Study of Affine Group

In order to simplify things, I'll use the block matrix multiplication rule (see https://en.wikipedia.org/wiki/Block_matrix#Block_matrix_multiplication).

For clarity, I will denote by $I$ (rather than by $1$) the identity matrix, and I'll use capital letters for $n \times n$ matrices, and lowercase letters for $n$-vectors.

To show that $M$ is a subgroup isomorphic to $\mathbb{R}^n$, first we define a map $\mathbb{R}^n \rightarrow M$ as \[ b \mapsto \begin{pmatrix} I & b \\ 0 & 1 \end{pmatrix}; \] it is obviously bijective, so all we are left to do is to show that this is a group homomorphism. For $b_1, b_2 \in \mathbb{R}$, we have \[ \begin{pmatrix} I & b_1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} I & b_2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} I \cdot I + b_1 \cdot 0 & I \cdot b_2 + b_1 \cdot 1 \\ 0 \cdot I + 1 \cdot 0 & 0 \cdot b_2 + 1 \cdot 1 \end{pmatrix} = \begin{pmatrix} I & b_1 + b_2 \\ 0 & 1 \end{pmatrix} \] which is indeed the image of $b_1 + b_2$. This proves that $M$ is a subgroup of $G$ isomorphic to $\mathbb{R}^n$ under addition.

To show that $N$ is a subgroup isomorphic to $GL_n(\mathbb{R})$, first we define a map $GL_n(\mathbb{R}) \rightarrow N$ as \[ A \mapsto \begin{pmatrix} A & 0 \\ 0 & 1 \end{pmatrix}; \] again, it is obviously bijective, so all we are left to do is to show that this is a group homomorphism. For $A_1, A_2 \in GL_n(\mathbb{R})$, we have \[ \begin{pmatrix} A_1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} A_2 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} A_1 \cdot A_2 + 0 \cdot 0 & A_1 \cdot 0 + 0 \cdot 1 \\ 0 \cdot A_2 + 1 \cdot 0 & 0 \cdot 0 + 1 \cdot 1 \end{pmatrix} = \begin{pmatrix} A_1 A_2 & 0 \\ 0 & 1 \end{pmatrix} \] which is indeed the image of $A_1 A_2$. This proves that $N$ is a subgroup of $G$ isomorphic to $GL_n(\mathbb{R})$ under multiplication.

Finally, to prove that $G = MN$, it is enough to take a generic element of $G$ and write it as product of an element in $M$ and an element in $N$. Given \[ \begin{pmatrix} A & b \\ 0 & 1 \end{pmatrix} \in G, \] we can take \[ \begin{pmatrix} A & 0 \\ 0 & 1 \end{pmatrix} \in M \quad \text{and} \quad \begin{pmatrix} I & A^{-1} b \\ 0 & 1 \end{pmatrix} \in N  \] and check that \[ \begin{pmatrix} A & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} I & A^{-1} b \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} A \cdot I + 0 \cdot 0 & A \cdot A^{-1} b + 0 \cdot 1 \\ 0 \cdot I + 1 \cdot 0 & 0 \cdot A^{-1} b + 1 \cdot 1 \end{pmatrix} = \begin{pmatrix} A & b \\ 0 & 1 \end{pmatrix} \] so any element of $G$ can be written as a product of an element in $M$ and an element in $N$, which implies $G < MN$. But since both $M < G$ and $N < G$, then $MN < G$ and so $MN = G$, as desired.