# General Linear Group Subgroup of Affine Group

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.

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...).

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

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 $AGL_n(K)/(I_n,K^n) \cong GL_n(K)$.

• 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) – P-S.D Jan 9 '17 at 17:29
• oh no, I understand now no worries. – P-S.D Jan 9 '17 at 17:30