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I have been trying to calculate the orders of the elements of $D_6/Z(D_6)$. For example, using $R_{60}$ to represent rotation by 60 degrees and $R_0$ to represent rotation by 0 degrees (the identity element), it seems to me that for $R_{60}Z(D_6)\in D_6/Z(D_6)$, we have,

$(R_{60}Z(D_6))(R_{60}Z(D_6))(R_{60}Z(D_6))(R_{60}Z(D_6))(R_{60}Z(D_6))(R_{60}Z(D_6))=(R_{60}*R_{60}*R_{60}*R_{60}*R_{60}*R_{60})Z(D_6)=R_0Z(D_6)$

This seems to imply that $|R_{60}Z(D_6)|=6$, since $(R_{60}Z(D_6))^6$ returns the identity. I have been told that this is wrong, however, but am having trouble seeing why.

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    $\begingroup$ It only shows that the order of $R_{60}Z(D_6)$ divides 6. $\endgroup$ – Jane Doé May 4 at 2:26
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Note that centre of $D_6$ has two elements, namely, identity and $R_{180}$. So order of $R_{60}Z(D_6)$ is three not six

Also the order of the quotient group is six. So the other way to see your calculation is wrong is by considering quotient group. If it we're cyclic then G is Abelian. But $D_6$ is non Abelian

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    $\begingroup$ Ah, I think I see. $(R_{60}Z(D_6))^6=(R_{60}Z(D_6))^3$, and 3 is a lesser $n$ for which this occurs. Correct? $\endgroup$ – JustSomeGuy716 May 4 at 3:06
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    $\begingroup$ Yes! $$(R_{60}Z(D_6))^3=R_{180}Z(D_6)=Z(D_6)$$ $\endgroup$ – Chinnapparaj R May 4 at 3:08

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