# Showing a presentation is a permutation group.

This is Exercise 1.2.8 of "Combinatorial Group Theory: Presentations of Groups in Terms of Generators and Relations," by Magnus et al.

## The Question:

Show that the group $$\langle a, b, c\mid a^3, b^2, ab=ba^2, c^2, ac=ca, bc=cb\rangle$$ has order $12$ and find a permutation group isomorphic to it.

## My Attempt:

By inspection, every word can be reduced to $a^\alpha b^\beta c^\gamma$, where $\alpha\in\{0, 1, 2\}, \beta\in\{0, 1\}, \gamma\in \{0, 1\}$, which gives a total of $3\times 2\times 2=12$ elements.

One can see that, since $$\mathcal S_3=\langle a, b\mid a^3, b^2, ab=ba^2\rangle$$ under the mapping $\theta: a\mapsto (123), b\mapsto (12)$ and $C_2=\langle c\mid c^2\rangle$, we have $$\mathcal S_3\times C_2=\langle a, b, c\mid a^3, b^2, ab=ba^2, c^2, ac=ca, bc=cb\rangle$$ is the dihedral group of order $12$. But that doesn't help since I can't add on, say, $\theta(c)=(14)$ because $(12)(14)=(142)\neq(124)=(14)(12)$; that is, $b$ and $c$ don't commute.

I'm stuck finding a permutation group the presentation in question is isomorphic to.

The easiest way for two permutations to commute is to have disjoint support. (To use different symbols.) So, once you've got $a$ and $b$ figured out, with support $\{1,2,3\}$, and you want $c$ to commute, the easiest way to do that is to set $c=(45)$.
Eg, the four right cosets of $\langle a\rangle$ are $\langle a\rangle$, $\langle a\rangle b$, $\langle a\rangle c$ and $\langle a\rangle bc$. Call these $1, 2, 3$ and $4$. Each element of $G$ permutes them by right multiplication, for example $b$ sends $\langle a\rangle$ to $\langle a\rangle b$, $\langle a\rangle b$ to $\langle a\rangle b^2$ = $\langle a\rangle$, $\langle a\rangle c$ to $\langle a\rangle cb$ = $\langle a\rangle bc$, and $\langle a\rangle bc$ to $\langle a\rangle bcb$ = $\langle a\rangle c$, so $b$ would be $(12)(34)$.
• You can use a coset enumeration method. Start with a subgroup, say $H=\langle a\rangle$, and then let AllCosets and NewCosets be {H}. Then repeat: (1) Choose an element to remove from NewCosets (2) Multiply it by each of the generators, adding any new cosets to NewCosets and to AllCosets (3) Repeat 1&2 steps, untill NewCosets is empty. Eg, given $H=\langle a\rangle$, the first pass adds new cosets $Hb$ and $Hc$ (but $Ha$ is not new), the second pass adds only $Hbc$, (since $Hba=Hba^4=Ha^2b=Hb$, $Hbb=H$, $Hca=Hac=Hc$, $Hcb=Hbc$ and $Hcc=H$), and that's all the cosets you get. – Michael Hartley Aug 1 '17 at 1:03
• In general, this gives you a permutation representation of a subgroup of $G$, but you can guarantee you get $G$ itself by choosing $H={1}$. This is Cayley's theorem. – Michael Hartley Aug 1 '17 at 1:04
• It looks like, for your example, cosets of $\langle a\rangle$ and $\langle c\rangle$ are no use, but $\langle b\rangle$ gives you a permutation representation on six points. – Michael Hartley Aug 1 '17 at 6:41