# Let $p<q$ be distinct prime numbers and $G$ be a group with $|G|=pq$ [duplicate]

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.

• Do you have any thoughts on this problem? Any theorems you think might be useful to show this?
– RKD
Mar 24 '16 at 19:07
• Can you think of a non-abelian group of order 6? Mar 24 '16 at 19:11

$S_3$, the symmetric group on $3$ letters, is not Abelian, hence $S_3 \not \cong \Bbb Z/6\Bbb Z$, although $|S_3| = 2\times 3$.
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 $$p\lt q$$. 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 $$Z_{pq}$$