Exercise 9, Section 2.5 of Hungerford’s Abstract Algebra

If $$|G| = p^nq$$, with $$p\gt q$$ primes, then $$G$$ contains a unique normal subgroup of index $$q$$.

Why does author mention both unique and normal? Subgroup of index $$q$$ is Sylow $$p$$-subgroup of $$G$$. So uniqueness $$\iff$$ normality. Showing uniqueness is relatively easy compared to normality. Here is proof of above exercise showing uniqueness.

• @citadel what do you mean? Nov 6, 2023 at 12:49
• Sorry! I missed the title! Nov 6, 2023 at 14:04

2 Answers

It is perhaps more subtle than you think: if a subgroup is unique for a property that is respected by automorphisms (like its order, abelianess, solvability, ...), then it is certainly normal (even characteristic), since conjugation is an automorphism. However, normality in general does not imply uniqueness (cf. for a non-abelian example, the quaternion group of order $$8$$ that has $$3$$ normal subgroups of order $$4$$). Being Sylow is of course a stronger condition and then uniqueness and normality are equivalent.

Perhaps the author has used both "unique" and "normal" because they considered part of the exercise to acknowledge that a subgroup of index $$q$$ it's just a $$p$$-Sylow subgroup. Once this is done, the claim follows from Sylow II, according to which the $$p$$-Sylow subgroups are conjugate.