When is a finite group prove that if is normal, then . Give an explicit example when this containment does not hold if is not normal in .
[With help from the GroupProps wiki.]
We begin with some lemmas.
Lemma 1: Let be a group, and let with be subgroups. Then . Proof: Let . Then where and , so that . Let . Then we have for some , , and . Then , so in fact . Thus .
Lemma 2: Let be a finite group. If is a proper subgroup, then is proper. Proof: is characteristic, hence normal, in , so that is a subgroup. Now is contained in some maximal subgroup since is finite, and by definition. Thus is proper.
Now to the main result.
Let be normal and suppose by way of contradiction that for some maximal subgroup . Since is characteristic and is normal, is normal. Thus is a subgroup, and in fact .
Now . However, since , then , so that is proper. But this implies (by Lemma 2) that is proper, a contradiction. Thus no such subgroup exists, and in fact for all maximal subgroups . Hence .
Now before we produce a counterexample, we prove another lemma.
Lemma 3: . Proof: is a proper normal subgroup, so that . Now contains a subgroup isomorphic to ; by Lagrange, we have that is prime, so that is maximal. Recall also that is maximal; thus divides both and ; hence divides 12, and we have .
Consider ; this subgroup is cyclic of order 4, and thus has a unique maximal subgroup which is isomorphic to . Thus but .