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

$$G/\ker(\phi) \simeq \operatorname{im}(\phi)$$

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

$$G\simeq \ker(\phi)\times \operatorname{im}(\phi)?$$

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 $ker(\phi)$. $\endgroup$ May 20, 2013 at 14:20
  • $\begingroup$ @TobiasKildetoft The conclusion does not necessarily follow if $G$ is infinite... $\endgroup$
    – user1729
    May 20, 2013 at 14:30
  • $\begingroup$ @user1729 Thank you, I forgot to mention that you also need the group to be generated by these two subgroups. $\endgroup$ May 20, 2013 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, 2013 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, 2013 at 14:37

3 Answers 3

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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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  • $\begingroup$ Please, does a map $p$ always exist if the group is abelian? I'm trying to prove this but up to no end $\endgroup$ Feb 25, 2021 at 11:53
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    $\begingroup$ @DaniloGregorinAfonso No, consider $0 \to \mathbb{Z}/2 \to \mathbb{Z}/4 \to \mathbb{Z}/2 \to 0$. $\endgroup$ Feb 25, 2021 at 12:15
  • $\begingroup$ Because in this case the only possible homomorphism would be $[0], [2] \mapsto [0]$ and $[1], [3] \mapsto [1]$? $\endgroup$ Feb 25, 2021 at 12:35
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    $\begingroup$ @DaniloGregorinAfonso Yes, or the zero homomorphism. Neither satisfy $p \circ s = \mathrm{id}$ where $s : \mathbb{Z}/2 \to \mathbb{Z}/4$ sends $1$ to $2$. $\endgroup$ Feb 25, 2021 at 12:40
  • $\begingroup$ Thank you, I will post another question with the exact statement of the example I'm trying to work out. Maybe I'm missing some detail $\endgroup$ Feb 25, 2021 at 12:41
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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, 2013 at 15:57
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In your special case you actually have a morphism $GL_n ^+ \to SL_n \times \mathbb{R}^+ $ given by

$$ M \mapsto (M/(\det M)^{1/n}, \det M) $$

the inverse being given by $(N, t) \mapsto t^{1/n} N$.

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  • $\begingroup$ The map is actually $A \mapsto (A \cdot \det(A)^{-1/n}, \det(A)^{1/n})$. $\endgroup$
    – Watson
    Oct 10, 2022 at 18:09
  • $\begingroup$ Thanks, I have edited accordingly $\endgroup$ Oct 12, 2022 at 7:44

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