$|G| =231$, all Sylow p subgroups are normal then, is G cyclic

Question

$|G|=231=3 * 7 * 11$.
Now it is given that all the Sylow p subgroups are normal so unique.
My attempt:
I have uniquely one Sylow 11 subgroup of order 11 which is isomorphic to $Z11$ say $P_1$, similarly I have uniquely one Sylow 7 subgroup which is isomorphic to $Z_7$ say $P_2$ , now say $P_1=<x>$
and $P_2=<y>$ now the order of $(xy)$ is $7,11 or 77$ , since there are uniquely one Sylow 7 and 11 subgroups so $|xy|=77$ otherwise $<xy>$ will be a group of order 7 or 11 otherthan $P_1$ and $P_2$.
So, suppose $<xy>= P_3$ where $P_3$ is isomophic to $Z77$ . Now say $h$ is a generator of $P_3$ and $k$ is a generator of the unique Sylow 3 subgroup say $P_4$, so, $P_4=<k>$ and $P_3=<h>$ where $|h|=77 , |k|=3$, so now the $|hk|= 3,77 or 231$ , it can't be 3 because there is uniquely one subgroup of order 3 otherwise $<hk>$ will be a group of order 3 otherthan $P_4$. Now $|hk|= 77 or 231$ , now if $|hk|=231$ then we are done because we have found a generator of this group so G is cyclic.
Now my question is if $|hk|=77$ can we get a contradiction or it is possible so in that case the G is not cyclic, then what is the examples of these type of groups.
Now does the generalisation hold?
$|G|=P_1*P_2*….*P_n$ where all $P_i$'s are primes and all Sylow $P_i$ subgroups are normal then is the group G cyclic? If not then give example.

Answer 1

(Note if $N$ is a normal subgroup and $H$ is any subgroup, $NH$ is also a subgroup. Moreover, if $H$ is also normal, then the subgroup $NH$ is normal too. Good exercise.)

Suppose $A=\langle a\rangle$ and $B=\langle b\rangle$ are normal cyclic subgroups. Then the commutator $[a,b]$ may be written as a product in two ways, $(aba^{-1})b^{-1}$ or $a(ba^{-1}b^{-1})$, showing it is in both $A$ and $B$. Thus, if $A$ and $B$ intersect trivially (e.g. when $|A|,|B|$ are coprime) we must have $[a,b]=e$, or in other words $a$ and $b$ commute. You can use this to show $AB=\langle a,b\rangle$ is an internal direct product of $A$ and $B$. (Actually, the hypothesis that $A$ and $B$ are cyclic isn't necessary - can you see why?)

(More generally, if $N_1,N_2\trianglelefteq G$ are normal subgroups, then $[N_1,N_2]\subseteq N_1\cap N_2$. Exercise.)

You can generalize this to multiple subgroups, not just two. If $N_1,\cdots,N_k$ are normal subgroups of $G$ with coprime order, then $N_1\cdots N_k$ is an internal direct product. Moreover, if $|G|=|N_1|\cdots|N_k|$ then all of $G$ is an internal direct product $N_1\times\cdots\times N_k$. In particular this applies for the collection of Sylow subgroups, and moreover if all the Sylow subgroups are cyclic we can use the Chinese Remainder theorem to show that $G$ is isomorphic to $\mathbb{Z}/n\mathbb{Z}$ (where $n=|G|$ is the product of its own prime factors).