Why is $\mathbb Z_8+\mathbb Z_{10}$ not a cyclic group? How I see it is that it is isomorphic to $\mathbb Z_2$ as $\gcd(8,10)=2$ which is prime and prime order groups are cyclic.
Investigation of a cyclic group
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cyclic-groups
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0There are various ways to show it is not cyclic. Just for fun, here is one way: Recall (or observe i.e. prove) that every subgroup of a cyclic group is cyclic. Now consider the given group's subgroup $\{0, 4\} \times \{0, 5\} \simeq \mathbb{Z}_2 \times \mathbb{Z}_2$, which is non-cyclic. – 2017-01-24
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1If you mean a direct sum of the two cyclic groups, a better notation might be $\mathbb{Z}_8 \oplus \mathbb{Z}_{10}$. – 2017-01-24
1 Answers
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I think you are confusing the order of a group with the order of an element. The order of a group is just the number of elements in the group. The order of an element is the minimum number of times you must add it to itself to get the identity element. The two ideas have the same name, but they are not related.
$\def\Z{\Bbb Z}$ It is true that if $a\in \Z_8$ is an element of order 8 and $b\in \Z_{10}$ is an element of order 10, then $\langle a, b\rangle \in \Z_8 \times \Z_{10}$ is an element of order 40, which is the least common multiple (not the greatest common divisor) of 8 and 10. But this is not what you argued. Since $\Z_8$ has 8 elements and $\Z_{10}$ has 10, $\Z_8\times \Z_{10}$ will have 80 elements, not 2.