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Let $p < q$ be distinct prime numbers and $G$ be a group with $|G| = pq$.

Give an example of $p$, $q$ and $G$ such that $G$ is not isomorphic to $mathbb{Z}_{pq}$.

Now suppose that $p = 5$ and $q = 7$. Show that $G$ is isomorphic to $mathbb{Z}_{35}$.

I know we are learning about Sylow's Theorems and group actions, but those are very confusing to me, so I'm not sure how to implement them really.

$S_3$, the symmetric group on $3$ letters, is not Abelian, hence $S_3 not cong Bbb Z/6Bbb Z$, although $|S_3| = 2times 3$.

If $|G| = 5 times 7$, then $s_5$ divides $7$ and since $5$ divides $s_5 -1$, we must have $s_5 = 1$. By the same reasoning $s_7 = 1$. Thus, $G$ has exactly one Sylow $5$-subgroup $W_5$ and one Sylow $7$-subgroup $W_7$, and these are normal in $G$, so their direct product $W_5 W_7 le G$. Also, $W_5 cap W_7 = {1}$ ($W_5 cap W_7$ is a subgroup of each, so its order must divide $5 land 7 = 1$). Also, $|W_5 W_7| = |W_5|times |W_7|/|W_5 cap W_7| = 35 = |G|$, hence $G = W_5 W_7$. Therefore $G = W_5 odot W_7$ and is then $cong W_5 times W_7 cong Bbb Z/5 Bbb Z times Bbb Z/7 Bbb Z cong Bbb Z/35 Bbb Z$.

(for a prime $p$, $s_p$ denotes the number of Sylow $p$-subgroups)

Use the fact that if $|G|=pq $ where both of them are distinct primes with $plt q $. If $p$ divides $q-1$ then there two groups of order $pq$ upto isomorphism, one being cyclic and the other being non abelian.
If $p$ doesn't divide $q-1$ then there is only $1$ group upto ismorphism and that it $Z_{pq}$