Fastest way to show that $D_6 \to S_5$ is an injective homomorphism I want to show that there is an injective homomorphism from $D_6 \to S_5$ where $D_6$ denotes the dihidral group of order 12 and $S_5$ the symmetric group. But I'm not sure how I can do this efficiently.
I define $f: D_6 \to S_5$ by $f(\sigma) = (12)$ and $f(\rho) = (123)(45)$, with $\sigma$ being a reflection and $\rho$ being a rotation.
I know that $D_6$ is generated by $\rho$ and $\sigma$ and that $S_5$ is generated $(12), (23), (34), (45)$. 
So what is the fastest way, for someone who is just starting with algebra, to show that this is a homomorphism? Do I have to show it explicitly for all 12 elements?
What confuses me is that you have to show for all $x,y \in D_6$ we have $f(xy)=f(x)(y)$, while $x$ and $y$ can be any combination of $\rho$ and $\sigma$.
Lastly, what is the fastest way to show it's kernel is trivial without going over all elements?
 A: This is really a response to arctic tern's comment that it would be fun to see the injective homomorphism as induced from a (presumably geometric) action. I thought so too, so here it is!

Label the diagonals of the regular hexagon $1, 2$, and $3$. Now, label the two inscribed equilateral triangles $4$ and $5$. 
It's visually clear that the third-of-a-turn rotation $\rho$ does what it should, inducing the permutation $f(\rho) = (123)(45)$. 
The reflection $\sigma$ is the one reflecting across the diagonal $3$, leading to $f(\sigma) = (12)$, as $\sigma$ does indeed fix both triangles.
A: An arbitrary element of $D_6$ can be written as $\sigma^i \rho^j$ for some unique choice of $i \in \{0,1\}$ and $j \in \{0,1,2,3,4,5\}$.
In $D_6$, the element $\sigma$ has order $2$ and $\rho$ has order $6$, and their product satisfies $\sigma \rho = \rho^{-1}\sigma$, and therefore $\sigma \rho^j = \rho^{-j} \sigma$ for any $j \in \mathbb Z$.
In $S_5$, define $h = (12)$ and $k = (123)(45)$. Note that $h$ has order $2$ and $k$ has order $6$, and their product satisfies $hk = k^{-1}h = (23)(45)$, so $hk^j = k^j h^{-1}$ for any $j \in \mathbb Z$.
So indeed it seems reasonable to set $f(\sigma) = h$ and $f(\rho) = k$.
If $f$ is to be a homomorphism, then there is no choice but to define $f(\sigma^i \rho^j) = f(\sigma)^i f(\rho)^j = h^i k^j$.
To verify that $f$ is in fact a homomorphism, take two arbitrary elements $x = \sigma^i \rho^j$ and $y = \sigma^m \rho^n$ of $D_6$. Then, using the property that 
$$xy = \sigma^i \rho^j\sigma^m \rho^n = \sigma^i \sigma^m \rho^{-j}\rho^n = \sigma^{i+m}\rho^{n-j}$$
we see that
$$f(xy) = f(\sigma^{i+m}\rho^{n-j}) = h^{i+m}k^{n-j} = h^i k^j h^m k^n = f(x)f(y)$$
To see that $f$ is injective, suppose that $f(\sigma^i \rho^j) = h^i k^j = 1$, with $i \in \{0,1\}$ and $j \in \{0,1,2,3,4,5\}$. Then $h^i = k^{-j}$, so $h^i \in \langle k \rangle$.
Now, $h = (12)$ is not in $\langle k \rangle$, because the only element of $\langle k \rangle$ with order $2$ is $k^3 = (123)^3(45)^3 = (45)$. This forces $h^i = 1$ and therefore $k^j = 1$. This in turn implies that $i=j=0$.
A: The homomorphism is clearly injective on the cyclic subgroup $C_6$ of rotations, so a nontrivial kernel would have to contain a reflection, but all the reflections are conjugate and kernels are normal so $\sigma$ would be in the kernel, a contradiction.
Also some discussion on how your construction of $f$ is good. Since $\rho$ has order $6$ it must be sent to a permutation of order $6$. The order of a permutation is the lcm of its cycle lengths (in its disjoint cycle representation). There are no $6$-cycles in $S_5$, which forces an element of order $6$ to have the cycle type you chose, the most obvious being $(123)(45)$. Then the reflection $\sigma$ must be sent to an involution in the symmetric group, and conjugating $(123)(45)$ must yield its inverse $(321)(45)$. Conjugating a permutation just relabels the cycle type according to the permutation conjugated by, so our permutation of order two can be chosen to be any $2$-cycle as that would reverse the cyclic ordering of $(123)$, hence $(12)$ works.
The defining relations of $\rho$ and $\sigma$ are $\rho^6=\sigma^2=e$ and $\sigma\rho\sigma^{-1}=\rho^{-1}$. There is a universal property that says we need only check $f(\rho)$ and $f(\sigma)$ satisfy the same relations to assume it extends to a homomorphism, although while this is "intuitive" to be fair it is indeed probably outside the scope of the elementary context of wherever this question came from, so perhaps take this to be some ideas or tangential discussion. 
This idea can of course be made elementary though: write down a $2\times2$ multiplication table for the dihedral group using symbols $\sigma\rho^i$ and $\rho^i$ (with $i$ an arbitrary integer), and this will cut down the amount of work required to check $f(xy)=f(x)f(y)$ to a manageable level.
A: Let $\sigma = (12)$ and $\rho = (123)(45)$ in $S_5$, and let $s$ be reflection and $r$ be rotation in $D_6$. Notice that $|\sigma| = 2$, $|\rho| = 6$, and $\rho\sigma\rho = \sigma$ which strongly suggests that $D_6$ is isomorphic to $<\sigma, \rho>$ ... recall that $D_6$ $=$ $<s,r : |s|=2, |r|=6, rsr = s>$. Define $\phi : D_6$ $\rightarrow$ $<\sigma, \rho>$ by $\phi(s^ir^j) = \sigma^i\rho^j$. It is simple to see that $\phi(s^ir^j) = \phi(s^mr^n)$ if and only if $i = m$ and $j = n$, that is $\phi$ is well-defined and injective. Because of the relationship $\rho\sigma\rho = \sigma$, every element in $<\sigma, \rho>$ can be expressed as $\sigma^i\rho^j$, so $\phi$ is surjective.
A: In my opinion, the fastest way to show $D_6\rightarrow S_5$ is injective, is to show $S_5$ has  subgroup isomorph by $D_6$, and so $D_6$ embed is $S_5$.
$$D_6\cong C_2 \times S_3\leq S_5$$
