# If $|G|=p^nq$, then $G$ contains a unique normal subgroup of index $q$

I'm trying to prove if $|G|=p^nq$ with $p\gt q$, primes, then $G$ contains a unique normal subgroup of index $q$.

I know by the first Sylow theorem that G has a Sylow p-subgroup $P$ with $[G:P]=q$. My problem is prove that it's unique.

I need help

• Possible duplicate of math.stackexchange.com/questions/33051/… though I assume this is meant to be done directly using Sylow's theorems, in which case, remember that the number of $p$-Sylows is congruent to $1$ mod $p$ and divides $q$. May 2, 2013 at 14:46

Use the third Sylow theorem: $n_p \equiv 1$ mod $p$ but $n_p$ | $q$. If $n_p = q$ then since $p > q$ we cannot have $n_p \equiv 1$ mod $p$. So $n_p = 1$. (Where $n_p$ is the number of Sylow p-subgroups).

By the first Sylow Theorem, a subgroup $$H$$ with $$|H|=p^n$$ exists.

Without further appeal to the Sylow theorems: use the classical fact (sometimes called the index factorial theorem) that if $$H$$ is a subgroup of $$G$$ with $$|G:H| = q$$, $$q$$ the smallest prime dividing $$|G|$$, then $$H$$ must be normal.

So why is such an $$H$$ unique? Assume there is another subgroup $$K$$ of index $$q$$ here. Since both $$H$$ and $$K$$ are normal, $$HK$$ is a subgroup. Observe that $$H \subseteq HK \subseteq G$$ and $$|G:H|=q$$. It follows that $$G=HK$$ or $$H=HK$$. In the latter case $$K \subseteq H$$ and since $$|H|=|K|$$ it follows that $$H=K$$. To refute the case $$G=HK$$, note that $$|HK|=\frac{|H|\cdot|K|}{|H \cap K|}$$, which nuber is a power of $$p$$, contradicting $$|G|=p^nq$$.

• Where does the uniqueness come from? Apr 26, 2020 at 19:55
• Normality of $H$, being a Sylow $p$-subgroup, implies its uniqueness by the Sylow theorems that will tell you that all Sylow $p$-subgroups are conjugate. Apr 26, 2020 at 20:18
• Your answer said is not using the Sylow Theorem Apr 26, 2020 at 20:20

Suppose that there exists two distinct subgroups $P,P'\leq G$ of order $p^n$. Since $P\neq P'$ then $|P\cap P'|\leq p^{n-1}$ and $p^nq=|G|\geq|PP'|=|P||P'|/|P\cap P'|\geq p^{2n}/p^{n-1}=p^{n+1}>p^nq$ - contradiction. Simplicity of $q$ is unnecessary.

There is a Lemma saying for a subgroup $$H$$ of a finite $$G$$, if $$[G:H]=q$$ where $$q$$ is smallest prime dividing $$|G|$$, then $$H$$ is normal.

In our case, there is a subgroup of index $$q$$, which is some Sylow $$p$$-subgroup $$P$$. Notice that $$p>q$$, which implies $$q$$ is the smallest prime dividing $$|G|$$. So P is normal in G. Now any other subgroup of index $$q$$ must be a Sylow $$p$$-subgroup too, and must equal $$P$$ because $$P$$ is the unique Sylow $$p$$-subgroup.

It seems we have a stronger case about any subgroup of index q, not just normal ones.