Let $\varphi: G \to G'$ be an injective homomorphism. If $G'$ is Abelian, then $G$ is Abelian? The context is fundamental groups.

Question

It is known that the image of a homomorphism of an abelian group is abelian (Here is a proof). I have a related conjecture.

Let $\varphi: G \to G'$ be an injective homomorphism. If $G'$ is Abelian, then $G$ is Abelian.

Proof: $$\varphi(xy)=\varphi(x)\varphi(y)=\varphi(y)\varphi(x)=\varphi(yx) \implies xy=yx$$

How do we generalize this? If we have

Let $\varphi: G \to G'$ be a surjective homomorphism. If $G$ has this property, then $G'$ has the same property.

then do we have

Let $\varphi: G \to G'$ be an injective homomorphism. If $G'$ has this property, then $G$ has the same property.

?

By the way, the context of this is Munkres Topology Lemma 60.5 and Theorem 60.6

Answer 1

For a given property $P$ your statements can be rewritten as:

(1) If $G$ has $P$ then so does a quotient of $G$

(2) If $G'$ has $P$ then so does a subgroup of $G'$

There are lots of examples of $P$ satisfying one of the above, two or none:

• (none) $P$ is "being infinite"

• (only (1)) $P$ is "being finitely genarated"

• (only (2)) $P$ is "being free"

• (both) $P$ is "being abelian"

etc.

There is no general method to check what kind of $P$ you are dealing with. You have to check case-by-case.