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?

  • $\begingroup$ I want it to have order 6, as I defined $D_6$ as the dihidral group of order 12. Also, this doesn't answer what the fastest way is to show that it is indeed an injective homomorphism. $\endgroup$ – user260710 Mar 23 '16 at 18:35
  • $\begingroup$ It'd be really cool to construct a homomorphism $D_6\to S_5$ as an induced action, but I don't see an obvious way to do that. (For example, the symmetry group of a 3D cube acts by permutations on vertices, edges, faces, space diagonals, etc.) If anybody does find a way to do that I'd like to see it. $\endgroup$ – arctic tern Mar 23 '16 at 18:55
  • $\begingroup$ Yes sorry I misinterpreted $D_6$! $\endgroup$ – Derek Holt Mar 23 '16 at 19:26
  • $\begingroup$ @arctictern I thought so too, so here it is :) $\endgroup$ – pjs36 Mar 23 '16 at 20:29

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!

the objects being acted on: 3 diagonals and 2 triangles

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.

  • $\begingroup$ Do note that this probably took 5 times as long as just bearing down and computing stuff! $\endgroup$ – pjs36 Mar 23 '16 at 20:31
  • $\begingroup$ Well, verifying the induced action is faithful is pretty straightforward, it's the thinking of it that might have took time. I was implicitly trying to think of an induced action in which $D_6$ acted transitively, so each "associated object" looked "the same," but it turns out if I had dropped this unconscious assumption the answer was easy! $\endgroup$ – arctic tern Mar 23 '16 at 20:41
  • $\begingroup$ Yeah, until I computed the various permutations and realized that $4$ and $5$ never showed up in a cycle with $1$ through $3$, I was looking for something more homogeneous as well. (But to be fair, this may show how to decompose $D_6$ as some kind of product, so that's nice at least). $\endgroup$ – pjs36 Mar 23 '16 at 21:06
  • 2
    $\begingroup$ Another way to look at this: there is a subgroup of $D_6$, with order $6$, which fixes triangle 4, namely rotations by 0, 120, 240 degrees along with three of the six reflections. This subgroup, call it $H$, is isomorphic to $D_3$, equivalently $S_3$. Moreover, $H$ has index $2$ in $D_6$, so it is normal. Then take the rotation $\tau$ by 180 degrees. The subgroup $K = \langle \tau \rangle$ has order $2$, is also normal, and $H \cap K = 1$. This shows that $D_6 = H \times K \simeq S_3 \times \mathbb Z_2$, which has an obvious embedding in $S_5$. $\endgroup$ – Bungo Mar 23 '16 at 21:09
  • $\begingroup$ @Bungo Perfect, that is something like what I was groping toward last comment. I definitely couldn't phrase it very well though, nice! $\endgroup$ – pjs36 Mar 23 '16 at 21:32

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 $i=j=0$.

  • $\begingroup$ This is much more elegant than doing it explicitly. I see what I was missing now. Thank you. $\endgroup$ – user260710 Mar 23 '16 at 19:54

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.

  • $\begingroup$ You need to show that $f$ is indeed a homomorphism first. $\endgroup$ – Cheerful Parsnip Mar 23 '16 at 18:46
  • $\begingroup$ @GrumpyParsnip That follows from the images of the generators satisfying the same defining relations. $\endgroup$ – arctic tern Mar 23 '16 at 18:48
  • $\begingroup$ This somewhat goes outside the scope of the material I've treated so far. $\endgroup$ – user260710 Mar 23 '16 at 18:50
  • $\begingroup$ @arctictern: ok, but that was the OP's main question. $\endgroup$ – Cheerful Parsnip Mar 23 '16 at 18:51
  • $\begingroup$ @GrumpyParsnip Yes, that's true. Updated post. $\endgroup$ – arctic tern Mar 23 '16 at 18:52

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.


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 $S_5$.
$$D_6\cong C_2 \times S_3\leq S_5$$

  • 1
    $\begingroup$ (+1) This just occurred to me too, when I was looking at the geometric answer given by pjs36, which makes it easy to see that $D_6$ contains $D_3$ (which is the same as $S_3$) as a subgroup - see my comment to his answer. $\endgroup$ – Bungo Mar 23 '16 at 21:26
  • $\begingroup$ @Bungo It's nice proof to show $D_6\cong C_2 \times S_3$. $\endgroup$ – user217174 Mar 23 '16 at 21:31

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