Two questions about groups and presentations i have two questions about group presentations:


*

*Given the presentation $\langle a,b:a^2=1=b^3,(ab)^3=1\rangle$. I know that this is the presentation of $A_4$ but how to deduce that. Should i give a homomorphism from the presentation to $A_4$ and conclude that this is an isomorphism? If so, how to prove that such a mapping is injective/surfective/homomorphism? For example we can make $a\mapsto (12)$ and $b\mapsto (123)$. But how to go further to get the result?

*Given the groupspresentation $G=\langle a_1,\cdots,a_g:\prod_{i=1}^g{a_i^2}=1\rangle$ (especially this is the presentation of the fundamentalgroup of closed non-orientable surfaces. I want to compute the abilization $G/[G,G]$. From my point of view it has to be $(\Bbb{Z}/2\Bbb{Z})\times\Bbb{Z}^{g-1}$. But how must i argument to make this clear?


Thank you for help, hints and solutions :)
 A: For (1), you know $A_4$ satisfies the presentation (just take $a = (12)(34), b = (123)$, and note that $a b = (134)$; please note that the element $(12)$ you consider is not in $A_{4}$), so the presented group
$$
G = \langle a,b:a^2=1=b^3,(ab)^3=1\rangle
$$ 
has order at least $12$, because it has $A_{4}$ as a homomorphic image. (For all we know at this stage, it might also be infinite.)
Now in $G$, note that the first relation  and the second relation mean
$$a^{-1} = a, \qquad b^{-1} = b^{2};\tag{pow}$$ also, note the consequences of the third relation:
$$
b a b = a b^{-1} a, 
\qquad 
a b a = b^{-1} a b^{-1}
.
\tag{cons}
$$
Now consider the elements
$$
a, b^{-1} a b,
$$
and the subgroup $V = \langle a, b^{-1} a b \rangle$ they span in $G$.
The two elements commute with each other, as (pow) and (cons) imply
$$
a (b^{-1} a b) = (a b^{-1} a) b = (b a b) b = b a b^{-1} = b^{-1} (b^{-1} a b^{-1}) = b^{-1} (a b a) = (b^{-1} a b) a.
$$
Moreover, again by (pow) and (cons),
$$
b^{-1} (b^{-1} a b) b = (b a b) b = (a b^{-1} a) b = a (b^{-1} a b) \in V.
$$
So $V$ is a normal subgroup of $G$, of order at most $4$, and the quotient group $G/V$ is generated by $b$, so $G$ has order at most $12$.
It follows that $G$ has order precisely $12$, and it is isomorphic to $A_4$.
For (2), if $H$ is your abelianization, then note that $H$ is generated by (the images of) $a_1, \dots, a_{g-1}$ and by $b = a_{1} a_{2} \dots a_{g}$. With respect to these generators, the relation becomes $b^{2} = 1$. So you are talking of the abelian group with presentation
$$
\langle a_1, \dots, a_{g-1}, b : b^2 = 1 \rangle
$$
which is indeed isomorphic to $\Bbb{Z}^{g-1} \times \Bbb{Z}_{2}$.
A: As for your second question, if you let $a=\prod^g_{i=1} a_i$ you can rewrite the abelianization of $G$ as
$$
G/[G,G]=\langle a_1,\dotsc,a_{g-1},a : a^2=1,[a_i,a_j]=1,[a_i,a]=1\rangle
$$
i.e. it is just $\Bbb Z^{g-1}\times C_2 \simeq \Bbb Z^{g-1}\times \Bbb Z/2\Bbb Z$.
