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$5$ has order $6$ in $Z_{30}$ and 3 has order 4 in $Z_{12}$, $\langle 5 \rangle \oplus \langle 3 \rangle$ is a subgroup of order $24$ in $Z_{30}\oplus Z_{12} $

Isn't that the order should be the $lcm(6,4)=12$. Why should it be $24$?

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    First, I think you should have meant $\;\langle\,(5,\,3)\,\rangle\le\Bbb Z_{30}\times\Bbb Z_{12}\;$ , since you mention "subgroup of direct product". Second, this subgroup I'm talking about has order $\;12\;$ in the big group $\;\Bbb Z_{30}\times\Bbb Z_{12}\;$ . Third, if you **really** meant $\;\langle 5\rangle\oplus\langle 3\rangle\;$, then this is in itself a direct product of order $\;24\;$ ... – 2017-01-31
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    Can I ask why $\langle\,(5,\,3)\,\rangle$ has order 12?Also, the reason that $\langle 5\rangle\oplus\langle 3\rangle$ has order 24 because it has 24 elements? – 2017-01-31
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    Because $\;12\cdot(5,3)=(12\cdot5,\,12\cdot3)=(60,36)=(0,0)$ . As for the second question: yes. – 2017-01-31
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    @DonAntonio can I see $\langle 5\rangle \bigoplus \langle 3\rangle$ as $\langle (5,0),(0,3)\rangle$?? – 2017-01-31
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    @JulioMaldonadoHenríquez I think you can indeed, *if* you want to see the left one as a subgroup of $\;\Bbb Z_{30}\times\Bbb Z_{12}\;$ . – 2017-01-31

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