Question: 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.

Answer: 
One way to see that $S\_3 \times \mathbb Z\_2$ is isomorphic to $D\_6$ is to exhibit an isomorphism.

$D\_6$ is the group of symmetries of a hexagon. So we want to find for each pair $(\sigma,k)$ with $\sigma \in S\_3$ and $ \in \mathbb Z\_2$ a symmetry of $D\_6$.

Denote the vertices of hexagon by the numbers $0$ to $5$. Let $S\_3$ act on the numbers $0,1,2$. Since $S\_3$ is generated by transpositions, we only need to say where the transposition $(ij)$ is sent: we send the pair $((ij),k)$ to the symmetry switching vertices $2i$ and $2j$ (and also $2i-1$ and $2j+1$) and keeping the others fixed) followed by a rotation of length $k$.

(draw a picture)

One must convince oneself that this is a well-defined homomorphism. 

A little thought shows that this function is injective (and thus surjective, since the groups are finite).
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 $.