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We define the following subgroups of the automorphism group of $k[x_1, \ldots, x_n]$

E(k,n) is the subgroup generated by elementary $k$-automorphisms of $k[x_1, \cdots, x_n]$ of the form $(x_1, x_2, \cdots, x_{i-1}, x_i + a, x_{i+1}, \cdots, x_n)$ where $a \in k[x_1,x_2, \cdots, x_{i-1},x_{i+1}, \cdots, x_n]$

Aff(k,n) is the subgroup of linear transformations

J(k,n) is the subgroup generated by elementary $k$-automorphisms of $k[x_1, \cdots, x_n]$ of the form $(x_1, x_2 + f_2, x_3 + f_3, \cdots, x_n+f_n)$ where $f_i \in k[x_1, \cdots, x_{i-1}]$ for $2 \leq i \leq n$.

I've been having trouble figuring out a way to write an element of E(k,n) as a composition of linear transformations and elements of J(k,n). I know it involves some permutation of the $x_i$s.

Any help would be appreciated. Thanks.

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You have to use that E(k,n) contains the automorphisms that permute the variables x1...xn.

If you pick a basic element $f_{i,a} = (x1,\dots,x_{i-1},x_i+a,x_{i+1},\dots,x_n) \in E(k,n)$, then conjugate with the automorphism $g_i \in Aff(k,n)$ that permutes $x_i$ with $x_n$, you get $g_i f_{i,a} g_i = f_{n,ag_i} \in J(k,n)$ :

$$\begin{array}{rcl}g_i f_{i,a} g_i (x_1 \dots x_i \dots x_n) & = & g_i f_{i,a} (x_1 \dots x_n \dots x_i) = \\ & = & g_i (x_1 \dots x_n + a(x_1 \dots \text{not} x_n \dots x_i) \dots x_i) \\ & = & (x_1 \dots x_i \dots x_n + a(x_1 \dots \text{not} x_n \dots x_i)) \\ & = & (x_1 \dots x_{n-1}, x_n + b(x_1 \dots x_{n-1}))\end{array}$$

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