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I am trying to solve this question:

"$\text{Prove that no two of the groups } C_2 \times C_2 \times C_2 , C_2 \times C_4 \text{ and } C_8 \text{ are isomorphic.} $"

I understand that to show they are not isomorphic I need to show that there is a different number of elements of order $ n $ for some $ n\in \mathbb{N} $. But how do I show this?

I am confused as to how you determine the order of each element in a cyclic group. I know that there is an element of order 4 in $ C_2 \times C_4 $ because it is the lowest common multiple of 2 and 4, however how do I know if there is an element of order 4 in $ C_8 $ for example?

Any help would be much appreciated.

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  • $\begingroup$ Note: $C_2\times C_2\times C_2$ has order $8$ (not $6$). All three groups have the same order. $\endgroup$ – paw88789 Mar 4 '15 at 22:11
  • $\begingroup$ Oh of course. Because it wouldn't be 2x2x2, but (2x2)x2? $\endgroup$ – kw3rti Mar 4 '15 at 22:13
  • $\begingroup$ Either of those expressions is correct for the order. Both work out to $8$. $\endgroup$ – paw88789 Mar 4 '15 at 22:14
  • $\begingroup$ Yeah sorry, don't know how I made that mistake! $\endgroup$ – kw3rti Mar 4 '15 at 22:19
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Hint: What is the highest element order in each group?

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  • $\begingroup$ So I could write out each group e.g. C8 as (e,a,...,a^7) and then calculate the order of each element in each group but is there not a faster way? The highest element order would be 8 but how does that help me? Sorry for my confusion. $\endgroup$ – kw3rti Mar 4 '15 at 22:51
  • $\begingroup$ The highest element order is not $8$ for each group. $\endgroup$ – paw88789 Mar 4 '15 at 22:59
  • $\begingroup$ Ok so is the maximum order in C8 = 8 but 4 in C2XC4 and 2 in C2XC2XC2. So since there is no element of order 8 in C2XC4 and C2XC2XC2 and similarly no element of order 4 in C2XC2XC2, they cannot be isomorphic? $\endgroup$ – kw3rti Mar 5 '15 at 7:45
  • $\begingroup$ That looks correct. $\endgroup$ – paw88789 Mar 5 '15 at 12:02

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