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Suppose I have two presentations for groups:

$\langle x,y|x^{7} = y^{3} = 1, yx = x^2y\rangle$ and $\langle x,y|x^{7} = y^{3} = 1, yx = x^4y\rangle$

What is the standard approach to deciding whether the presentations are isomorphic?

I'm working through an application of Sylow Theory which classifies groups of order $21$.

In the text it says that these two presentations above are isomorphic, but I cannot see how to prove it or even suspect it.

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Well, you'd want to find a set of generators of the first group that satisfied the relations of the second group. If we rewrite $yx=x^2y$ as $yxy^{-1}=x^2$ (which turns this somewhat abstract equality into something a bit more concrete), we see immediately that $y^2xy^{-2}=x^4$ and indeed since $y^2$ is of order 3, $x,y^2$ are the generators you're looking for.

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I think you mean $yx^2y^{-1}=x^4$, and that you would mean $y^2$ is of order $3$ but that instead you'll change that based on the other typo to $x^2$ is of order $7$. –  Kevin Carlson Oct 12 '12 at 23:19
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No, that's not what I meant. If you take $x^2$ and $y$ as generators, you get the same relations back. The stated equality is correct (although maybe I shouldn't have used the word "immediately"); the idea is simply that if conjugating by $y$ corresponds to squaring, then conjugating twice should correspond to taking a fourth power. –  anonymous Oct 12 '12 at 23:31
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Yep, you're right on all material counts-my deleted answer looks to have constructed the inverse of what I meant it to. Don't you still want to say $y^2$ is of order $3$ to guarantee you can use it as a generator? –  Kevin Carlson Oct 12 '12 at 23:46
    
That was the intended implication but the argument is certainly clearer (and equally short) if I say $y^2$ is of order 3, so I have edited my answer accordingly. –  anonymous Oct 12 '12 at 23:50
    
Thanks for the great answer. –  Kyle Schlitt Oct 13 '12 at 0:56

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