Free group on $\{x,y\}$ and Group $\mathbb Z\times \mathbb Z$ Let $F$ be free group on $\{x,y\}$ and Let $G=\mathbb Z\times \mathbb Z$ be a group with operation coordinatewise addition.
So there exists an unique homomorphism namely: $$\varphi:F\to G\\with\\\varphi(x)=(1,0), \varphi(y)=(0,1)$$
Since $$\langle (1,0),(0,1) \rangle=\mathbb Z\times \mathbb Z$$
I want to identify $$\ker(\varphi)=\{z\in F | \varphi(z)=(0,0)\}$$
Now observe $$\varphi(xy)=\varphi(x)+\varphi(y)=(1,1)\\ \varphi(yx)=(1,1)\\ \varphi(xyx^{-1}y^{-1})=(0,0)$$
I claim $$\langle xyx^{-1}y^{-1} \rangle = \ker(\varphi)$$
To prove it I take general object from $z\in \ker(\varphi)$ and I want to show $z=(xyx^{-1}y^{-1})^n$ for some $n$ -integer.
First of all I cannot think general terms of $F$ properly, and cannot reduce it. Any hint, answer would be appreciated.
 A: Your claim is incorrect. Certainly $xyx^{-1}y^{-1}\in\ker\varphi$, but it is not true that $\langle xyx^{-1}y^{-1}\rangle=\ker\varphi$. For example, $x^{-1}y^{-1}xy\in\ker\varphi$ but $x^{-1}y^{-1}xy\neq(xyx^{-1}y^{-1})^n$ for any $n\in\mathbb{Z}$.
The confusion seems to be with presentations. Yes, $\langle x, y\mid xyx^{-1}y^{-1}\rangle$ is a presentation for $\mathbb{Z}^2$, but the notation "$\langle x, y\mid xyx^{-1}y^{-1}\rangle$" means $$\frac{F(x, y)}{\langle\langle xyx^{-1}y^{-1}\rangle\rangle}$$ where $\langle\langle xyx^{-1}y^{-1}\rangle\rangle$ denotes the smallest normal subgroup containing the element $xyx^{-1}y^{-1}$, and not the smallest subgroup (that is, not $\langle xyx^{-1}y^{-1}\rangle$). The subgroup $\langle\langle xyx^{-1}y^{-1}\rangle\rangle$ is much bigger than the other one, as for example it is closed under conjugacy so we see that the element given above is contained in this subgroup: $$x^{-1}y^{-1}xy=(xy)^{-1}\cdot xyx^{-1}y^{-1}\cdot xy\in\langle\langle xyx^{-1}y^{-1}\rangle\rangle.$$

More generally, it can be proven that a normal subgroup $N$ of a free group $F$ is finitely generated if and only if it has finite index in $F$. The subgroup $\langle xyx^{-1}y^{-1}\rangle$ is finitely generated, but does not have finite index in $F(x, y)$ (as it is contained in the infinite-index subgroup $\ker\varphi$), so with this result in mind we immediately know it is not normal in $F(x, y)$.
