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Let's say we have a commutative group G that's specified by generators and relations. We find that the group G normal form is: $Z_2\times Z_6\times Z_{12}$ and that the elementary form is $Z_2\times Z_2\times Z_{2^2}\times Z_3\times Z_3$. I need to find out how many elements are of order 12?

Also how many elements of order 12 are there in the group: $Z_2\times Z_{2^3}\times Z_{2^4}\times Z_3$?

Sorry if I misused some of the terms but English isn't my native language.

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Hint: in a direct product of abelian groups, the order of an element is the lowest common multiple of the orders of the "coordinates". Look at $C_2 \times C_6 \times C_{12}$. $C_{12}$ has 4 elements of order 12, which gives you 48 elements of order 12 for the direct product. $C_6$ has 2 elements of order 3, $C_{12}$ has 2 elements of order 4, which gives you 8 elements of order 12 for the direct product. So 56 in total.

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  • $\begingroup$ Thanks for the tip, I knew that, but the question is how many elements are of that order. $\endgroup$ – Pavle Joksović Aug 24 '15 at 13:19
  • $\begingroup$ You should now be able to find out that yourself. One of the coordinates should have order $3$, and another order $4$. So figure out how to construct those elements. $\endgroup$ – Nicky Hekster Aug 24 '15 at 13:43

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