# Quotient of free group with normal subgroup

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?

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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

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 $Ax^5B=Ay^2B$.