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

Suppose $G$ abelian and $f:G\rightarrow \mathbb Z$ is surjective with kernel $K$.

  • Show that $G$ has a subgroup $H$ such that $H \cong \mathbb{Z}$
  • Show that $G \cong H\bigoplus K$

To get started:

There exists $g \epsilon G$ such that $f(g)=1$. Now set $H=<g>$. Clearly $H \cong \Bbb Z$.

share|cite|improve this question
This is false. Consider $\mathbb{Z}/p^2 = G$ and $f$ multiplication by $p.$ Then $G \not\cong ker(f) \oplus im(f) \cong \mathbb{Z}/p \oplus \mathbb{Z}/p.$ – jspecter Oct 31 '12 at 2:20
$Z$ was changed to $\Bbb Z$...should work...formatting lapse on my part. – KUSH Oct 31 '12 at 3:00
up vote 3 down vote accepted

If you knew anything about projective modules, this result is pretty much trivial.

If you don't, I think it's still not bad. Let $h\in G$ be arbitrary (but fixed) such that $f(h)=1$ (such an $h$ exists because $f$ is onto). It is easy to show $|h|$ (the order of $h$) is infinite (otherwise, $0=f(0)=f(|h|h)=|h|f(h)=|h|\cdot 1=|h|>0$). Thus, $G\geq\langle h\rangle\cong\mathbb{Z}$. Denote $\langle h\rangle$ by $H$.

If, $g\in G$ is arbitrtary, say $f(g)=n\in\mathbb{Z}$. Write $g=nh+g-nh=nh+(g-nh)$. Obviously $nh\in H$, and $f(g-nh)=f(g)-nf(h)=n-n\cdot 1=n-n=0$, so $g-nh\in \mathrm{ker}f=K$. This shows $G=H+K$. To show this is a direct sum, it suffices to show $H\cap K=\{0\}$. This is clear because no multiple of $h$ may map to $0$ (by the same argument from the previous paragraph).

share|cite|improve this answer
Thanks...I don't know projective modules but that makes sense! – KUSH Oct 31 '12 at 3:14

Since $f$ is surjective, there exists $a \in G$ such that $f(a) = 1$. Let $g\colon \mathbb{Z} \rightarrow G$ be a unique homomorphism such that $g(1) = a$. Since $fg = 1$, $g$ is injective. Let $H = g(\mathbb{Z})$. Then $H$ is isomorphic to $\mathbb{Z}$. For $x \in G$, $f(x - gf(x)) = f(x) - f(x) = 0$. Hence $x - gf(x) \in K$. Hence $G = H + K$. Suppose $y \in H \cap K$. There exists $n \in \mathbb{Z}$ such that $y = g(n)$. Since $0 = f(y) = fg(n) = n$, $y = 0$. Hence $G = H \oplus K$.

share|cite|improve this answer
Thanks, very clear! – KUSH Oct 31 '12 at 3:15

Let us look at the short exact sequence

$$0\rightarrow \ker f\stackrel{i}\rightarrow G\stackrel{f}\rightarrow \Bbb Z\rightarrow 0$$

with $\,i\,$ injection.

Let us use the universal property of the free abelian group $\,\Bbb Z=\langle\,1\,\rangle\,$ : choose $\,x\in G\setminus\ker f\,$ and put $\,\overline g(1):=x\,$ , then there exists a unique homomorphism $\,g:\Bbb Z\to G\,$ s.t. $\,g(1)=\overline g(1)=x\,$.

Since $\,\overline 0\neq x+\ker f\in G/\ker f\cong\Bbb Z\,$ , clearly $\,x\,$ has infinite order and thus

$$H:=g(\Bbb Z)=\langle\,x\,\rangle\,\cong\Bbb Z$$

Now : $$y\in H\cap \ker f\Longrightarrow y=mx\wedge f(y)=0\;\;,\;m\in\Bbb Z\Longrightarrow$$

$$ 0=f(mx)=mf(x)\Longrightarrow m=0\vee x\in \ker f$$

Since $\,x\notin\ker f\,$ by choice, it must be $\,m=0\Longrightarrow y=0\,$

Can you take it from here now and end the argument?

share|cite|improve this answer
Since the full answer has been accepted, here's a great exposition of the end of the argument: – Alexei Averchenko Oct 31 '12 at 5:19

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.