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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 $f(x)$.

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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! –  Kyle Schlitt Sep 7 '12 at 2:24
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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 $\operatorname{im}(f)$

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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 $|\ker(f)|$.

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