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1) Let $F$ be the free group on the three generators $x,y,z$. For non-zero integers $r,s,t$

then CLAIM: the subgroup of $F$ generated by $x^r , y^s , z^t$ is freely generated by these elements.

2) Let $H$ be the subgroup of $F(\{x,y\})$ generated by $x^2 , y^2 , xy , yx$

then CLAIM: $H$ is not freely generated by these elements.

Can anyone help me?

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Typo in there - x^"? – Thomas Andrews Nov 2 '12 at 15:27
For some basic information about writing math at this site see e.g. here, here, here and here. – Julian Kuelshammer Nov 2 '12 at 15:43
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. – Julian Kuelshammer Nov 2 '12 at 15:44
@thomas yes, thank you! – Dexter Nov 2 '12 at 15:56
@Julian thank you, I will investigate – Dexter Nov 2 '12 at 15:58

2) there is a nontivial relation among them: $$(x^2)^{-1}(xy) = x^{-1}y = (yx)^{-1}(y^2) $$ 1) Consider the homomorphism $\phi$ from $F$ to this subgroup $G:=\langle x^r,y^s,z^t\rangle$, mapping $x\mapsto x^r$, $y\mapsto y^s$, $z\mapsto z^t$, and conclude that $\ker\phi=\{1\}$, thus $im\phi\cong F$, free with those generators.

Anyway, the elements of $G$ look like words with strings of letters $X:=\overbrace{xx..xx}^r$, $Y:=\overbrace{yy..yy}^s$ and $Z=\overbrace{zz..zz}^t$ and their inverses $X^{-1}$, $Y^{-1}$, $Z^{-1}$, and $\phi^{-1}$ thus is just replacing the strings $X$ to $x$, etc.

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thank you so much! – Dexter Nov 2 '12 at 15:44

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