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In the book Representation Theory, A First Course by William Fulton and Joe Harris, they state the following:

Proposition 1.8 For any representation $V$ of a finite group $G$, there is a decomposition $$ V=V_1^{\oplus a_1}\oplus\cdots\oplus V_k^{\oplus a_k}, $$ where the $V_i$ are distinct irreducible representations. The decomposition of $V$ into a direct sum of the $k$ factors is unique, as are the $V_i$ that occur and their multiplicities $a_i$.

Proof. It follows from Schur's lemma that if W is another representation of G, with a decomposition $W=\oplus W_j^{\oplus b_j}$, and $\phi:V\rightarrow W$ is a map of representations, then $\phi$ must map the factor $V_i^{\oplus a_i}$ into that factor $W_j^{\oplus b_j}$ for which $W_j\cong V_i$; when applied to the identity map of $V$ to $V$, the stated uniqueness follows.

I don't quite follow the proof. I think for the first part, he uses that if $\phi$ is a linear map between representations, then $f$ restricted to $W_i$ is a map between irreducible representations and Schur's lemma tells us that those representations are isomorphic (correct?). Also, I am not sure about the last part "when applied to the identity map of $V$ to $V$, the stated uniqueness follows".

Edit: I found another proof in REPRESENTATION THEORY FOR FINITE GROUPS by Shaun Tan, but I also do not find it easy to understand that one. Specifically, I do not understand why he says that "If $j=i$, then $\phi\left(V_{i}^{\oplus a_{i}}\right) \neq 0$ for any $i$."

Theorem 4.3. For any finite-dimensional representation $(\rho, V)$ of a finite group $G$ there is a unique decomposition $V=V_{1}^{\oplus a_{1}} \oplus V_{2}^{\oplus a_{2}} \oplus \ldots \oplus V_{i}^{\oplus a_{i}}$ where the $V_{i}$ are inequivalent and irreducible with unique multiplicities $a_{i}$.

Proof. We suppose $V=W_{1}^{\oplus b_{1}} \oplus W_{2}^{\oplus b_{2}} \oplus \ldots \oplus W_{j}^{\oplus b_{j}}$. Then we let $\phi: V \rightarrow V$ be the identity map. We use Schur's Lemma. For each irreducible $V_{i}^{\oplus a_{i}}$, we restrict the domain of $\phi$ to that component. Then, either $\phi=0$ or $\phi$ is an isomorphism. If $j=i$, then $\phi\left(V_{i}^{\oplus a_{i}}\right) \neq 0$ for any $i$. For each component, $\phi$ is an isomorphism such that $V_{i}^{\oplus a_{i}}$ maps to $W_{j}^{\oplus b_{j}}$ where $V_{i}$ is isomorphic to $W_{j}$.

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  • $\begingroup$ Whose book? "They" indeed... $\endgroup$ Commented May 6, 2023 at 16:58
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    $\begingroup$ @ArturoMagidin, William Fulton and Joe Harris. $\endgroup$
    – Jakamay
    Commented May 6, 2023 at 17:05
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    $\begingroup$ Put it in the post, not the comments. $\endgroup$ Commented May 6, 2023 at 17:08
  • $\begingroup$ @ArturoMagidin, done $\endgroup$
    – Jakamay
    Commented May 6, 2023 at 17:11

1 Answer 1

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For any representation $V$ of a finite group $G$ there is a decomposition $$V \cong V_1^{\oplus a_1} \oplus \cdots \oplus V_k^{\oplus a_k},$$ where the $V_i$ are distinct irreducible representations. The decomposition of $V$ into a direct sum of the $k$ factors is unique, as are the $V_i$ that occur and their multiplicities $a_i$.

Note there are two big parts of the statement here: existence and uniqueness (the uniqueness part has several subparts). Both proofs you quote take the existence part for granted, so I will too (it is a corollary of Maschke's theorem).

Suppose we have an isomorphism $$\phi: V = V_1^{\oplus a_1} \oplus \cdots \oplus V_k^{\oplus a_k} \to W = W_1^{\oplus b_1} \oplus \cdots \oplus W_\ell^{\oplus b_\ell},$$ where $V_1, \dots, V_k$ are distinct irreducible representations and so are $W_1, \dots, W_\ell$, and $a_1, \dots, a_k, b_1, \dots, b_\ell$ are positive integers. For $1\le i \le k$ and $1 \le j \le \ell$, let $\phi_{ij}$ denote the composition $$V_i^{\oplus a_i} \to V \stackrel{\phi}{\longrightarrow} W \to W_j^{\oplus b_j},$$ where the first map is inclusion and the last map is projection. By further restricting to one of the $a_i$ $V_i$-factors and further projecting to one of the $b_j$ $W_j$-factors, we get a $G$-module map $V_i \to W_j$. Now Schur's lemma applies and tells us that this map is zero unless $V_i \cong W_j$. Since this holds for each of the $a_i$ $V_i$-factors and each of the $b_j$ $W_j$-factors, it follows that $\phi_{ij} = 0$ unless $V_i \cong W_j$. Therefore $\phi$ must map $V_i^{\oplus a_i}$ into $W_{i'}^{\oplus b_{i'}}$ for some unique index $i'$ such that $V_i \cong W_{i'}$. The same argument applied to $\phi^{-1}$ shows that $\phi^{-1}$ maps $W_{i'}^{b_{i'}}$ into $V_i^{\oplus a_i}$, so $\phi$ restricts to an isomorphism $V_i^{\oplus a_i} \stackrel{\sim}\to W_{i'}^{\oplus b_{i'}}$. Now by comparing dimensions we get $a_i \dim V_i = b_{i'} \dim W_{i'}$, so $a_i = b_{i'}$ since $\dim V_i = \dim W_{i'}$.

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  • $\begingroup$ To be sure, is it correct that $\phi_{ij}=\pi_j\circ \phi \circ \varphi_i$, where $\varphi_i$ is the inclusion map that takes a vector in $V_i^{\oplus a_i}$ and sends it to $V$ as a direct summand and $\pi_j$ is the projection map that takes a vector in $W$ and projects it onto $W_j^{\oplus b_j}$? And restricting it further would mean that we consider $\psi_j\circ\phi_{ij}\circ \chi_i$ where $\chi_i$ takes a vector in $V_i$ and sends it to $V_i^{\oplus a_i}$ and $\psi_j$ takes a vector in $W_j^{\oplus b_j}$ and projects it onto $W_j$? Or am I misunderstanding something? $\endgroup$
    – Jakamay
    Commented May 7, 2023 at 15:20
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    $\begingroup$ @Jakamay Yes that's correct. But do note that there $a_i$ such maps $\chi_i$ (and similarly $b_j$ such maps $\psi_j$). They are given by $v \mapsto (v,0\dots,0)$, $v \mapsto (0, v,0, \dots,0)$, etc. $\endgroup$ Commented May 7, 2023 at 18:51

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