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 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).