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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

Thanks in advance

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  • $\begingroup$ 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 $q$. $\endgroup$ May 2, 2013 at 14:46

4 Answers 4

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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).

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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 $|G|=p^nq$.

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  • $\begingroup$ Where does the uniqueness come from? $\endgroup$ Apr 26, 2020 at 19:55
  • $\begingroup$ 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. $\endgroup$ Apr 26, 2020 at 20:18
  • $\begingroup$ Your answer said is not using the Sylow Theorem $\endgroup$ Apr 26, 2020 at 20:20
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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.

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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.

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