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Trying to establish a simpler proof for a theorem in a special case I arrived at the following question:

Let $c:\operatorname{GL}(\mathbb R^d) \to \mathbb R \setminus \{0\}$ an arbitrary group homomorphism. Can we show in general that $\operatorname{SO}(\mathbb R^d) = \operatorname{ker}(\operatorname{det}) \subseteq \operatorname{ker}(c)$ by giving simple algebraic arguments?

I showed that statement for a special case I was interested in and thought about the general situation for a time but in this case I don't see a potential point to attack this question.

Yaddle
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The second answer to this question here: Is the determinant the "only" group homomorphism from $\mathrm{GL}_n(\mathbb R)$ to $\mathbb R^\times$? shows that for every homomorphism $c$ of your form, there is a homomorphism $f$ such that $$c = f \circ \det.$$ Thus, the kernel of $c$ always contains the kernel of the determinant.

Dirk
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  • This is right, but since I want to prove exactly this statement in general using the one above, it is circular reasoning. I maybe should have added this information. – Yaddle Nov 23 '17 at 11:41
  • Well, there is a really elementary proof in the link, why not use it? – Dirk Nov 23 '17 at 11:46
  • I wondered if I could proof the statement purely structure theoretic by using the homomorphism theorem without having to deal with these calculations involving elemental matrices. That was my motivation to think about the question :) – Yaddle Nov 23 '17 at 11:49