Quotient of a non-elementary nilponent group

Question

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?

Answer 1

(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 $\mathbb{Z}_p\times\mathbb{Z}_p\times\mathbb{Z}_q\times\mathbb{Z}_q\simeq\mathbb{Z}_{pq}\times\mathbb{Z}_{pq}$.

Answer 2

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 $p_1p_2$.