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.
 A: Let $GL_n(K)$  be the General Linear group over $K$ and $AGL_n(K)$ be the affine group that $$AGL_n(K)=\{(A,C); A\in GL_n(K), C\in K^n \}$$
Every $(A,C)\in AGL_n(K)$ can be written in the form of a $(n+1)\times (n+1)$ matrix $ \left(\array{A &C \\0 &1 }\right)$. This matrix is invertible since $A$ is invertible. Therefore, $AGL_n(K) < GL_{n+1}(K)$. And also $GL_n(K)$ can be embedded as a subgroup of $AGL_n(K)$. To see this, we can define a map $$g:GL_n(K)\to AGL_n(K)$$ $$ A\mapsto g(A)=(A,0)$$ we have $g$ a homomorphism as an inclusion map. This shows $GL_n(K)$ can be embedded as a subgroup of $AGL_n(K)$. In conclusion,$$GL_n(K) < AGL_n(K) < GL_{n+1}(K)$$ Additionally, we can define a homomorphism $h$ from $AGL_n(K)$ to $GL_n(K)$,$$h:AGL_n(K)\to GL_n(K)$$ $$ (A,C)\mapsto h(A,C)=A$$ and $\ker(h)=(I_n,K^n)$ and by the first isomorphism theorem of group we have $$ AGL_n(K)/(I_n,K^n) \cong GL_n(K)$$.
A: 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...).
