cosets of $N$ in $G$. They form a group of order 4 (since there are exactly four of them) using the multiplication they inherit from $G$. If $M$ is a subgroup of order 2 of $G/N$, then $M$ must consist of two of the four cosets that make up $G/N$ one of which is $N$ itself which acts as the identity. To show that the union of the two cosets form a subgroup $N'$ of G you need to show that if $a$ and $b$ are in the union then $ab$ is also. This can be divided into cases: both $a$ and $b$ are in $N$, one is in $N$, and neither is. $\endgroup$ Nov 9, 2017 at 22:35

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