Is this a valid group structure (of order 12)?

Question

Let $$ G = \big\langle \, a, b, c \colon a^3 = b^2 = c^2 = (ab)^2 = (bc)^2 = 1, ac = ca \big\rangle. \tag{0} $$ That is, $G$ is a group that has elements $a$, $b$, and $c$ (though these are not the only elements in $G$ of course) and the binary operation of $G$ is such that $$ a^{-1} = a^2, \ b^{-1} = b, \ c^{-1} = c, \tag{1} $$ and also $$ (ab)^{-1} = ab, \ (bc)^{-1} = bc, $$ that is, $$ b^{-1} a^{-1} = ab, \ c^{-1} b^{-1} = bc, $$ which together with (1) above yields $$ ba^2 = ab, \ cb = bc. \tag{2} $$

Thus $G$ has at least the following 13 distinct elements: $$ 1, a, b, c, ab, bc, ca, a^2 b, a^2c, a^2 bc, abc, ba, bac. $$

Is what I've stated above correct? If so, then what is the order of $G$?

If what I've stated is incorrect, then where have I erred?

Answer 1

SAGE provides a very powerful public domain environment for doing all calculations in finite group theory in particular, but also in many many many other branches of mathematics involving formal specifications.

For the case at hand, just type

https://sagecell.sagemath.org/

A window will open. Type in it :

F.<A,B,C> = FreeGroup()
G = F / [A^3, B^2, C^2, (A*B)^2,(B*C)^2,(A*C)/(C*A)]
G.list()

Then press button "Evaluate". The result will be the list of all constituents of the generated group in terms of $A,B,C$ :

$$(1, A, A^{-1}, B, C, AB, AC, A^{-1}B, A^{-1}C, BC, ABC, A^{-1}BC)\tag{1}$$

The generated group has therefore $12$ elements.

Please note

• how the group is defined as a quotient group of the free group on $3$ elements.

• how the generating relations are described : $A^3$ alone means $A^3=1$ (neutral element), and in particular $(AC)/(CA)$ meaning $(AC)(CA)^{-1}=1$.

You can also ask :

G.cayley_table()

and you will get :

*  a b c d e f g h i j k l
 +------------------------
a| a b c d e f g h i j k l
b| b a e f c d i j g h l k
c| c e a g b i d k f l h j
d| d j g h l b k a e f c i
e| e c b i a g f l d k j h
f| f h i j k a l b c d e g
g| g l d k j e h c b i a f
h| h f k a i j c d l b g e
i| i k f l h c j e a g b d
j| j d l b g h e f k a i c
k| k i h c f l a g j e d b
l| l g j e d k b i h c f a

Remark : the neutral element is denoted a.

#G.order()

G.as_permutation_group(limit=1000)

(please note that a line beginning by # is unevaluated).

The previous instruction gives you the 3 generators of $G$ as a subgroup of the permutation group $S_{12}$, each of them being written as a product of disjunct cycles :

 (1,2,3)(4,8,6)(5,7,9)(10,12,11), 
 (1,4)(2,6)(3,8)(5,10)(7,11)(9,12), 
 (1,5)(2,7)(3,9)(4,10)(6,11)(8,12)

No doubt that the first one corresponds to $A$ (whose order is $3$) and the second and third ones to $B$ and $C$.

There are many other things that you can ask. Refer to the abundant litterature on SAGE.

Edit : among them, here is this one :

G.is_isomorphic(DihedralGroup(6))

giving the answer : "True" !

Answer 2

Using the relations, every element of your group can easily be written $$a^ic^jb^k,\quad i\in\{0,1,2\},j,k\in\{0,1\}.$$ Therefore, $|G|\le12.$

Consider the morphism $\varphi$ from $G$ to the dihedral group $D_6$ of order $12$ generated by a rotation $r$ of order $6$ and a reflection $s$ of order $2,$ given by $$\varphi(a)=r^2,\varphi(c)=r^3,\varphi(b)=s.$$ It is:

• well-defined because these three images satisfy the given relations between $a,b,c:$ $$(r^2)^3 =(r^3)^2 =s^2 = (r^2s)^2 = (sr^3)^2 = 1,(r^2)(r^3)=(r^3)(r^2),$$
• surjective since its range contains the two generators $r=\varphi(ca^{-1})$ and $s=\varphi(b)$ of $D_6.$

Since moreover $|G|\le|D_6|,$ $\varphi$ is an isomorphism.

Alternatively, notice that your presentation amounts to: $G$ is the semidirect product $N \rtimes H$ of $$N:=\langle a,c\mid a^3=c^2=1,ac=ca\rangle\cong C_6$$ and $$H=\langle b\mid b^2=1\rangle\cong C_2,$$ the action of $H$ on $N$ being given by $$bnb^{-1}=n^{-1}$$ (for $n=$ each of the two generators $a,c$ hence for every $n\in N$), and such semidirect products $C_n\rtimes C_2$ are one of the definitions of the dihedral group of order $2n$.