# example of a non-abelian group $G$ and a non-trivial maximal normal subgroup $N$ so that $[G : N] ≥ 3$.

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 $[G:H]=3$?

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

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 $$|S|$$.

• 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? – Arturo Magidin May 5 at 16:31
• Donwvoting without a reason is vandalism – James May 5 at 23:59
• @James: That’s... extreme. – Arturo Magidin May 6 at 0:22