2
$\begingroup$

Suppose that $|G| = pq$ where $p$ and $q$ are distinct primes such that $p$ does not divide $q-1$.
Prove that G has a normal Sylow $p$-subgroup .

I know what by Sylow's Theorem, either $n_p=1$ or $n_p=q$.

I am stuck on how to proceed here, can anyone give me a hint?

$\endgroup$
3
$\begingroup$

The 3rd Sylow theorem also tells you that $n_p\equiv 1\bmod p$. Thus $n_p\neq q$, since $q\not\equiv 1\bmod p$. So the only possibility for $n_p$ is it's equal to $1$, which means the $p$-Sylow subgroup is normal, since all $p$-Sylow subgroups are conjugate to each other, by the 2nd Sylow theorem.

$\endgroup$
1
$\begingroup$

This can be proved without an appeal to the Sylow theorems.

WLOG, assume $p \lt q$. By Cauchy's theorem, $G$ has an element of order $q$, let $Q$ be the subgroup generated by this element. Hence $|Q|=q$. Observe that $Q$ has to be normal. For if $Q^*$ is a conjugate of $Q$ and $Q \neq Q^*$, then $Q \cap Q^*=1$, since $q$ is prime. But then $|QQ^*|=\frac{|Q||Q^*|}{|Q \cap Q^*|}=q^2 \gt |G|$, which is absurd.

Since $Q$ is abelian, we have $Q \subseteq C_G(Q) \subseteq N_G(Q)=G$. But $|G|=pq$, so $|G:C_G(Q)|$ equals $1$ or $p$. In the latter case, we have $Q=C_G(Q)$, and we apply the $N/C$ theorem: $G/Q$ embeds homomorphically in $Aut(Q) \cong C_{q-1}$. However, this obstructs $p$ not dividing $q-1$.

We conclude that $G=C_G(Q)$, which is equivalent to $Q \subseteq Z(G)$, the center of $G$.
Finally, again by Cauchy's theorem we can find a subgroup $P$ of order $p$. But then $|PQ|=\frac{|P||Q|}{|P \cap Q|}=pq$, so $G=PQ$, and since $Q$ is central, $P$ is certainly normal in $G$.

Note: in fact we proved that $G$ is cyclic of order $pq$.

$\endgroup$
0
$\begingroup$

I'm assuming that $n_p$ is the index of the normalizer of the $p$-Sylow subgroup. If $n_p=q$,then there are $q$ $p$-Sylow subgroups. But this implies $q\equiv 1\pmod p$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.