-1
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I know this factor group is isomorphic to $C_2$, but I have tried calculating it and I only get one coset.

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  • $\begingroup$ We have the coset $e C_6=C_6$ and another one. $\endgroup$ – Dietrich Burde Apr 3 at 9:31
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$C_6$, as a subgroup of $C_{12}$, contains $6$ of the $12$ elements. The remaining $6$ must make up the second coset.

Specifically, if $$ C_{12} = \{0,1,2,3,4,5,6,7,8,9,10,11\} $$ with addition modulo $12$ as the group operation, then the two cosets of $C_6$ are $$ C_6 = \{0,2,4,6,8,10\}\\ 1 + C_6 = \{1,3,5,7,9,11\} $$

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-1
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$C_{12}/C_6\cong C_2$, as the order is $2$ and there is only one $2$-element group.

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