Let $n$ be fixed. Then, can we say that a group of order $n$ attains the supremum on maximal normal subgroups iff it is abelian?
I think yes. This is because, if $N$ be a maximal normal subgroup, then $G/N$ is simple. This implies that the group has the highest number of simple subgroups for a given $n$. This is, I think, only possible when the group $G$ is abelian. Am I right? Or are there counterexamples? Thanks beforehand.
Best Answer
We can suppose that the intersection of all maximal normal subgroups is trivial. But then the group is a direct product $G_1 \times \cdots \times G_k$, where each $G_k$ is either an elementary abelian $p$-group for some prime $p$, or a direct product of isomorphic nonabelian simple groups.
Now an elementary abelian group of order $p^n$ has $(p^n-1)/(p-1)$ maximal normal subgroups. The only normal subgroups of $S^k$ with $S$ with nonabelian simple are the direct products of subsets of the factors, so it has only $k$ maximal normal subgroups.
So the abelian $G_i$ have many more maximal normal subgroups than the nonabelian $G_i$, for a given group order $n$, we get the largest number of maximal normal subgroups when the group is a direct product of elementary abelian groups.
The topic of the total number of maximal subgroups of a finite group is more interesting!