Why the Euclidean Group $E(n)=\left\{Ax+b|A\in O(n),b \in T(n) \right\}$? I'd like to show
$$E(n)=\left\{ f |f(x)=Ax+b \ \text{where} \ A\in O(n),b \in T(n) \right\}$$ 
To be more precise, $E(n)$ is the isometry group of $\mathbb{R}^n$, which is also known as $I(\mathbb{R}^n)$.
without utilizing the semidirect product that is utilized in the detailed discussion part of here (because I haven't learned it yet).
I believe it is possible, but I need some assistance.
So far, I've shown that if $f \in E(n)$, then $f(\lambda a + (1-\lambda)b)=\lambda f(a)+(1-\lambda)f(b)$ for $a,b \in \mathbb{R}^n, \lambda \in \mathbb{R}$ by the fact that $f \in GL(n,\mathbb{R})$. As the wikipedia article pointed out, this indicates that $E(n)$ is a subgroup of an affine group.
I think $E(n)$ being a subgroup of an affine group implies that we can let $f(0) = c$, so $g(0) = T_{-c}f(0)=0$ for some $g = T_{-c}f$ where $T_{-c}(x)=x-c$.
And, I think I should show that $g$ is also an isometry, then show that $g$ is in $O(n)$ so that $f(x)=T_{c}g(x)=g(x)+c$ to get a desired result.
Is my strategy correct? Then I wonder how to show that $g \in O(n)$. In addition, I'm not sure if $T(n) \triangleleft E(n)$ should be considered in this approach.
Any kind of help would be greatly appreciated.
 A: This is a rather lengthy proof based on the assumption that $E(n)$ is just the set of isometries of $\mathbb{R}^n$, i.e., $E(n)=\{f:\mathbb{R}^n\to\mathbb{R}^n |\, dist(f(x),f(y))=dist(x,y) \text{ for all }x,y\in\mathbb{R}^n\}$. If you assume that the isometries in $E(n)$ are affine, i.e., of the form $f(x)=Ax+b$ for some matrix $A$ and vector $b$, the proof is much simpler as you only need to show that $b=f(0)$ and that $A$ maps the standard basis to another ONB of $\mathbb{R}^n$.
Let $f\in E(n)$. Then $f$ is an isometry and since the composition of two isometries is also an isometry and all translations are isometries, it follows that $g(x):f(x)-f(0)$ is also an isometry and it satisfies $g(0)=0$.
Let $x\in\mathbb{R}^n\setminus\{0\}$ and $\lambda\in\mathbb{R}\setminus\{0\}$. The triangle with vertices $g(0)$,$g(x)$ and $g(\lambda x)$ has side of lengths $\|x\|,|\lambda|\|x\|$ and $|1-\lambda|\|x\|$ and is therefore degenerate, i.e., the points $g(0),g(x),g(\lambda x)$ are colinear. Checking the lengths, it follows that $g(\lambda x)=\lambda g(x)$.
Observe that the norm is preserved by $g$ since $\|x\|=dist(x,0)=dist(g(x),g(0))=dist(g(x),0)=\|g(x)\|$. In other words, if we denote by $C(z,r)$ the circle of radius $r$ centered at $z$, we have that $g(C(z,r))\subset C(g(z),r))$.
Now take $x,y\in\mathbb{R}^n\setminus\{0\}$ such that $x\perp y$. Considering the triangle with vertices $g(x),g(y),g(0)$, we see that it has edges of lenths $\|x\|,\|y\|$ and $dist(g(x),g(y))=\|x-y\|$. However, $\|x-y\|^2=\langle x-y,x-y\rangle=\langle x,x\rangle-2\langle x,y\rangle+\langle y,y\rangle=\|x\|^2+\|y\|^2$, where $\langle\cdot,\cdot\rangle$ is the usual dot product in $\mathbb{R}^n$, since $x\perp y$ is equivalent to $\langle x,y\rangle=0$. Therefore the triangle with vertices $g(0),g(x),g(y)$ is right-angled, so $g(x)\perp g(y)$.
We can therefore extend the pair $(g(x),g(y))$ to an orthogonal basis of $\mathbb{R}^n$, say $\{g(x),g(y),\xi_3,\xi_4,\dots,\xi_n\}$. We can write $g(x+y)$ in terms of this basis in a unique way, say $g(x+y)=ag(x)+bg(y)+\sum_{k=3}^n\alpha_k\xi_k$, for $a,b,\alpha_k\in\mathbb{R}$.
Now, for arbitrary $\lambda\in[0,1]$,
\begin{equation*}
\begin{split}
\|g(x+y)-\lambda g(x)\|^2 & =\|(a-\lambda)g(x)+bg(y)+\sum_{k=3}^n\alpha_k \xi_k\|^2 \\
& =(a-\lambda)^2\|x\|^2+b^2\|y\|^2+\|\sum_{k=3}^n\alpha_k\xi_k\|^2,
\end{split}
\end{equation*}
since the vectors $g(x),g(y),\sum_{k=3}^n\alpha_k\xi_k$ are pairwise perpendicular. On the other hand, note that $\|g(x+y)-\lambda g(x)\|^2=\|g(x+y)-g(\lambda x)\|^2=\|(x+y)-\lambda x\|^2=(1-\lambda)^2\|x\|^2+\|y\|^2$ (again $x\perp y$), so subtracting these two equations we get
\begin{equation*}
(a^2-2\lambda a+2\lambda -1)\|x\|^2+(b^2-1)\|y\|^2+\|\sum_{k=3}^n\alpha_k\xi_k\|^2=0\quad\quad (*)
\end{equation*}
Playing the same game for $g(x+y)-\mu g(y)$ for arbitrary $\mu\in[0,1]$, we obtain
\begin{equation*}
(a^2-1)\|x\|^2+(b^2-2\mu b+2\mu-1)\|y\|^2+\|\sum_{k=3}^n\alpha_k\xi_k\|^2=0\quad\quad (**)
\end{equation*}
Subtracting $(*)$ from $(**)$ and dividing by $2$ we obtain $\lambda(a-1)\|x\|^2=\mu(b-1)\|y\|^2$, and this must hold for any $\lambda,\mu\in[0,1]$. Since $\|x\|,\|y\|\neq 0$, this implies $a=b=1$. But then, from equation $(*)$ for example, we see that $\|\sum_{k=3}^n\alpha_k\xi_k\|=0$.
Therefore, $g(x+y)=g(x)+g(y)$ whenever $x\perp y$.
Now let $e_1,e_2,\dots,e_n$ be the standard basis of $\mathbb{R}^n$, i.e. $e_i$ is the vector containing all zeros except the $i$-th entry which is $1$. Let $g_i:=g(e_i)$ for $1\leq i\leq n$ and note that, by the above, for any $1\leq i\neq j\leq n$, $g_i\perp g_j$ and therefore $g_1,g_2,\dots,g_n$ is also an ONB for $\mathbb{R}^n$.
Taking an arbitrary $x\in\mathbb{R}^n$, we can uniquely write it as $x=\sum_{k=1}^n a_k e_k$ and therefore, using scaling and linearity with respect to perpendicular vectors, we see that $g(x)=\sum_{k=1}^n a_k g_k$. Hence, $g$ is a linear map that sends the standard basis of $\mathbb{R}^n$ to another ONB of $\mathbb{R}^n$. By definition, this means $g(x)=Ax$ for some orthogonal matrix $A$ (in fact the columns of $A$ are just the $g_i$) and therefore $f(x)=Ax+f(0)$, i.e., $f$ is the translation of some orthogonal transformation.
