TITLE: For $p,q$ prime and $p \neq q$ show that $C_p \times C_q \cong C_{pq}$
QUESTION [2 upvotes]: Here are my thoughts so far: 

$C_p = \langle x \mid x^p =e\rangle$
$C_q = \langle y \mid y^q =e\rangle$
$C_p \times C_q$ has elements of the form $(x^a,y^b)$
There are $p$ possible values for $x^a$ and $q$ possible values for $y^b$. So there are $pq$ possible elements in $C_p \times C_q$.
$C_{pq} = \langle z\mid z^{pq} = e\rangle$ and there are $pq$ elements in this group.
For an isomorphism from $C_p \times C_q$ to $C_{pq}$ we send $(e,e)$ to $e$ to ensure that the identities are mapped to each other. However, I'm not sure how to define an isomorphism for the other elements.

REPLY [1 votes]: You know $C_p\simeq \Bbb Z/p\Bbb Z$ so you might as well work with that. Then use $(q,p)=1$ and the Chinese Remainder Theorem.
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