A presentation of a group of order 12 
Show that the presentation $G=\langle a,b,c\mid a^2 = b^2 = c^3 = 1, ab = ba, cac^{-1} = b, cbc^{-1} =ab\rangle$ defines a group of order $12$.

I tried to let $d=ab\Rightarrow G=\langle d,c\mid d^2 =c^3 = 1, c^2d=dcdc\rangle$. But I don't know how to find the order of the new presentation. I mean I am not sure how the elements of new $G$ look like. (For sure not on the form $c^id^j$ and $d^kc^l$ otherwise $|G|\leq 5$).
Is it good step to reduce the number of generators or not necessary?
 A: Direct proof.
$N:=\langle a,b \rangle$ is clearly a normal subgroup of $G$ with $G/N = \langle c : c^3 = 1 \rangle = C_3$, and $N$ is a quotient of the Klein four-group $V_4 = \langle a,b : a^2=b^2=1, ab=ba \rangle$. Therefore $|G|$ divides $12$. Now one checks that the permutations $a=(12)(34)$, $b = (14)(23)$, $c = (123) \in A_4$ satisfy the relations which define $G$, and that they generate $A_4$. Hence we get a surjective homomorphism $G \to A_4$. Since $|G|$ divides $12$, this shows that $G \cong A_4$ is an isomorphism and $|G|=12$.
A more conceptual proof using semidirect products.
The automorphism group of the Klein four-group $V_4$ is $S_3$ (since you can permute $a,b,ab$ freely, you can also see this from linear algebra applied to $V_4 \cong \mathbb{F}_2^2$). In particular there is an automorphism $c$ of order $3$, namely the one mapping $a \mapsto b, b \mapsto ab, ab \mapsto a$. This means that the cyclic group $C_3 = \langle c : c^3 = 1 \rangle$ of order $3$ acts on this group. The corresponding semidirect product $V_4 \rtimes C_3$ has the desired presentation $\langle a,b,c : a^2=b^2=c^3=1, ab=ba, c a c^{-1} = b, c b c^{-1} = ab \rangle$. See MO/96078 for the universal property and the resulting group presentation of a semidirect product. But the usual construction of the semidirect product $N \rtimes H$ shows that $|N \rtimes H|= |N| |H|$. In particular, $G \cong V_4 \rtimes C_3$ has order $12$.
A: Although my approach is not the way you expected, it is a nice way for your group. This way is called Coset enumeration or Todd-Coxeter Algorithm. You have see $12$ rows completed as follows. In fact, I found $[G:\langle e\rangle]=12$:

A: Consider $D := \langle a, b\rangle$ and $C = \langle c \rangle$. Note that
$$ab = ba \implies D = \langle a \rangle \langle b \rangle$$
so $|D| \leq 4$. Further, the last two relations tell you that $c \in N_{G}(D)$ (the normalizer in $G$ of $D$), so $C \leq N_{G}(D)$. It follows that $DC \leq G$, but $DC$ contains every generator of $C$; therefore, $DC = G$. We then have
$$|G| = |DC| = \frac{|D||C|}{|D \cap C|} \leq \frac{|D||C|}{1} \leq \frac{4 \cdot 3}{1} = 12$$
It just remains to find a group of order $12$ satisfying the given relations - so how many groups of order $12$ do you know?
A: If you note in the original presentation that the exponents of $a,b,c$ are $2,2,3$ whose product is $12$ a natural strategy would be to try to put a general element in the form $a^pb^qc^r$  and then show that these elements are distinct.
A: We can construct a directed graph called the Cayley diagram of the group $G := \langle a,b,c | a^2=b^2=c^3=1, ab=ba, ca=bc, cb=abc \rangle$ with respect to the generators $a,b,c$.  The vertices of this digraph will be the group elements.  The set of arcs will be of of the form $\{(g,gs): g \in G, s \in \{a,b,c\} \}$.  After this digraph is constructed completely, each vertex $g \in G$ will have 3 outgoing arcs, labeled with colors $a,b,c$, and 3 incoming arcs with colors $a,b,c$, and the number of vertices will be the order of the group. 
To construct this digraph, start with the identity vertex 1. Draw an arc from 1 to $a$, and label the arc as $a$. This is an edge-colored arc, denoted $(1,a,a)$, where the third coordinate refers to the color of the arc $(1,a)$.  Add a new arc from vertex $a$ to vertex $a^2$ with color $a$, and notice that the relation $a^2=1$ forces us to identify the vertex $a^2$ and $1$.  So we really have a directed cycle of length 2 at this point.  Another directed cycle of length 2 arises due to arcs $(1,b,b)$ and $(b,1,b)$.  Next we draw a directed cycle of length 3 on vertices $1,c,c^2$.  
Next start at vertex $a$ and draw its 3 outgoing arcs, one of which will $(a,b,ab)$.  Draw three outgoing arcs from $b$, one of which will be $(b,a,ba)$.  Since $ab=ba$, we identify (i.e. merge) these two vertices $ab$ and $ba$.  In addition, we start at the existing vertices (say the vertex 1), and enforce each relation at each vertex.  Thus, starting at 1, the directed path $ab$ of length 2 and the directed path $ba$ of length 2 must end up at the same vertex.  Similarly, $ca$ and $bc$ end up at the same vertex, as do $cb$ and $abc$.  
Continue this process until every vertex has three outgoing and three incoming arcs, and until each relation is enforced at each vertex.  We identify vertices whenever a relation forces us to do so.  The number of vertices will then end up being the order of the group.  I constructed this digraph in your example and verified it has order 12. 
A: The presentation can be rewritten as  $\langle d,c\mid d^2 =c^3 = (cd)^3=1 \rangle$.
This is the standard presentation for the symmetry group (rotations)  of the regular tetrahedron where $d$ represents a rotation about an edge, $c$ represents a rotation about a face and $cd$ represents a rotation about a vertex.
