I have a doubt about free group in Dummit's Abstract Algebra on page220 :
More generally, suppose $G$ is presented by, say, generators $a, b$ with relations $r_1 , . . . , r_k$ . If $a', b'$ are any elements of a group $H$ satisfying these relations, there is a homomorphism from $G$ into $H$ . Namely, if $\pi : F(\{a, b\}) \to G$ is the presentation homomorphism, we can define $\pi' : F(\{a, b\}) \to H$ by $\pi ' (a) = a'$ and $\pi' (b) = b'$ . Then ${\color{red}{\ker \pi \leq \ker \pi'}}$ so $\pi'$ factors through $\ker \pi$ and we obtain
$$ G\cong F(\{a,b\})/\ker\pi\to H\ . $$
In the text, $F(\{a,b\})$ means the free group on the set $\{a,b\}$ . My doubt is that how can we get " $\ker\pi\leq\ker\pi'$ ", or I think it is more exactly to write " $\ker\pi=\ker\pi'$ " : $\pi(a)=a,\pi(b)=b,\pi'(a)=a',\pi'(b)=b'$ together with the fact that $a,b$ and $a',b'$ have the same relations, must imply that $\ker\pi=\ker\pi'$ . Is my thought correct or false?
Here $H$ is an arbitrary group with some elements that satisfy at least the relations that define $G$. It isn't just another version of $G$. It can have other elements/relations outside of the patterns that define $G$.
For example, suppose $G$ is presented by a single generator $a$ with the single relation $4a = 0$ (I'm using additive notation rather than multiplicative notation for what follows). Then $G$ is a cyclic group of order 4 (i.e. $G \cong \mathbb{Z}/4\mathbb{Z}$).
Let $H = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/9\mathbb{Z}$, and consider the element $a' = (1, 1, 0) \in H$. Note that this element satisfies the relation in $G$'s presentation.
The claim is that the homomorphism $\pi': \langle a \rangle \rightarrow H$ given by $a \mapsto a'$ factors through $\ker \pi$, which works because $4\mathbb{Z}a = \ker \pi \le \ker \pi' = 2\mathbb{Z}a$.
So we get a homomorphism from $G\cong \mathbb{Z}/4\mathbb{Z}$ to $H = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/9\mathbb{Z}$ (with $1 \mapsto (1, 1, 0)$) just because the given element of $H$ satisfies the defining relation of $G$.
My choice of $H$ here was unnecessarily complicated - I just wanted to demonstrate that $H$ doesn't have to have the same number of generators (mine isn't cyclic like $G$) and can have parts that are totally ignored by the image of the homomorphism (like the $\mathbb{Z}/9\mathbb{Z}$ component here).