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I'm trying to make up an example of a quotient of a free group to check if I understand quotients properly. I do for the usual cases but I've not seen free groups before. So here I go:

Let $F = \langle x, y \rangle$ and let $N$ be the group that is generated by the element $x^5 y^{-2} = 1$, i.e. the relation $x^5 = y^2$.

Is it right then that

$F/N = \{ x, x^2, \dots , x^5, \dots, x^6y^3x^6, \dots, x^6y^{-3}, \dots\} / N = $ $\{ x, x^2, \dots , x^5, \dots, x^6y^3x^6, \dots, xy^{-1}, \dots\}$, i.e. all the elements in $N$ get reduced to $1$, even if they're sandwiched between two expressions?

Thanks for your help!

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Yes. When you mod out a relation such as $x^5=y^2$, you really mod out $N$ that is the smallest normal subgroup of $F$ containing $r=x^5y^{-2}$. In other words, your $N$ contains all conjugates of $r$ like $y^{-1}ry=y^{-1}x^5y$, their products $r(xrx^{-1})$ et cetera. –  Jyrki Lahtonen Jul 7 '11 at 9:00
    
An exponent $-1$ is missing from one of the $y$s. Sorry about that. The good answer by PseudoNeo gives more information anyway. –  Jyrki Lahtonen Jul 7 '11 at 17:23

1 Answer 1

up vote 2 down vote accepted

Yes. I'll let you see how you can write that completely using the very definition, following Jyrki's comment.

What you have to understand, is that in the quotient group, the relations become equalities (in a sense, the quotient group is the largest group in which those relations are equalities). And of course, in a group, if you have $x^5 = y^2$, you have various consequences, among which all the equalities $Ax^5B=Ay^2B$.

Remark: In general, the difficulty isn't to prove equalities in the quotient group (that is, to derive consequences from the relations) but to prove inequalities. It's in general very hard to prove that the group is "not too small", that is to find things that aren't consequences of the relation.

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