Proving Injectivity of Group Homomorphism Given Relations

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Homework Statement



Let G = \langle x,y \ | \ x^2, y^3, (xy)^3 \rangle, and f: G \rightarrow A_4 the unique homomorphism such that f(x) = a, f(y) = b, where a = (12)(34) and b = (123). Prove that f is an isomorphism. You may assume that it is surjective.

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N/A

The Attempt at a Solution



The previous stages of the question proved that the stated homomorphism was in fact unique, and considered the subgroup H = \langle y \rangle along with the set of cosets A = \{H, xH, yxH, y^2xH\}, from which I have proven that gH \in A, \ \forall g \in G. I'm not sure what relevance those stages have to the proof of isomorphism though I'd assume there is some relevance since otherwise the question would be a little disjoint; my only thought so far is that it relates to the first isomorphism theorem for groups, since we have a group homomorphism and a set of cosets, but I can't see how to make it work in this example. We did a similar question in lectures involving proving that D_{2n} \cong \langle x,y \ | \ x^n, y^2, (xy)^2 \rangle where injectivity was proven by rewriting elements of the group as words of the form x^ky^l using the given relations, and concluding that \#G \leq 2n. I have tried something similar with this question but am struggling to rewrite elements of G in a similar form (though I'd guess there'd be an irreducible factor of (xy)^m in the word somewhere) which is useful.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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