For every subgroup $H$ of a finite abelian group $ G$ there exist an endomorphism $\phi$ with $\text{Im}(\phi)=H$

Question

This is exercise 7 from "The Theory of Finite Groups" by Hans Kurzweil, page 49.

For every subgroup $H$ of a finite abelian group $ G$ there exist an endomorphism $\phi$ with $\text{Im}(\phi)=H$

I believe it can be tackled using the classification theorem of abelian groups, which states that every abelian group is a direct product of cyclic groups. However, I am unable to proceed further with the given information. I know that $G$ is a product of cyclic groups and $H$ and $G$ is a product of some of these factors.

Answer 1

Step 1. If $G$, $H$ are groups of relatively prime order (even non abelian) then every subgroup of $G\times H$ is a direct product of subgroups of $G$ and $H$.

Step 2. Given an abelian $p$-group $G$, it is isomorphic to product of cyclic $p$-groups. Then a subgroup of $G$ is isomorphic to direct product of subgroups of those cyclic groups (as you can see here: Subgroups of abelian $p$-groups). Note: unlike step 1 here we have only isomorphism.

Step 3. For cyclic group $\mathbb{Z}/p^n\mathbb{Z}$ its subgroups are simple: each is isomorphic to $\mathbb{Z}/p^m\mathbb{Z}$ for some $m\leq n$. Therefore there is an easy construction of an epimorphism $\mathbb{Z}/p^n\mathbb{Z}\to \mathbb{Z}/p^m\mathbb{Z}$.

Combine all the steps with direct product of homomorphisms and classification of finite abelian groups to obtain required homomorphism.

Answer 2

You want to see that for any finite abelian group $G$ and subgroup $H$ that there is a group homomorphism $G \to G$ with image $H$, or equivalently that there is a surjective group homomorphism $G \to H$. If we had instead asked for an injective group homomorphism $H \to G$ then the answer would be obvious: the inclusion map. Well, an answer to your question can be created from that, by using the dual of the inclusion map $H \to G$ in the setting of characters of finite abelian groups!

When $A$ is a finite abelian group, let $A^* = {\rm Hom}(A,S^1)$ be the character group of $A$: all homomorphisms from $A$ to the unit circle, which is a group under pointwise multiplication of functions.

The relation between $A$ and its character group $A^*$ is similar in many ways to that between a finite-dimensional vector space and its dual space. In particular, $A^* \cong A$ as groups and to each homomorphism $f : A \to B$ of finite abelian groups there is a homomorphism $f^* : B^* \to A^*$ going in the other direction between their dual groups, where $f^*(\chi) = \chi \circ f$, which is defined just like the dual to a linear map in linear algebra.

Okay, now I can address your question. When $G$ is a finite abelian group and $H$ is a subgroup, the dual of the inclusion homomorphism $\iota : H \to G$ is a homomorphism $\iota^* : G^* \to H^*$. Concretely, $\iota^*$ is nothing other than the restriction operation, sending a character $\chi$ on $G$ to its restriction $\chi|_H$ to the subgroup $H$. It can be shown that every character of a subgroup $H$ can be extended (non-canonically) to a character of $G$, so every $\psi \in H^*$ is some $\chi|_H$ and thus $\psi = \iota^*(\chi)$: the map $\iota^* : G^* \to H^*$ is surjective. (That $\iota$ is injective and $\iota^*$ is surjective is part of a general pattern: if a homomorphism $f : A \to B$ of finite abelian groups is injective then its dual homomorphism $f^*$ is surjective and if $f$ is surjective then its dual $f^*$ is injective by arguments that resemble the similar properties of dual linear maps in linear algebra.) Since $G^* \cong G$ and $H^* \cong H$ as groups, composing $\iota^* : G^* \to H^*$ with these isomorphisms gives us a homomorphism $G \to G^* \to H^* \to H$ that is surjective.

Remark. This method may appear to avoid needing the cyclic decomposition of finite abelian groups, but I left out details for the properties I mentioned, and the simplest way to see $G^* \cong G$ when $G$ is a finite abelian group is to use the cyclic decomposition of finite abelian groups to reduce the argument to the cyclic case.