# Sylow's Theorems And Normal Subgroups of prime order

$$Theorem$$-If $$G$$ is a group of order $$pq$$ where $$p$$ & $$q$$ are prime , $$p>q$$ and $$q$$ does not divide $$p-1$$ then there is a normal subgroup $$H$$ in $$G$$ which is of order $$q$$.

$$Proof$$-By $$Sylow's \ first \ theorem$$ there exists a subgroup $$H$$ of order $$q$$ and it is a sylow-$$q$$ subgroup, since it is the largest subgroup whose order is of the form $$\ q^{n}$$

By $$Sylow's \ third \ theorem$$ we know that number of such sylow $$q$$ subgroup is of the form $$kq+1$$ for some integer $$k \geq 0$$ and divides $$|G|=pq.$$ If $$kq+1$$ divides $$pq$$ then it either divides $$p$$ or $$q$$, since it divides $$q$$ only when $$kq+1=1$$ and if it divides $$p$$ then $$kq+1=p \ or \ 1$$ if $$kq+1=p$$ then $$k=\frac{p-1}{q}$$ but since $$q$$ doesnot divide $$p-1,$$ $$k=\frac{p-1}{q}$$ is not an integer hence $$kq+1=1$$ and hence number of sylow $$q$$ subgroups is $$1$$. Thus $$H$$ is the only subgroup of order $$q$$.

By $$Sylow's \ second \ theorem$$ all sylow $$q$$ subgroups are conjugate to each other and since $$H$$ is the only sylow $$q$$ subgroup. $$xHx^{-1}=H \ \forall \ x \in G$$. Thus H is a unique normal subgroup of order $$q$$

I am assuming this proof is correct. Please correct me if there are any mistakes.

My doubt here is that in the theorem we include the condition $$q$$ does not divide $$p-1$$ to insure $$kq+1\neq p$$ and we know that $$kq+1\neq q$$ for any integer $$k\geq 0$$. But $$p$$ and $$q$$ are not the only divisors of $$pq, pq$$ is its own divisor too, so don't we need to include the condition $$q$$ doesnot divide $$pq-1$$ to insure that $$kq+1\neq pq$$

$$q$$ cannot be a divisor of $$pq - 1$$ since we know it is a divisor of $$pq$$. Or to phrase it differently, $$pq - 1 \equiv -1\ (\mathrm{mod}\ q)$$