How do I prove that $D_3\times \Bbb Z_2$ is not a cyclic group?

Question

I have been told to 'show it's isomorphic to $D_{6}$, since we know $D_6$ is not cyclic'. I'm not sure how to do that, or how can I prove it using a different approach.

Answer 1

Subgroups of cyclic groups are cyclic groups. However, $D_3=S_3$ is a group which is not cyclic, because it is not even abelian. Hence $D_3\times C_2$ is not cyclic.

Answer 2

Observe that $(\rho,\bar{0})\cdot(\sigma,\bar{0})=(\rho\circ\sigma,\bar{0})\neq(\rho^{-1}\circ\sigma,\bar{0})=(\sigma\circ\rho,\bar{0})=(\sigma,\bar{0})\cdot(\rho,\bar{0})$, thus $D_3\times C_2$ is not commutative and thus not cyclic.