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Suppose you have a group isomorphism given by the first isomorphism theorem:

$G/ker(\phi) \simeq im(\phi)$

What can we say about the group $ker(\phi)\times im(\phi)$? In particular, when does the following hold:

$G\simeq ker(\phi)\times im(\phi)$?

I ask this question because i want to prove that $GL_n^+(\mathbb{R}) \simeq SL_n(\mathbb{R}) \times \mathbb{R}^*_{>0}$, with $GL_n^+(\mathbb{R})$ the group of matrices with positive determinant. I proved that $SL_n(\mathbb{R})$ is a normal subgroup and that $GL_n^+(\mathbb{R})/ SL_n(\mathbb{R}) \simeq \mathbb{R}^*_{>0}$, using the surjective homomorphism $det(M)$. I tried something with semidirect products but I got stuck.

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    $\begingroup$ You need to prove that $G$ has a normal subgroup isomorphic to $im(\phi)$ and which intersects trivially with $ker(\phi)$. $\endgroup$ – Tobias Kildetoft May 20 '13 at 14:20
  • $\begingroup$ @TobiasKildetoft The conclusion does not necessarily follow if $G$ is infinite... $\endgroup$ – user1729 May 20 '13 at 14:30
  • $\begingroup$ @user1729 Thank you, I forgot to mention that you also need the group to be generated by these two subgroups. $\endgroup$ – Tobias Kildetoft May 20 '13 at 14:33
  • $\begingroup$ (Counter-example: Take the group $\langle a, t; a^t=a^{-1}\rangle$, then this is not a cross-product. Take the map which kills $a$ in the presentation (and so maps onto $\mathbb{Z}$). Then note that $\langle t^2\rangle$ is normal and intersects trivially with the kernel of this map.) $\endgroup$ – user1729 May 20 '13 at 14:35
  • $\begingroup$ @TobiasKildetoft: I do not think that even that is enough! My point is that in infinite groups the image can contain an isomorphic copy of itself which can be normal in the whole group without the kernel itself being normal. I do not think generation should be too difficult to get around. $\endgroup$ – user1729 May 20 '13 at 14:37
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Let $N = ker(\phi)$ and $K = im(\phi)$, then you're asking when, given an exact sequence $1 \to N \to G \to K \to 1$ is trivial.

  • First you need the extension to be split, that is, there must exist a morphism $s : K \to G$ such that the composition $\phi \circ s$ is the identity. In this case $G \simeq N \rtimes K$, the semidirect product of $N$ and $K$ : this is the splitting lemma (for non-abelian groups).
  • Now you want this semidirect product to be direct; this is true iff $K$ is also normal in $G$, or equivalently that there exists a morphism $G \to N$ which is the identity on $N$.

I don't include proofs here, as they're found in any basic group theory notes.


In fact you can get away with the first condition. Indeed, if there exists a map $p : G \to N$ which is the identity on $N$, then a section of $\phi$ automatically exists and the isomorphism $G \cong \operatorname{im}(\phi) \times \operatorname{ker}(\phi) = K \times N$ holds. The required isomorphism is $(\phi, p) : G \to K \times N$ (it's not hard to check that this is in fact an isomorphism).

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Even if it can't be applied to your example, I would like to point out that in the abelian case (more generally in any abelian category) it's equivalent to have a split exact sequence: $0 \to \ker(\phi) \to G \to Im(\phi) \to0$

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  • $\begingroup$ (The key word being split.) $\endgroup$ – user1729 May 20 '13 at 15:57

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