When does the isomorphism $G\simeq \ker(\phi)\times \operatorname{im}(\phi)$? hold?

Question

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.

Answer 1

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.

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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).

Answer 2

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: $0 \to \ker(\phi) \to G \to Im(\phi) \to0$

Answer 3

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 $(N, t) \mapsto t^{1/n} N$.