If $G$ is a group of order $pq$, where $p$ and $q$ are distinct primes, prove that $G$ has subgroups of order $p$ and $q$?

Question

This question confuses me greatly. How do we even know that there are subgroups in the first place if all we know about $G$ is that it is a group of order $pq$, where p and q are distinct primes?

I know that by Lagrange's theorem the order of any subgroup must divide the order of the group itself (and we get subgroups of order $p$ and $q$ from that) but how do we know these subgroups exist in the first place?

This is easy to prove for cyclic groups. if $G$ is a cyclic group of order $pq$, then it has subgroups of order $p$ and $q$. This one follows because $G$ has a generator element, say $w$. The first subgroup would be ${x: x=w^{qn}; 0<n<p}$ and the second one ${x: x=w^{pn}; 0<n<q}$ But for a more complicated groups you can refer to Sylow's theorem. — Aven Desta, Feb 17 '20 at 05:58

score 4 · Accepted Answer · answered Feb 17 '20 at 06:02

You are right, the converse of Lagrange's theorem is false in general. If $G$ is a finite group such $d$ divides the order of the group, then $G$ does not necessarily contain a subgroup of order $d$.

However, the converse is true if $d$ is prime. You can search for proof of Cauchy's theorem as Geoffrey Trang commented or try to prove it by yourself.

Hint : $ |G| = |H| [G:H]$

This result also follows from Sylow theorems as well but that will come much further in the book.

If $G$ is a group of order $pq$, where $p$ and $q$ are distinct primes, prove that $G$ has subgroups of order $p$ and $q$?

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