For example, let's find the Cayley permutation representation of $\mathcal D_3$ in $S_6$.

$\mathcal D_3 = \left<r,s \mid r^3=s^2=1, rs=sr^{-1}\right>$.


\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 1 & r & r^2 & s & rs & r^2s \end{pmatrix}

By multiplying on the right, we find:







What is the formal process behind finding a Cayley permutation representation. Is this a group action from $\mathcal D_3$ on $\mathbb Z_6$? A group action wouldn't give you cycles like this.

  • $\begingroup$ This is a group action on a set. In your example, the set is the elements of the group itself. In general, permutation representations come from the group action on cosets of a given subgroup. $\endgroup$
    – Josh B.
    Apr 12 '15 at 3:24
  • $\begingroup$ So, this is $\mathcal D_3 \times \mathcal D_3 \to \text{Sym}(\mathcal D_3)$? $\endgroup$ Apr 12 '15 at 3:30

This all looks good to me. You were looking for an injective homomorphism $\phi : \mathcal{D}_3 \to \operatorname{Sym}(\mathcal{D}_3) \cong S_6$ by letting $\mathcal{D}_3$ act on itself; and that's exactly what you've found.

In general, you've followed the process:

  • Set up a bijection, $L : \{1, 2, \ldots, \lvert G\rvert \} \to G$, labeling the elements of your group
  • For each group element $g \in G$, construct the permutation $\varphi_g = L^{-1} \circ r_g \circ L,$ which is a permutation in $S_{\lvert G \rvert}$ (here, the map $r_g : G \times G \to G$ is the right multiplication by $g$ map; I don't know if you used right- or left-multiplication, but it would only be a superficial difference leading to isomorphic representations)
  • Your injective homomorphism is then $\phi : G \to S_{\lvert G \lvert}$, given by $\phi(g) = \varphi_g$

I'm not sure I understand your comments about cycles, or $\Bbb Z_6$. You're just writing permutations in $S_6$ in cycle notation; of course there will be cycles. It doesn't mean that your group is cyclic (but again, I'm really not sure what that comment means).

As far as "a group action on $\Bbb Z_6$", I guess you could call it that, but it would be misleading. You could consider this a group action on any set of $6$ elements (by setting up a bijection like the $L$ used here), but traditionally you'll pick the integers from $1$ to the order of your group (or just the original group elements themselves). However, just because your group acts on a set in bijection with $\Bbb Z_6$, that doesn't mean this permutation representation ties them together in any way. It's strictly about cardinality here; groups act on sets, you need a homomorphism to $\Bbb Z_6$ (which this isn't) if you want to relate their algebraic properties.


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