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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.
tomasz
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user112773
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    Hint: can you find an element of order $pq$ in $C_p \times C_q$? – universalset Dec 06 '13 at 02:43
  • It would be really easy to go for a map if you can actually look at $C_p$ as ${\bar{0},\bar{1},\dots,\bar{p-1}}$ and $C_q$ as ${\bar{0},\bar{1},\dots,\bar{q-1}}$... first obvious map would then be... $(\bar{a},\bar{b})\rightarrow ??$ –  Dec 06 '13 at 02:45
  • To add to @universalset's hint: once you find an element of order $pq$, then you can confidently say that $C_p\times C_q$ is cyclic. Cyclic groups of the same order are isomorphic. – under_the_sea_salad Dec 06 '13 at 02:49
  • @universalset $(x,y)$ has order $pq$ – user112773 Dec 06 '13 at 02:49

2 Answers2

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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.

Pedro
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Hint: if you have $g\in G$ of order $n$, $h\in H$ of order $m$, then what's the order of $(g,h)$ in $G\times H$?

tomasz
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