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 $D\_6$.
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.