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I am looking at non-abelian and non-trivial maximal normal subgroups whose indexes are greater than or equal to $3$. I can't find any examples of this anywhere.

Could someone give me an example of a non-abelian group $G$ and a non-trivial maximal normal subgroup $N$ so that $[G : N] \ge 3$.

A normal subgroup $N$, of $G$, is said to be maximal in $G$ if the only normal subgroups of $G$ which contain $N$ are $N$ and $G$ themselves.

The following link looks similar to my question but I still didn't clarify much to me

Can a nonabelian group $G$ have a normal abelian subgroup $H$ with $[G:H]=3$?

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    $\begingroup$ Try groups of the form $G \times (\mathbb{Z}/(p\mathbb{Z}))$ $\endgroup$ – Brian Moehring May 4 at 19:58
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    $\begingroup$ There are trivial examples, simply by taking an arbitrary nonabelian group $G$, and then taking the direct product with an abelian group $A$ that has a maximal subgroup $M$ of index $n$. Then $G\times M$ is a maximal normal subgroup of $G\times A$, nontrivial, of index $n$. Presumably, you want something a little more interesting, in which case you should say so in your question. (These examples only exist when $n$ is prime, though) $\endgroup$ – Arturo Magidin May 4 at 20:16
  • $\begingroup$ @ArturoMagidin I would prefer non-trivial examples..but thank you for your feedback $\endgroup$ – Student146 May 4 at 20:33
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A group $\mathfrak{G}$ with a maximal normal subgroup $N$ and $[\mathfrak{G}:N]=n$ exists if and only if there is exists a simple group of order $n$.

Indeed, if such a group exists, then $|\mathfrak{G}/N|=n$, and the maximality of $N$ means, by the correspondence theorem, that $G/N$ has no proper nontrivial normal subgroups; i.e., $G/N$ is simple of order $n$.

For the converse, while one can just take a nonabelian group $G$ and then consider $G\times S$, let’s look for a slightly more interesting example, one in which $S$ is not (usually) also normal.

to that end, let $S$ be as simple group of order $n$; let $G$ be any nontrivial group. We construct a group $\mathfrak{G}$ with a normal subgroup $N$ such that $\mathfrak{G}/N\cong S$. $\mathfrak{G}$ will be nonabelian.

The group is the standard (unrestricted) wreath product of $G$ by $S$, $\mathfrak{G}=G\wr S$, constructed as follows:

Let $B=G^S$, the set of all set theoretic functions from $S$ to $G$, endowed with the pointwise product. This is isomorphic to the direct product of $|S|$ copies of $G$, indexed by $S$.

Let $S$ act on $B$ on the right by letting $S$ act on the indices via the regular right action. That is, given $\mathbf{g}=(g_s)_{s\in S}\in B$, we let $$\mathbf{g}\cdot t = (g_{st})_{s\in S}.$$ This action allows the construction of the semidirect product $B\rtimes S$. The standard unrestricted wreath product is given by that semidirect product, $G\wr S=B\rtimes S$.

This group is nonabelian whenever $G$ and $S$ are nontrivial. In addition, a theorem of Kaloujnine and Krasner shows that any group that is an extension of $G$ by $S$ can be realized as a subgroup of $G\wr S$.

The subgroup $B$ is normal in $G\wr S$ with $(G\wr S)/B\cong S$. Since $S$ was chosen to be simple, the correspondence theorem guarantees that $B$ is a maximal normal subgroup of $G\wr S$. The index of $B$ is $|S|$.

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    $\begingroup$ Someone is going to have to explain what the downvote is for. What is wrong, or how does it not provide examples the poster requested? $\endgroup$ – Arturo Magidin May 5 at 16:31
  • $\begingroup$ Donwvoting without a reason is vandalism $\endgroup$ – James May 5 at 23:59
  • $\begingroup$ @James: That’s... extreme. $\endgroup$ – Arturo Magidin May 6 at 0:22

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