9
$\begingroup$

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?

$\endgroup$
2
  • 2
    $\begingroup$ Just take a simple example, for instance, images of $\mathbb Z$. $\endgroup$ Feb 18, 2013 at 0:13
  • 2
    $\begingroup$ From a purely set theoretic standpoint, the fact that $\phi$ and $\psi$ are surjections does not guarantee that $f$ is a surjection. $\endgroup$ Feb 18, 2013 at 1:24

1 Answer 1

19
$\begingroup$

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 $\mathbb{Z_4} \cong \mathbb{Z_2}\times\mathbb{Z_2}$.

$\endgroup$
3
  • 3
    $\begingroup$ 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, 2013 at 0:20
  • 8
    $\begingroup$ 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! $\endgroup$ Feb 18, 2013 at 0:22
  • 3
    $\begingroup$ @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, 2013 at 0:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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