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The multiplicative group $\mathbb Z_2:=\{1,-1\}$ acts on a vector space $V$ with a basis $\{v_1,v_2,v_3,v_4,v_5,v_6\}$ over the finite field $\mathbb F_2=\{0,1\}$ by fixing the basis vectors $v_1,v_2,v_3,v_4$ pointwise and interchanging $v_5$ and $v_6$. What is the decomposition of this representations as direct sum of indecomposable representations ? Looks like $V=4 \rho_1 \oplus \rho_2$ where $\rho_1$ is the trivial representation of $\mathbb Z_2$ and $\rho_2$ is the 2-dimensional indecomposable represnetation of $\mathbb Z_2$ generated by $v_5$ and $v_6$. Then $-1$ acts trivially on $v_5+v_6$ and it generates a one dimensional representation of $\mathbb Z_2$. Then where does this element belong in the above decomposition ?

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Well, $v_5+v_6$ is an element of the subspace spanned by $v_5$ and $v_6$. So this one-dimensional representation is just a subrepresentation of $\rho_2$ in your decomposition $4\rho_1\oplus\rho_2$.

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  • $\begingroup$ The operator is not diagonalizable but is there any other decomposition that separates the eigen space of the eigen value 1 ? $v_1,v_2,v_3,v_4,v_5+v_6$ are all in the eigen space of 1. $\endgroup$ – icmes imrf May 31 '17 at 2:03

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