# a doubt on free group in Dummit&Foote's Abstract Algebra

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?

• $H$ could satisfy additional relations, not just $r_1,\dots, r_k$, and then the two kernels could be different. Commented Jun 4, 2023 at 4:15

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).

• Your answer helps me realize this fact: the dihedral group $D_4$ has the same relations with $D_8$ but since a presentation means the largest group which generated by $s,r$ satisfying the relations $s^2=r^4=srsr=1$ , $\langle s,r\ |\ s^2=r^4=srsr=1\rangle\neq D_4$ . In the past I misunderstood the relation $r^n$ in the presentation refered to $|r|=n$ , and hence I thought that $|G|\leq|H|$ and then $\ker\pi$ must equal $\ker\pi'$ . Your example correct my mistake. But now I still do not know how to show $\ker\pi\leq\ker\pi'$ in a strict way. Thanks for your reply! Commented Jun 4, 2023 at 8:07
• Could you help me check if this idea can reach the desired result: $\ker\pi=\langle frf^{-1}\ |\ f\in F(S),r\in R\rangle$ , and since $H$ has additional relations, so $\ker\pi\leq\ker\pi'$ . Commented Jun 4, 2023 at 8:20
• That seems like the right idea to me. The point is that $\ker \pi$ is the normal closure of the relations defining $G$ (which is what you wrote). Since the specified elements of $H$ satisfy at least those relations, $R \subset \ker \pi'$. But $\ker \pi'$ is normal, so it contains the normal closure of $R$ (i.e. $\ker \pi$). Commented Jun 4, 2023 at 21:15