Question about a homomorphism from free group to $\mathbb{Z}/2\mathbb{Z}$ I am currently reading through the solution to the following problem: 

Let $F$ be the free group on the set $\{a,b\}$ and let $\alpha: F \rightarrow \mathbb{Z} / 2 \mathbb{Z}$ defined by $\alpha(a) = \alpha(b) = 1$. Show $\ker \alpha = \langle a^2 , b^2, ab \rangle$. 

It is easy to show that $b^2, a^2, ab \in \ker \alpha$ and so $\langle a,b,ab \rangle \subseteq \ker \alpha$, but the reverse containment is what is giving me trouble. The approach taken to show that $H = \langle a ,b : a^2 , b^2 , ab \rangle$ has at most 2 elements. While I understand how to show that $H$ has at most 2 elements, I do not see precisely why this implies that $a^2, b^2, ab$ generate $\ker \alpha$.  
 A: There are a couple of different angles you can look at this situation from. One is that you have nested normal subgroups
$$ \langle a^2, b^2, ab \rangle \subseteq (\ker \alpha) \subseteq F $$
which implies that you also have nested quotients
$$ F \to F / \langle a^2, b^2, ab \rangle \to F / (\ker \alpha) $$
You have that the inclusion $\langle a^2, b^2, ab \rangle \subseteq \ker \alpha$ is an equality if and only if the quotient map $F/\langle a^2, b^2, ab \rangle  \to F / (\ker \alpha)$ is an isomorphism.
The quotient map is surjective since it's a quotient map, so you have a surjective map from one group of order either 1 or 2 to another group known to be order 2.
A: Since $\langle a^2, b^2, ab \rangle \subseteq \ker \alpha$ there is a surjective map
$$ F/\langle a^2, b^2, ab \rangle \to F/\ker \alpha \tag{1} $$
Thus
$$ |F/\langle a^2, b^2, ab \rangle| \ge |F/\ker \alpha| = |\mathbf{Z}/2\mathbf{Z}| $$
where $H = F/\langle a^2, b^2, ab \rangle$ by definition. Thus if $|H| = 2$ then the map $(1)$ is an isomorphism. That implies $\langle a^2, b^2, ab \rangle = \ker \alpha$.
