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Let $G$ be a group of order $pq$, where $p$ and $q$ are primes with the property $q \equiv 1 \pmod{p}$. Prove: if $G$ is not Abelian, the Sylow $q$-subgroup of the group $G$ is normal subgroup in $G$, and there are exactly $q$ Sylow $p$-subgroups of $G$.

I observe that $q>p$, because otherwise the property $q \equiv 1 \pmod{p}$ will not hold. My main problem is where to use that $G$ is not Abelian. By kind answer from Kan't, I now was able to prove the first statement: the Sylow $q$-subgroup of the group $G$ is normal subgroup in $G$ (by simple contradiction and here i didn't use anywhere that $G$ is not Abelian). Any idea how to prove second: there are exactly $q$ Sylow $p$-subgroups of $G$ (and where to use the fact that $G$ is not Abelian)? Thanks for all your help!

Note: I don't understand links given in the comments: Non-abelian groups of order $pq$ with $p\mid (q-1)$ and Structure of a group, $G$, of order $pq$ where $p, q$ are prime., because they mentioned $Aut$. So, can this exercise be solved with basic knowledge of Sylow theorem.

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    I think your inequality is backward. $q \equiv 1 \pmod p \Rightarrow q \gt p$. – Robert Shore Jan 19 '25 at 00:55
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    relevant: https://math.stackexchange.com/questions/2684164/non-abelian-groups-of-order-pq-with-p-mid-q-1 – Randall Jan 19 '25 at 01:03
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    also relevant: https://math.stackexchange.com/questions/881971/structure-of-a-group-g-of-order-pq-where-p-q-are-prime – Randall Jan 19 '25 at 01:04
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    Also, I don't think you can conclude it must be dihedral. That might work when $p=2$. – Randall Jan 19 '25 at 01:06
  • The classification of groups of order pq is a duplicate and answers this question. – Martin Brandenburg Jan 19 '25 at 03:27
  • If there were more than one $q$-Sylow subgroup, say $Q$ and $Q'$, then $QQ'$ would be a subset of $G$ of cardinality $q^2$, a contradiction as $q^2>pq$. – Kan't Jan 19 '25 at 08:07
  • If there was one $p$-Sylow subgroup only, say $P$, then: $P\unlhd G$, $Q\unlhd G$, $G=PQ$, $P\cap Q=1$. Hence, $G\cong P\times Q$ would be abelian. – Kan't Jan 19 '25 at 08:21
  • @Kan't, I understand your first argument. For your second comment, why is then $G=PQ$ and $P\cap Q=1$. Also, why does from here follow that $G$ is isomophic to $P \times Q$. Unfortunatelly, I don't understand listed links above. –  Jan 19 '25 at 08:54
  • I agree they are normal subgroups in $G$, but how did you conclude what I mentioned. Can you provide what did you use here? –  Jan 19 '25 at 08:59
  • @Kan't Also, you need to show there are exactly $q$ Sylow $p$-subgroups. If you provide an answer in more details, I will accept it. –  Jan 19 '25 at 09:04

2 Answers2

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This is in case you haven't got Sylow's theory at hand. By a counting argument, $G$ has one subgroup only of order order $q$ (hence normal), namely $q-1$ elements of order $q$. Since $G$ is nonabelian, it is in particular noncyclic; so, it has $pq-1-(q-1)=$ $q(p-1)$ elements of order $p$, which are necessarily grouped into $q$ subgroups of order $p$.

Kan't
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For the group you are considering, the Sylow theorems imply: $(1)$ A Sylow $q$-subgroup is normal in $G$ iff it is the only Sylow $q$-subgroup, $(2)$ The number $N_p$ of Sylow $p$-subgroups satisfies: $N_p|q$.

Given the first result (The Sylow $q$-subgroup is normal in $G$), try to work out possible values of $N_p$ bearing in mind $G$ has $pq$ elements.

ranto
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  • Thanks for your answer. Here $N_p\in {1,q}$. But as I already mentioned, how can I show that $N_p \neq 1$ using the fact that $G$ is NOT abelian? @ranto –  Jan 19 '25 at 09:56
  • Sorry, I should have paid attention, it already uses that $G$ is not abelian since that's how you got the first result: $N_p$ is not $1$ because $N_q$ is $1$ because the Sylow $q$-subgroup is normal because $G$ is not abelian. – ranto Jan 19 '25 at 10:00
  • But why $N_p$ can't be equal to $1$? I still don't understand. –  Jan 19 '25 at 10:04
  • The idea is to check case-by-case until you find something that contradicts your previous results or hypothesis: suppose $N_p = 1$, so the situation is we have 1 Syl $q$-subgroup and 1 Sylow $p$-subgroup: is that coherent with the fact that $G$ has in total $pq$ elements? (Hint: those two subgroups are obviously not enough to fill $G$, where are the other elements, can they be outside of a Sylow subgroup?) – ranto Jan 19 '25 at 10:15
  • I understood that you might need to get very comfortable with isomorphism theorems and group products first before understanding the solution Kan't offers and therefore I proposed a more elementary way of solving your problem, have you even tried it?

    Anyways: $PQ$ = $G$ is because the product theorem and the second part is direct application of [https://math.stackexchange.com/questions/2333666/group-is-isomorphic-to-direct-product-of-its-subgroups], elementary theorems.

     – ranto Jan 19 '25 at 11:07