If $G, H$ are groups and $\phi : G \to H$ is a homomorphism, is it true that $G/\ker \phi \times \ker \phi \cong G$?

I am pretty sure this is right, but I can't remember how to prove it.

We can think of $\phi$ as a surjection of $G$ into $G/\ker \phi$, so I was thinking that for $\phi$ there ought to be a surjection $\psi : G \to \ker \phi$, such that $\psi$ maps an element of $G$ into its "position" in its coset of $\ker \phi$. Then then isomorphism between $G$ and $G/\ker \phi \times \ker \phi$ would be $f(g) = (\phi(g), \psi(g))$.

To show $f$ is an isomorphim we only need to show it is injective since $\phi, \psi$ are both surjective. $f(g) = f(g')$ implies $\phi(g) = \phi(g')$ so they are in the same coset of $\ker\phi$ and $\psi(g) = \psi(g')$ so they are in the same "position" in that coset. Therefore $g = g'$.

If my intuitive notion of position works, I am still not sure how I define $\psi$. Can anyone point me in the right direction?

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    $\begingroup$ Just take a simple example, for instance, images of $\mathbb Z$. $\endgroup$ – Thomas Andrews Feb 18 '13 at 0:13
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    $\begingroup$ From a purely set theoretic standpoint, the fact that $\phi$ and $\psi$ are surjections does not guarantee that $f$ is a surjection. $\endgroup$ – Jason DeVito Feb 18 '13 at 1:24

This is not true. $\mathbb{Z_4}$ has the normal subgroup$\{0,2\} \cong\mathbb{Z_2}$ with quotient $\mathbb{Z_2}$ but it is certainly not true that $\mathbb{Z_4} \cong \mathbb{Z_2}\times\mathbb{Z_2}$.

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    $\begingroup$ It's not even true that every group is the semidirect product of a normal subgroup and the quotient group (I can't think of a counterexample, but this came up on the big list of mathematical false beliefs over on MathOverflow). $\endgroup$ – Kris Feb 18 '13 at 0:20
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    $\begingroup$ It is very disappointing that it's not true, and it is a misconception mostly due to misleading notation! Who thought it would be a good idea to have a product and a quotient that aren't inverses! $\endgroup$ – Tom Oldfield Feb 18 '13 at 0:22
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    $\begingroup$ @Kris, A standard counter-example is the group of quaternions, which is not a semidirect product but it has normal subgroups. $\endgroup$ – Damien L Feb 18 '13 at 0:23

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