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: 
If you know that it must be isomorphic to either $D\_6$ or $A\_4$, then if we can rule out $A\_4$, then we're done.

Now $A\_4$ has 8 elements of order $3$ (the 3-cycles), whereas $D\_6$ only has $2$. So we only need to work out how many order $3$ elements $G=S\_3 \times \mathbb{Z}\_2$ has.

To get an order $3$ element in $G$, it must have the form $(a,b)$ where $a^3$ and $b^3$ are both the identity, ie have order dividing 3. Now $S\_3$ has $2$ elements of order 3, whereas $\mathbb{Z}\_2$ has none. This means to get an order $3$ element, it must have one of the order $3$ elements from $S\_3$ and the identity in $\mathbb{Z}\_2$. Hence there are only $2$ such elements.

Since an isomorphism preserves orders, this shows that $G$ can't be isomorphic to $A\_4$ and therefore must be $D\_6$.
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 $.