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: 
Notice that $S\_3$ is isomorphic to $D\_3$ and hence $S\_3$ has presentation $\langle \tau,\sigma\,|\, \tau^3 = \sigma^2=1, (\tau\sigma)^2=1\rangle$. Similarly, $D\_6 = \langle r,s\,|\, r^6=s^2=1, (rs)^2=1\rangle$. We obtain a group homomorphism
$$ D\_6\longrightarrow S\_3\oplus \mathbb Z/2\mathbb Z,\qquad r\mapsto (\tau,1),\; s\mapsto (\sigma,0).$$

Well-definedness is clear, since $(\tau,1)^6 = (\tau^3,6\cdot 1) = (1,0)$ which is the identity element, and similarly $(\sigma,0)^2 = (\sigma^2,0) = (1,0)$ and $(\tau \sigma,1)^2 = ((\tau\sigma)^2,2\cdot 1) = (1,0)$.

Surjectivity is easy because $(\tau,1)$ and $(\sigma,0)$ generate $S\_3\oplus \mathbb Z/2\mathbb Z$: This follows from $(\tau,0) = (\tau,1)^4$, $(1,1) = (\tau,1)^3$ and the defining relations in $S\_3$.

Since both groups $D\_6$ and $S\_3\oplus \mathbb Z/2\mathbb Z$ have order 12, it follows that the above homomorphism is bijective.
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 $.