Let $G$ be a group of order $pqrt$, where $p>qrt$ and $p,q,r,t$ are prime ($\underline{\text{not necessarily distinct}}$).
I have to show that $\exists$ $M,N,L \le G$ of order $pq,pr$ and $pt$.
The condition that I underlined is my doubt in this exercise because if I 'd start considering all the possible cases it would become really long and tedious.
I wanted to share on MSE my ideas about this exercise because I found several observations which I think are plausible.
The first way we can follow to show the property is using the fact that every group of a prime order is cyclic.
In $G$, the product of two cyclic subgroups of order $k\in\{p,q,r,t\}$ will be a cyclic group whose order is the product of the two prime numbers that we chose.
The second method is the following:
I noticed that, since $|G|=pqrt$ with $p>qrt$, $G$ must have only one $p$-Sylow group. All $p$-Sylow are conjugated, so we have that $\exists! P\trianglelefteq G$ of order $p$.
For the condition $P\trianglelefteq G$, I considered the quotient group $G/P$, whose order is $qrt$.
Let $n_q$ be the number of $q$-Sylow in $G/P$ $\implies n_q\equiv 1\text{ (mod q)}$ and $n_q\in\{1,\not r,\not t,rt\}$. The possibilities are $n_q=1,rt$.
If $n_q=rt\implies\exists Q^{(1)},\dots,Q^{(rt)}\le G/P$ of order $q$. Since their order is a prime, they are all cyclic groups, so they are abelian and $\forall i,j=1,\dots,rt$ ($i\ne j$) $Q^{(i)}Q^{(j)}=Q^{(j)}Q^{(i)}\implies Q^{(i)}Q^{(j)}\le G$. The order of the product of the $q$-Sylow is
$$|Q^{(i)}Q^{(j)}|=\dfrac{|Q^{(i)}||Q^{(j)}|}{|Q^{(i)}\cap Q^{(j)}|}\Big|qrt\text{ and } Q^{(i)}\cap Q^{(j)}\le Q^{(i)},Q^{(j)}\implies |Q^{(i)}\cap Q^{(j)}|\in\{1,q\}.$$
If $|Q^{(i)}\cap Q^{(j)}|=1\implies|Q^{(i)}Q^{(j)}|=q^2\Big|qrt\implies q\Big|rt$ but $q$ is a prime, hence $q\Big|r$ or $q\Big|t$, which is a contradiction.
So we state that $|Q^{(i)}\cap Q^{(j)}|=q$, then $\exists!Q\trianglelefteq G/P$.
For the correspondence theorem $\exists\mathcal Q\trianglelefteq G$ such that $|\mathcal Q|=pq$.
We can apply this idea for the other subgroups in $G/P$ and then use the correspondence theorem.
What abou the underlined Hypothesis??
Hint: by Cauchy's Theorem, $G$ possesses an element $g$ of order $q$. Now $\langle g \rangle P$ is a subgroup since $P$ is normal! And it has order $pq$. Same goes for the other primes.