Let $G$ be a group of order $pq$, where $p$, $q$ are distinct primes and $pThis is an exercise in Serge Lang’s Algebra in the first chapter. I am wondering why q $\not\equiv$ 1 mod $p$ is assumed considering it is unnecessary.
Indeed, if that is excluded from the requirements, then let $H_q$ and $H_p$ be the Sylow subgroups of orders $q$ and $p$, respectively. Then they are cyclic and thus have trivial intersection. Since they have trivial intersection, the product of groups $H_qH_p$(which is a group since $H_q$ is normal) is isomorphic to $H_q$$\times$$H_p$ which has order $pq$ and so it is equal to $G$. Considering $p$ and $q$ are coprime, $G$ is cyclic.
Is this solution correct/an acceptable answer to this problem? If so, why is the aforementioned requirement provided? Note that all of the information used in my proof is either in the exercises preceding this one or in the chapter on Sylow subgroups.
 A: It is not true that $H_p\cap H_q=\{1\}$ implies $H_pH_q\simeq H_p\times H_q$. This is only true when both $H_p$ and $H_q$ are normal in $G$.
Indeed, for a counterexample you can take $G=S_3$.
Edit: to add to this, let's explain how you should proceed with a correct proof. What $H_p\cap H_q$ and $H_q\trianglelefteq G$ does imply is that $G$ is isomorphic to a semidirect product $H_q\rtimes_\theta H_p$ for some homomorphism $\theta:H_p\to\operatorname{Aut}(H_q)$, and you can use the condition $q\not\equiv1\pmod p$ to narrow down the possibilities for $\theta$.
A: Other answers showed that $q\not\equiv 1\pmod p$ cannot be dropped completely wit the counterexample $p=2$, $q=3$. However, it is always the case that $q\equiv 1\pmod p$ allows us to come up with a non-abelian $G$.
Indeed, $\Bbb F_q$ is a field with cyclic additive group $\Bbb Z/q\Bbb Z$ and multiplicative group$^1$ $\Bbb F_q^\times$ of order $q-1$. As $p\mid q-1$, Cauchy's theorem guarantees us that there exists $a\in \Bbb F_q^\times$ of order $p$. This allows us to define a non-trivial action of $H_p=\Bbb Z/p\Bbb Z$ on $H_q=\Bbb Z/q\Bbb Z$, by letting the generator of $H_p$ act as multiplication with $a\bmod q$. In other words, on the set $\Bbb Z/q\Bbb Z\times \Bbb Z/q\Bbb Z$ we can define the operation
$$ (x+q\Bbb Z,y+p\Bbb Z)*(u+q\Bbb Z,v+p\Bbb Z)=(x+a^yu+q\Bbb Z,y+v+p\Bbb Z) $$
and the result is not abelian(e.g., $(0+q\Bbb Z,1+p\Bbb Z)*(1+q\Bbb Z,0+p\Bbb Z)=(a+q\Bbb Z,1+p\Bbb Z)$, whereas
$(1+q\Bbb Z,0+p\Bbb Z)*(0+q\Bbb Z,1+p\Bbb Z)=(1+q\Bbb Z,1+p\Bbb Z)$)

$^1$ Incidentally, $\Bbb F_q^\times$ is cyclic - this is a nice exercise of its own.
A: There is a non-abelian group of order 6. So, it is certainly not cyclic. $6=pq=(2)(3).$
A: It is not unnecessary; consider $S_3$.
The issue is that $G=H_qH_p$ does not mean $G=H_q\times H_p$. That is, $H_qH_p$ is the group generated by all of the elements $h_1h_2$ with $h_1\in H_p$, $h_2\in H_q$. The two subgroups can still "see each other," so the group operation is, in general, different from $H_p\times H_q$.
Also, if $G\cong H_p\times H_q$, this would imply both $H_p$ and $H_q$ are normal, which, as you noted, is not necessarily true when $q\equiv 1 \pmod p$.
A: Let $n_p$ := no of Sylow p subgroup $n_q$:= no of Sylow q subgroup. Let $n_q=kq+1$ for some k=0,1,... Then (kq+1)| pq implies kq+1 divides p implies k=0 [as p< q]
Again let $n_p=k'p+1$ for some k'=0,1,...Then (k'p+1)| pq implies k'p+1 divides q implies k'p+1=1 or p i.e. k'=0 [as p does not divide (q-1)]
Thus there is unique Sylow p subgroup and Sylow q subgroup and so they are normal and hence $G= H_p X H_q$ so G is cyclic.
