The isotypic decomposition of a completely reducible representation is unique

By [L3], choose irreducible subrepresentations $U_1,\dots ,U_n$ with $V=U_1\oplus \cdots \oplus U_n\mathrm{.}$ Group these summands by equivalence class: for each class represented by $S_i$, let $W_i$ be the direct sum of those $U_j$ equivalent to $S_i$. Then $V=W_1\oplus \cdots \oplus W_r,$ and each $W_i$ is contained in $V_{(S_i)}$ by [L1].

Fix $i$ and let $U\le V$ be any irreducible subrepresentation equivalent to $S_i$. Write ${\pi }_j:V\to U_j$ for the projection attached to step 1.1. If $U_j$ is not equivalent to $S_i$, then ${\pi }_j{\mathrm{\mid }}_U:U\to U_j$ is an intertwiner between non-equivalent irreducibles, so [L2] makes it zero. Hence the projection of $U$ onto ${⨁}_{j:U_j\cong ̸S_i}{U}_j$ is zero, and therefore $U\le W_i$. Since this holds for every such $U$, the defining sum [L1] satisfies $V_{(S_i)}\le W_i$. Together with step 1.1, this gives $V_{(S_i)}=W_i$.

Step 2.1 shows that each grouped block $W_i$ is exactly the isotypic component $V_{(S_i)}$, so it depends only on the equivalence class of $S_i$, not on the chosen irreducible splitting. Since the $W_i$ already form a direct sum in step 1.1, the displayed isotypic decomposition is unique.