# Quotient of a non-elementary nilponent group

Here is a (seemingly) simple problem in group theory. Given a non-elementary finite nilpotent group $$N$$, show there exist $$p \neq q$$ primes such that $$N$$ has a quotient $$\Bbb Z_{pq}^{2}$$.

Here, an elementary group is defined to be a direct product of a $$p$$ group and a cyclic group of order coprime to p. That is, $$E$$ elementary $$\iff \exists P, C : E = P \times C$$ where $$|P| = p^{k}$$, and $$C$$ is cyclic such that $$(|C|, p) = 1$$.

A nilpotent group is defined in the standard way, a group $$N$$ is nilpotent $$\iff$$ the central series of $$N$$ is finite.

I'm not sure how to approach this one-- does anyone have any pointers?

• I've added the definition used. The context was during a lecture on representation theory, after talking about Brauer theorem. May 14 at 8:34
• The quotient by its Frattini subgroup is abelian and we have this. Perhaps that's helpful. May 14 at 8:41
• To achieve this all you need is that two of the Sylows are non-cyclic. If that is not so then the nilpotent group is what you call elementary. May 14 at 9:33

(i) A finite nilpotent group is the direct product of its Sylow-subgroups.

(ii) The quotient of a finite $$p$$-group by its Frattini subgroup is elementary abelian.

(iii) If the quotient of a finite group by its Frattini subgroup (the set of non-generators) is cyclic then the group itself is cyclic.

(iv) The direct product of two finite cyclic groups of coprime orders is itself cyclic.

From these standard results it is clear that a finite nilpotent group which is not "elementary" must have at least two Sylow subgroups which are non-cyclic, for primes $$p,q$$ say. Hence we get a quotient $$\mathbb{Z}_p\times\mathbb{Z}_p\times\mathbb{Z}_q\times\mathbb{Z}_q\simeq\mathbb{Z}_{pq}\times\mathbb{Z}_{pq}$$.

A finite group is nilpotent if and only if it is a direct product of its Sylow subgroups.

So say you have a nilpotent group $$G$$, then $$G = P_1 \times P_2 \times \cdots \times P_t$$ with $$P_i$$ a $$p_i$$-group for all $$i$$ and $$p_1, p_2, \ldots, p_t$$ are the prime divisors of $$|G|$$.

If $$t \geq 2$$, prove that there is a normal subgroup $$N = N_1 \times N_2 \times P_3 \times \cdots \times P_t$$ with $$G/N$$ cyclic of order $$p_1p_2$$.

• Order $p_1 p_2$ is not what is requested. May 14 at 9:25
• @ancientmathematician That's right, but I found the notation ${\mathbb Z}^2_{pq}$ very confusing, so I don't think this misinterpretation merits a downvote. May 14 at 17:03
• @ancientmathematician: Right, I must have misread it as $\mathbb{Z}_{pq}$. But the same idea works, as seen in your answer.
– spin
2 days ago