Let $F_2$ be the free group with generators $x_1$,$x_2$, and let $F_3$ be the free group with generators $y_1$,$y_2$,$y_3$.
We define a group homomorphism $\phi:F_3\rightarrow F_2$ by $\phi(y_1):=x_1^2$, $\phi(y_2):=x_1x_2$, $\phi(y_3):=x_2^2$, and another group homomorphism $\psi:F_2\rightarrow \mathbb{Z}/2\mathbb{Z}$ by $\psi(x_1):=1$, $\psi(x_2):=1$.
Show that the kernel of $\psi$ is equal to the image of the homomorphism $\phi$, and hence is isomorphic to a free group on three generators.
Until yet I showed that $\phi$ is injective (relevant?); then I proceeded with:
$\ker(\psi)=\{x\in F_2:\psi(x)=0\}$
We look for $x \in F_2$ such that $\psi(x)=0$ (identity element in $\mathbb{Z}/2\mathbb{Z}$), then, working with the generators, $0=1+1=$
- $=\psi(x_1)+\psi(x_2)=\psi(x_1x_2)$
- $=\psi(x_1)+\psi(x_1)=\psi(x_1^2)$
- $=\psi(x_2)+\psi(x_2)=\psi(x_2^2)$
So these compositions of generators of $F_2$ are mapped to zero by $\psi$, and in particular their equal to, respectively, $\phi(y_2)$,$\phi(y_1)$,$\phi(y_3)$.
Is it enough to say that $\ker(\psi)={\rm Im}(\phi)$ ?
From this can I conclude immediately that it is isomorphic to a free group on three generators?
Best Answer
$\;\;\;\;$ The kernel $\ker(\psi)$ is precisely the normal $F_2$ subgroup $F_2'$ of words of even length, each of which we may unambiguously refer to as $\it{even\;words}$.
$\;\;\;\;$ The $F_2$ subgroup $\text{Im}(\phi)=\langle x_1^2,x_1x_2,x_2^2\rangle$ is generated by even words and therefore $\text{Im}(\phi)\subseteq F_2'$ with $x_2x_1^{-1}=(x_1x_2x_2^{-2})^{-1}\in\text{Im}(\phi)$ whereas $F_2'=\langle x_1^2,x_1x_2,x_2x_1,x_2^2\rangle$ is generated together by all the possible words of length $2$. Therefore, to see $\text{Im}(\phi)=F_2'$ it suffices to show $x_2x_1\in\text{Im}(\phi)$ ; $x_2x_1=x_2x_1^{-1}x_1^2\in \text{Im}(\phi)$.