Obviously
^{\mathbb{Z}_{18}}\!/\!_{<6>}= \mathbb{Z}_6=\{0,1,2,3,4,5\}
There are no elements of order 5 here because ord(Z6)=6ord(\mathbb{Z}_6)=6ord(Z6)=6 and order of any element must be a divisor of the group order, but 5∤65\nmid65∤6
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