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 I have homomorphism $f:A\to B$ of abelian groups.

Is it possible in general to find a bijection between $A$ and $\operatorname{im}(f)\times \operatorname{ker}(f)$?

The canonical isomorphism between $im(A)$ and $A/\operatorname{ker}(f)$ makes me think the answer should be yes.

In the example of $g:\mathbb{Z}\to\mathbb{Z}_{2}$, where $g(x)$ is the remainder when $x$ is divided by $2$, $\operatorname{im}(g) = \{0,1\}$ and $\operatorname{ker}(g) = 2\mathbb{Z}$.

In this case I can easily decompose $x\in \mathbb{Z}$ into a unique sum of $y\in 2\mathbb{Z}$ and $z\in \{0,1\}$, and then use the mapping $x\mapsto (y,z)$. e.g.: pairing $17$ with $(16,1)$, $98$ with $(98,0)$ etc.

I need to show that the number of elements of $A$ is the product of the number of elements in $\operatorname{im}(f)$ and $\operatorname{ker}(f)$, and I think this is the idea but I am having difficulty extending this to general groups.

Any advice on how I can extend this generally?

My problem in general is choosing the element of $\operatorname{ker}(f)$. Given $x\in A$, the obvious choice for the element of $\operatorname{im}(A)$ is $f(x)$.

share|cite|improve this question
In any group $G$ with normal subgroup $N$, if $T$ is a transversal (an exhaustive collection of distinct coset representatives), then every $g\in G$ can be written uniquely in the form $tn$ for some $t\in T$ and $n\in N$. Can you prove this? Can you see how to use this? – anon Sep 7 '12 at 2:14
Yes i see it. So I don't need to know the choice of element of $ker(f)$, but just know that it's there and unique. Thanks! – roo Sep 7 '12 at 2:24
up vote 5 down vote accepted

I think it is easier than this. You know that $A/\ker(f)\cong\operatorname{im}(f).$ Elements of $A/\ker(f)$ are cosets with cardinality $\lvert\ker(f)\rvert$ which partition $A$, and which are in one-to-one correspondence with elements of $\operatorname{im}(f)$

share|cite|improve this answer

By the First Isomorphism Theorem

$$\frac{A}{\ker(f)} \sim Im(f) \,.$$

So you are trying to build a bijection between $A$ and $\frac{A}{\ker(f)} \times \ker(f)$.

Hint For each class $\overline{x} \in \frac{A}{\ker(f)}$ pick one representative $y_{\overline{x}}$. $x- y_{\overline{x}} \in \ker(f)$.

P.S. It should be pretty obvious that the set $\frac{A}{\ker(f)} \times \ker(f)$ has the same cardinality as $A$. $\frac{A}{\ker(f)}$ can be viewed as a partition of $A$ into $\left| \frac{A}{\ker(f)} \right|$ sets of cardinality $|\ker(f)|$.

share|cite|improve this answer

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.