2
$\begingroup$

Let $GL_n(K)$ be the General Linear group over $K$ (i.e. all invertible matrices with components of $K$). Let $A$ be an element of $GL_n(K)$ and $C$ be an element of $K^n$. We denote the affine map determined by $(A,C)$ by $f(A,C):X\mapsto AX+C$. Show that $GL_n(K)$ is a subgroup, perhaps normal, of the affine group (group of all affine maps).

I have already proved myself that the set of all affine maps forms a group called the affine group and that $GL_n(K)$ is itself a group. So then all I have to prove is that $GL_n(K)$ is a subset of the affined group and show if it is a normal subgroup or not. Could anyone help me on these two parts? Thank you.

$\endgroup$
0
$\begingroup$

Let me denote the affine group of $K^n$ by $A_n(K)$.

If you've shown that $A_n(K)$ is a group, you must understand how composition works. It shouldn't be hard to guess what subgroup of $A_n(K)$ must be isomorphic to $GL_n(K)$; if things in $A_n(K)$ look like $(A, C)$ where $A \in GL_n(K)$, it's most likely the case that $H = \{(A, 0) : A \in GL_n(K)\}$ is isomorphic to $GL_n(K)$. You should be able to verify the map

\begin{align*} \varphi : H &\to GL_n(K)\\ (A, 0) &\to A \end{align*} is an isomorphism (there's almost nothing to it).

The straightforward approach to see if (the subgroup isomorphic to) $GL_n(K)$ is normal in $A_n(K)$ is to conjugate the elements of (the subgroup isomorphic to) $GL_n(K)$ by something in $A_n(K)$ and see if you wind back up in (the subgroup isomorphic to) $GL_n(K)$.

Thus, we pick $(A, C) \in A_n(K)$ and $(B, 0) \in H \cong GL_n(K)$ and compute $(A, C)^{-1}(B, 0)(A, C)$. Let's see what $(A, C)^{-1}(B, 0)(A, C)$ does to $X \in K^n$, working right-to-left:

\begin{align*} (A, C)^{-1}(B, 0)(A, C)X &= (A,C)^{-1}(B, 0)(AX + C) \\ &= \ldots \\ &= (A^{-1}BA, A^{-1}BC)X. \end{align*}

But $(A^{-1}BA, A^{-1}BC) \in H$ if and only if $A^{-1}BC = 0$; this certainly doesn't hold for all $(A, C) \in A_n(K)$ (in fact it holds for very few of them: since $A$ and $B$ are invertible...).

$\endgroup$
2
  • $\begingroup$ So H (and therefore GLn(K)) is not a normal subgroup right? $\endgroup$ – P-S.D Jan 9 '17 at 18:36
  • $\begingroup$ Yes, correct! The subgroup $H \cong GL_n(K)$ is not fixed by conjugation, and so is not normal in the whole affine group. $\endgroup$ – pjs36 Jan 9 '17 at 18:40
0
$\begingroup$

Let be $GL_n(K)$ General Linear group over K and $AGL_n(K)$ affine group that $$AGL_n(K)=\{(A,C); A\in GL_n(K), C\in K^n \}$$ Every $(A,C)\in AGL_n(K)$ we can write by form matrix $ (\array{A &C \\0 &1 }) $ then $AGL_n(K) \subseteq GL_n(K)$ and $AGL_n(K)$ subgroup in $GL_n(K)$, and we can define a map $$g:AGL_n(K)\to GL_n(K); g((A,C))=A$$ we have $g$ is surjective homomorphism and $kerg=(I_n,K^n)$ and $ AGL_n(K)/(I_n,K^n) \cong GL_n(K)$.

$\endgroup$
2
  • $\begingroup$ Thank ou for this proof. How do you show that every (A,C) in GLn(K) can be written by form matrix (A,C,0,1) $\endgroup$ – P-S.D Jan 9 '17 at 17:29
  • $\begingroup$ oh no, I understand now no worries. $\endgroup$ – P-S.D Jan 9 '17 at 17:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.