Is $S_3 \oplus \Bbb Z_2$ isomorphic to $A_4$ or to $D_6$? 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.
 A: 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). 
A: 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 $D_6$.
A: 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.
A: 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 $.
