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 $[G:H]=3$?

  • 1
    $\begingroup$ Try groups of the form $G \times (\mathbb{Z}/(p\mathbb{Z}))$ $\endgroup$ May 4, 2020 at 19:58
  • 4
    $\begingroup$ 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) $\endgroup$ May 4, 2020 at 20:16
  • $\begingroup$ @ArturoMagidin I would prefer non-trivial examples..but thank you for your feedback $\endgroup$
    – Student146
    May 4, 2020 at 20:33

1 Answer 1


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 $|S|$.

  • 1
    $\begingroup$ 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? $\endgroup$ May 5, 2020 at 16:31
  • $\begingroup$ Donwvoting without a reason is vandalism $\endgroup$
    – James
    May 5, 2020 at 23:59
  • $\begingroup$ @James: That’s... extreme. $\endgroup$ May 6, 2020 at 0:22

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.