Free groups, generators and group homomorphisms

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?

• I think you've shown $\operatorname{Image }(\phi) \subseteq \ker (\psi)$. You also need the opposite inclusion to show equality. Commented Nov 27, 2019 at 21:28
• And you need to show that $\phi$ is injective, else it is not clear that Im$(\phi)$ is isomorphic to the free group on three generators.
– lulu
Commented Nov 27, 2019 at 21:58

$$\;\;\;\;$$ 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 $$x_2x_1\in\text{Im}(\phi)$$ ; $$x_2x_1=x_2x_1^{-1}x_1^2\in \text{Im}(\phi)$$.