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i have two questions about group presentations:

  1. 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?
  2. 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 :)

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2 Answers 2

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 $\Bbb{Z}^{g-1} \times \Bbb{Z}_{2}$.

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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 $\Bbb Z^{g-1}\times C_2 \simeq \Bbb Z^{g-1}\times \Bbb Z/2\Bbb Z$.

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Why is this a presentation of $G/[G,G]$? –  Darrick59 Mar 20 '13 at 11:23
    
A presentation for $G/[G,G]$ is just a presentation for $G$ plus relations for the commutators of the generators. Then by commutativity $\left(\prod^g_{i=1}a_i\right)^2=\left(\prod^g_{i=1}a_i^2\right)$. Moreover $a_1,\dotsc,a_{g-1},a$ is a system of generators, since $a_g=a(\prod^{g-1}_{i=1}a_i^{-1})$. –  A.P. Mar 20 '13 at 11:29

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