Kernel of the homomorphism of an affine group onto the general linear group is the translation space

Question

Let $E$ be an affine space attached to a $K$-vector space $T$. Let $G$ be the group of all bijective affine mappings of $E$ onto itself. Write $\phi$ for the group homomorphism associating the $K$-linear isomorphism $\phi_u$ to $u\in G$ s.t.
$$u(t+x)=\phi_u(t)+u(x)$$
for all $t\in T$ and $x\in E$. Note that this means that $\phi$ is a surjection from $G$ onto $\mathbb{GL}(T)$. For each $t\in T$, write $h_t$ for the bijection of $E$ onto itself sending $x$ to $t+x$. Clearly $\{h_t\ |\ t\in T\}$ is a subset of $\text{Ker}(\phi)$. How can I show that $\text{Ker}(\phi)\subset\{h_t\ |\ t\in T\}$?

Let $u\in G$ such that $\phi_u=1_T$. Then $u(t+x)=t+u(x)$ for all $t\in T$ and $x\in E$. How can I show that $u=h_s$ for some $s\in T$?

Answer 1

Fixing an $x\in E$, note that $t+x$ runs through all elements of $E$ when $t\in T$.
Then $u(t+x)=t+u(x)=(t+x)+(u(x)-x)$, where $u(x)-x\in T$, so for all $y\in E$, we get $$u(y)=y+(u(x)-x)$$ thus $u=h_{u(x)-x}$.