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I am reading Bruce Sagan's The Symmetric Group, and my question below is concerning Section 1.7.

Here, and in the book, every group is assumed finite, and every base field is assumed to be $\Bbb C$. Also, the group $\{ A \in Mat_{d \times d} : \det A \neq 0 \}$ is denoted by $GL_d$. ($d>0$)

Given a group $G$ and a matrix representation $X: G \to GL_d$, the correseponding commutant algebra is defined as Com$(X)=\{A \in Mat_{d \times d} : TX(g)=X(g)T $ for all $g \in G \}$.

Using Maschke's Theorem and Schur's lemma, it has been proved in the book that

(1) If $X = \oplus _{i=1} ^k X_i$ where the $X_i$ are pairwise inequivalent irreducibles with $\deg X_i =d_i$, then Com$(X)=\{\oplus _{i=1} ^k c_i I_{d_i} : c_i \in \Bbb C \}$.

(2) If $X$ is irreducible, then Com$(mX)=\{M_m \otimes I_d : M_m \in Mat_{m \times m} \}$ .

Then Sagan says that, in the general case $X= \oplus _{i=1} ^k mX_i$ where the $X_i$ are pairwise inequivalent irreducibles with $\deg X_i =d_i$, we obtain Com$(X)=\{\oplus _{i=1} ^k (M_{m_i} \otimes I_{d_i} ):M_{m_i} \in Mat_{m_i \times m_i} \}$ by combining (1) and (2).

The $\oplus$ notation on matrices mean the block diagonal matrix, and for two matrices $X=(x_{ij})$ and $Y$, the matrix $(x_{ij}Y)$ is denoted by $X \otimes Y$.

I tried to prove the general case by induction on $k$, but I got stuck; I think (1) and (2) are not enough to prove the final result. How (1) and (2) imply the final assertion?

In this book, Maschke's Theorem and Schur's lemma are written as follows:

Maschke's Theorem. Every representation of a finite group having positive dimension is completely reducible.

Schur's Lemma. If $\theta : V \to W$ is a $G$-homomorphism between two irreducible $G$-modules, then either $\theta$ is an isomorphism or the zero map.

(Matrix version) If $X,Y$ are two irreducible matrix representations of $G$, and if $T$ is any matrix such that $TX(g)=Y(g)T$ for all $g \in G$, then either $T$ is invertible or $T=0$.

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With $X=\bigoplus_{i=1}^k mX_i$, we need to find matrix $T=(T_{i,j})$ where we $T_{i,i}$ is the block diagonal matrix of $T$ with the same size as $mX_i$. Then $(XT)_{i,j}=T_{i,j}(mX_j)$ since $X$ is block diagonal. Similarly, $(TX)_{i,j}=(mX_i)T_{i,j}$. We have $T_{i,j}(mX_j)=(mX_i)T_{i,j}$.

If $i\ne j$, you can argue that $T_{i,j}$ is seen as $G$-module homomorphism between $mX_i$ and $mX_j$. Since $X_i,X_j$ are nonisomorphic irreducible representation of $G$, using Schur's lemma, we can find $T_{i,j}=0$.

If $i=j$, this is essentially (2), we you find $T_{i,i}=M_m\otimes I_{d_i}$.

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