$\mathbb{Z}_2 \times \mathbb{Z}_3 \cong \mathbb{Z}_6$?
How can I show that the groups are isomorphic? (Or not?)
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1In general, $\mathbb Z_m\times \Bbb Z_n \cong \Bbb Z_{mn}$ iff $(m,n)=1$. I'm sure this has been asked many times here before. – Oct 23 '14 at 23:20
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1For example here. – Oct 23 '14 at 23:21
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But Z2 X Z2 is not isomorphic to Z4, as Z4 is cyclic and Z2 is not.... – lifeofjuds Oct 23 '14 at 23:32
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@lifeofjuds $(2,2) \not = 1$ – Oct 23 '14 at 23:48
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If you are looking to construct an isomorphism between the two groups why not start by listing the elements of $\mathbb{Z}_2 \times \mathbb{Z}_3$ and figuring out if any of them have order $6$? If you can find one then it generates a cyclic subgroup of order $6$ (a subgroup isomorphic to $\mathbb{Z}_6$) inside your group of order $6$ .
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I made a table of all of the 36 elements in ℤ2 × ℤ3, but I don't know how to figure out if any of them have order 6. How do I do that? – lifeofjuds Oct 24 '14 at 04:14
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You have too many elements in $\mathbb{Z}_2 \times \mathbb{Z}_3$. There are only 6, have you looked at the other question at all? – Oct 24 '14 at 14:20