Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to understand why $\mathbb{Q}$ is an injective abelian group but $\mathbb{Z}$ is not.

We say that an abelian group $A$ is injective if it has the following property: Given an abelian group $Q$, a subgroup $P<Q$ and a homomorphism $f:P\rightarrow A$ then there exists a homomorphism $g:Q\rightarrow A$ whose restriction to $P$ is $f$.

$\mathbb{Z}$ is not injective and this can be seen by considering the group $Q=(\mathbb{Q},+)$ with $P = \mathbb{Z}$ and $f:\mathbb{Z} \rightarrow \mathbb{Z}$ the identity map.

The following is a proof that $\mathbb{Q}$ is injective(from some of my course notes):

Let $Q$ be an abelian group with subgroup $P$ emedded within $Q$ via an injective map $i:P\rightarrow Q$. Since $P$ and $Q$ are both abelian they are direct products of cyclic groups. Let $G$ be a set of generators for $P$ and $H$ a set of generators for $Q$ containing $i(G)$. Notice that the image $f(P)$ is free abelian, since torsion elements must be mapped to $0\in\mathbb{Q}$

Define a map $g:H\rightarrow \mathbb{Q}$ by $g(x)=f(y)$ if there exists $y\in G$ so $i(G)=x$, otherwise define $G(x)=0$. Extend this map in the natural unique way to make a homomorhism $g:Q\rightarrow \mathbb{Q}$. This is well defined as $i$ is injective. Moreover, for all $x\in G$, we have $g(x)=f(y)=f(i(x))$; so as the image of $P$ under both maps is free abelian we have $g(x)=f(i(x))\forall{x}\in P$. Hence $\mathbb{Q}$ is injective.

I understand the counterexample for $\mathbb{Z}$ and I understand the proof. However I cannot understand why this proof doesn't apply to $\mathbb{Z}$. Having thought about this for a while I am now not sure if the proof is valid, since it assumes all abelian groups (not just finitely generated) are the product of cyclic groups.

Thanks for any help

share|cite|improve this question
you might want to rewrite "Z is not injective and this can be seen by considering the group Q=(Q,+) with P=Z and f:Z→Z the identity map.".... – Praphulla Koushik Oct 23 '13 at 9:37
In fact, as Rotman says, $\mathbb Q$ is divisible while $\mathbb Z$ is not. And that's why it has not the injective property. – Babak S. Oct 23 '13 at 9:39
@BabakS. thanks, so where in the proof that $\mathbb{Q}$ is injective do we use the fact that $\mathbb{Q}$ is divisible? – hmmmm Oct 23 '13 at 9:54
up vote 2 down vote accepted

The proof is bogus. Take for example $Q=\mathbb Z$, $P=2\mathbb Z$, $G=\{2\}$, $H=\{1,2\}$. If $f\ne0$, there is no homomorphism $g$ extending $f$ with $g(1)=0$.

share|cite|improve this answer
Why does this show that the proof that $\mathbb{Q}$ is injective is bogus? – hmmmm Oct 23 '13 at 10:23
Because, if I read it correctly, this is exactly what the construction in the proof yields. Did I misunderstand it? – Carsten S Oct 23 '13 at 10:48
@hmmmm Carsten is right. The proof is bogus, because the map $g$ it defines is not a group homomorphism. If $y\in Q$ is not in $P$, but $2y$ is, then you cannot just send $y$ to 0, since the image of $2y$ is determined for you by $f$. You have to send $y$ to "half that image", i.e. to some element of $A$ that doubles to $f(2y)$. This also shows why the fact that $\mathbb{Q}$ is divisible is crucial for it to be injective. – Alex B. Oct 23 '13 at 10:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.