Show that $S_3$ is presented by $\langle a,b\mid a^3, b^2,ab=ba^2\rangle$ 
Show that group $S_3$ of the objects $x,y,z$ is presented by $\langle a,b \mid a^3,b^2,ab=ba^2\rangle$ under the mapping $a \to (xyz)$ , $ b\to (xz)$

I'm confused to what is to be shown in these type of problems, I've started learning Combinatorial Group theory  from book by Magnus, and having a hard time to understand the subject! 
 A: In this case, you need to show two things: (1) that the given relations hold for the specific realization of the generators that you're given, and (2) that they actually generate all the elements of the group, and no elements that aren't in the group.
Towards (1), the relations $a^3$ (i.e., $a^3=e$) and $b^2$ should be easy; to show that $ab=ba^2$ just find the permutation representation of both sides and show that they're equal.
(2) is a bit trickier, and in general it's an impossible problem; even showing from a given presentation of a group whether it's the trivial group or not is (literally) undecidable in the general case!  But here, there's enough structure to make things relatively straightforward.  Firstly, you can use $ab=ba^2$ to move any instances of $b$ in a word all the way to the left, and so show that all words in the group are equivalent to a word in the form $b^ia^j$; then the relations $a^3=e$ and $b^2=e$ should let you give bounds on the exponents $i$ and $j$.  Finally, just enumerate all of the words of this form and use the realizations of $a$ and $b$ you're given to map them to the elements of $S_3$, showing that you get all the permutations this way.
A: When you want to show that a group $H$ has a presentation $\langle G,R\rangle$ where $G$ is a set of generators and $R$ is a set of relators (or relations), you need to show two things:


*

*Each relation is satisfied in $H$. More precisely, the image under the epimorphism $\varphi:F\to H$ (in your case $φ(a)=(xyz),φ(b)=(xz)$) of every relator is trivial. This implies that there is a well-defined epimorphism $\tildeφ:F/⟨R⟩\to H$, where $F$ is the free group generated by $G$ and $⟨R⟩$ is the smallest normal subgroup of $F$ containing the relators.

*In order to show that $\tildeφ$ is injective, we have to show $ker(φ)\subseteq⟨R⟩$. This means that whenever some product of generators in $F$ is mapped to the identity in $H$, then this string can be turned into the empty word by an application of finitely many relations.
Consider for example the word $ab^{-1}ab$, this is mapped to $(xyz)(xz)(xyz)(xz)$. But this is the same as $(xz)(xzy)(xz)(xzy)=(xz)(xz)(xyz)(xzy)=e$. So the image of $ab^{-1}ab$ is trivial. On the other hand, $ab^{-1}ab=abab=ba^2ba^2=baba^4=bba^6=e$.
The idea is to move all instances of $(xz)$ to the left by the formula $(xyz)(xz)=(xz)(xzy)$ and $(xzy)(xz)=(xz)(xyz)$. These modifications correspond to the relations $ab=ba^2$ and $a^2b=ba$ (or equivalenty $bab=a^2$). Then since the sign of the identity is $1$, there must be an even number of transposition $(xz)$ on the left (which reduce to zero) followed by $(xyz)$'s whose number is a multiple of $3$.
