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Find an element of $\mathbb Z_{6}\times\mathbb Z_{10}$ having the largest possible order.

The largest possible order is lcm$(6,10) = 30$, correct? So how would I go about finding an element with this order? Not sure how to continue. Thanks for the help.

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Reall that for the direct product $G \times H$ of groups we have for $g\in G$, $h \in H$ that $\def\ord{\mathop{\rm ord}}$ $$ \ord (g,h) = \mathrm{lcm}(\ord g, \ord h) $$ hence taking an element of order $6$ from $\mathbb Z_6$, say $1 + 6\mathbb Z$ and an element of order $10$ from $\mathbb Z_{10}$, say $1 + 10\mathbb Z$ gives an element $$ (1 + 6 \mathbb Z, 1 + 10 \mathbb Z) \in \mathbb Z_6 \times \mathbb Z_{10} $$ of order 30.

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    $\begingroup$ Ah so simply (1, 1) would have order 30? $\endgroup$ – user1282637 Mar 4 '15 at 16:09
  • $\begingroup$ Exactly.${}{}{}$ $\endgroup$ – martini Mar 5 '15 at 8:32
  • $\begingroup$ It's easy to check the order of $(1,1)$ is not $1,2,3,4,5,6$ or $15$: we see that $(2,2),(3,3),(4,4),(5,5)$ are not the identity, nor is $6(1,1) = (0,6)$, or $15(1,1) = (3,5)$. $\endgroup$ – David Wheeler Mar 12 '15 at 9:42

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