[Math] A group homomorphism from $GL(n,\Bbb {R})$ such that $SL(n,\Bbb {R})$ is the kernel

abstract-algebragroup-homomorphismlinear-groups

I have a confusion about the answer to this question: Every normal subgroup is the kernel of some homomorphism.

I was working on the following problem:

Define a group homomorphism from $GL(n,\Bbb{R})$ to a suitable group
so that the kernel is $SL(n,\Bbb{R})$.

I think one easy homomorphism would be simply $f:GL(n,\Bbb{R})\to \Bbb{R}^{\times}$ (i.e. we just take the determinant of the invertible matrices in the general linear group)

But according to this answer, since $SL(n,\Bbb{R})$ is a normal subgroup of $GL(n,\Bbb{R})$ we could also consider the homomorphism $\pi:GL(n,\Bbb{R})\to GL(n,\Bbb{R})/SL(n,\Bbb{R})$ defined by $\pi(x)=xSL(n,\Bbb{R})$ such that $x\in GL(n,\Bbb{R})$.

Questions:

  1. Is my interpretation of $\pi$ correct? I'm not completely sure what the members of the quotient group $GL(n,\Bbb{R})/SL(n,\Bbb{R})$ would be in this case. Would it (i.e. $GL(n,\Bbb{R})/SL(n,\Bbb{R})$) simply be the "group of equivalence classes of $n\times n$ matrices having the same determinant"?

  2. Could also someone clarify whether $f$ and $\pi$ are actually the same thing?

  3. Are more such possible group homomorphisms from $GL(n,\Bbb R)$ to a suitable group such that $SL(n,\Bbb R)$ is the kernel?

P.S: Please note that I'm only a beginner in abstract algebra, so it would help if you would keep the language simple and avoid unnecessary jargon while answering.

Best Answer

  1. Yes, your intepretation of $\pi$ is correct.
  2. They are basically the same thing, since $GL(n,\mathbb{R})/SL(n,\mathbb{R})$ and $\mathbb{R}\setminus\{0\}$ under the map $[M]\mapsto\det M$.
  3. Consider $\det^3$, for instance.
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