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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.

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  • $\begingroup$ 1. Your interpretation of $\pi$ is correct. 2. They're not the same thing, but there exists an isomorphism $u:\operatorname{GL}(n,\mathbf R)/\operatorname{SL}(n,\mathbf R)\simeq \mathbf R^\times$ such that $\; f=\det=u\circ \pi$. 3. Any group which is isomorphic to $\mathbf R^\times$. $\endgroup$
    – Bernard
    Commented May 21, 2018 at 14:17
  • $\begingroup$ @Bernard Sorry, I don't know what $ f=\det=u\circ \pi$ means? Could you explain that notation? $\endgroup$
    – user563280
    Commented May 21, 2018 at 14:21
  • $\begingroup$ I mean $f=u\circ \pi$, and incidentally I recall the standard name of $f$ is the determinant map ($\det$). $\endgroup$
    – Bernard
    Commented May 21, 2018 at 14:24
  • $\begingroup$ @Bernard Okay, but I still have one confusion. Isn't "taking determinant" just defined for matrices? How can we define it for "equivalence classes of matrices having the same determinant"? That is, it doesn't make sense to take the determinant of an "equivalence class" (?) Perhaps I need to re-frame the definition of $f$ somehow, but not sure $\endgroup$
    – user563280
    Commented May 21, 2018 at 14:26
  • $\begingroup$ @Bernard Also, I don't know what $f=u\circ \pi$ means. What does the $\circ$ stand for? $\endgroup$
    – user563280
    Commented May 21, 2018 at 14:28

1 Answer 1

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  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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  • $\begingroup$ I can't understand your second point. Could you please elaborate? Are you telling me that $GL(n,\mathbb{R})/SL(n,\mathbb{R})$ is isomorphic to $\mathbb{R}\setminus\{0\}$? Could you please add the proof for that? $\endgroup$
    – user563280
    Commented May 21, 2018 at 14:18
  • $\begingroup$ Moreover, what does $M$ stand for? I'm not completely sure how you're defining the isomorphism $\endgroup$
    – user563280
    Commented May 21, 2018 at 14:23
  • $\begingroup$ Yes, $GL(n,\mathbb{R})/SL(n,\mathbb{R})$ and $\mathbb{R}\setminus\{0\}$ are isomorphic. Let $f\colon GL(n,\mathbb{R})/SL(n,\mathbb{R})\longrightarrow\mathbb{R}\setminus\{0\}$ defined by $f\bigl(M.SL(2,\mathbb{R})\bigr)=\det M$. And let $g\colon\mathbb{R}\setminus\{0\}\longrightarrow GL(n,\mathbb{R})/SL(n,\mathbb{R})$ be the map defined by$$g(x)=\begin{pmatrix}x&0&0&\ldots&0\\0&1&0&\ldots&0\\\cdots&\cdots&\cdots&\ddots&\cdots\\0&0&0&\ldots&1\end{pmatrix}.$$They are group homomorphisms and they are the inverse of each other. $\endgroup$ Commented May 21, 2018 at 14:28
  • $\begingroup$ But $g(x)=\begin{pmatrix}x&0&0&\ldots&0\\0&1&0&\ldots&0\\\cdots&\cdots&\cdots&\ddots&\cdots\\0&0&0&\ldots&1\end{pmatrix}$ does not produce "equivalence classes of matrices". It simply produces a single matrix for every element of $\mathbb{R}\setminus\{0\}$. We need it to produce "equivalence classes of matrices having the same determinant $x$". $\endgroup$
    – user563280
    Commented May 21, 2018 at 14:32
  • $\begingroup$ @Blue Sorry. What I meant is that $g(x)$ is the equivalence class of that matrix. $\endgroup$ Commented May 21, 2018 at 14:36

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