Is $S_3 \oplus \Bbb Z_2$ isomorphic to $A_4$ or to $D_6$? I know $S_3 \oplus \Bbb Z_2$ is isomorphic either to  $A_4$ or to  $D_6$, where $S_3$ is the symmetric group of degree $3$, $A_4$ is the alternating group of degree $4$, $D_6$ is the dihedral group of order $12$, and $\oplus$ is the external direct product.  
Without writing out the tables for each group and comparing, is there an easier way to show which one of the two groups it is?  I've tried comparing a few elements but haven't made any progress.

One way to easily see the isomorphism is to note that we may identify $ C_6 $ in $ S_3 \times C_2 $ as the normal subgroup $ N = A_3 \times C_2 $, and if we denote $ H = \langle ((12), e) \rangle \cong C_2 $ and let $ ((12), e) = h $ then $ NH = S_3 \times C_2 $ and $ N \cap H $ is trivial. It is easily checked that conjugation by $ h $ induces the automorphism $ x \to x^{-1} $ on $ N $, therefore $ S_3 \times C_2 = N \rtimes_\varphi H = C_6 \rtimes_\varphi C_2 $ where $ \varphi : C_2 \to \textrm{Aut}(C_6) $ is given by $ g \to (x \to x^{-1}) $ where $ g $ is the generator of $ C_2 $. This semidirect product is clearly the group $ D_6 $.
