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5
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edited Jan 15 2012 at 15:40 |
This happens if and only if $V$ U$ is expressible as a sum of pairwise non-isomorphic irreducible $G$-modules. If, for example $V U \cong U V \oplus U$ V$ for an irreducible $G$-module $U,$ V,$ then the natural invariant submodule $ \{(u,0): u {(v,0): v \in U V \}$ has at least two complements: one is the natural choice $\{(0,u): u \{(0,v): v \in U V \}.$ Another is $\{(u,u): u \{(v,v): v \in U V \}.$ On the other hand, if $V U \cong \bigoplus_{i=1}^{n} U_i,$ V_i,$ where $U_i V_i \not \cong U_j$ V_j$ for $i \neq j,$ then the only $G$-submodules of $V$ U$ are of the form $U_I V_I = \bigoplus_{i \in I} U_i$ V_i$ for a subset $I$ of $\{1,2,\ldots ,n\},$ and each such $G$-submodule $U_I$ V_I$ has the unique complement $U_{I^{\prime}},$ V_{I^{\prime}},$ where $I^{\prime} = \{ 1,2,\ldots ,n \} \backslash I.$
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4
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edited Jan 15 2012 at 12:24 |
This happens if and only if $V$ is expressible as a sum of pairwise non-isomorphic irreducible $G$-modules. If, for example $V \equiv cong U \oplus U$ for an irreducible $G$-module $U,$ then the natural invariant submodule $ \{(u,0): u \in U \}$ has at least two complements: one is the natural choice $\{(0,u): u \in U \}.$ Another is $\{(u,u): u \in U \}.$ On the other hand, if $V \equiv cong \bigoplus_{i=1}^{n} U_i,$ where $U_i \not \equiv cong U_j$ for $i \neq j,$ then the only $G$-submodules of $V$ are of the form $U_I = \bigoplus_{i \in I} U_i$ for a subset $I$ of ${1,2,\ldots \{1,2,\ldots ,n},$ n\},$ and each such $G$-submodule $U_I$ has the unique complement $U_{I^{\prime}},$ where $I^{\prime} = \{ 1,2,\ldots ,n \} \backslash I.$
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3
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edited Jan 15 2012 at 11:53 |
This happens if and only if $V$ is expressible as a sum of pairwise non-isomorphic irreducible $G$-modules. If, for example $V \equiv U \cong oplus U$ for an irreducible $G$-module $U,$ then the natural invariant submodule $ \{(u,0): u \in U \}$ has at least two complements: one is the natural choice $\{(0,u): u \in U \}.$ Another is $\{(u,u): u \in U \}.$ On the other hand, if $V \equiv \bigoplus_{i=1}^{n} U_i,$ where $U_i \not \equiv U_j$ for $i \neq j,$ then the only $G$-submodules of $V$ are of the form $U_I = \bigoplus_{i \in I} U_i$ for a subset $I$ of ${1,2,\ldots ,n},$ and each such $G$-submodule $U_I$ has the unique complement $U_{I^{\prime}},$ where $I^{\prime} = \{ 1,2,\ldots ,n \} \backslash I.$
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2
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edited Jan 15 2012 at 11:40 |
This happens if and only if $V$ is expressible as a sum of pairwise non-isomorphic irreducible $G$-modules. If, for example $V \equiv U \cong U$ for an irreducible $G$-module $U,$ then the natural invariant submodule $ {(u,0): \{(u,0): u \in U }$ \}$ has at least two complements: one is the natural choice ${(0,u): \{(0,u): u \in U }.$ \}.$ Another is ${(u,u): \{(u,u): u \in U }.$ \}.$ On the other hand, if $V \equiv \bigoplus_{i=1}^{n} U_i,$ where $U_i \not \equiv U_j$ for $i \neq j,$ then the only $G$-submodules of $V$ are of the form $U_I = \bigoplus_{i \in I} U_i$ for a subset $I$ of ${1,2,\ldots ,n},$ and each such $G$-submodule $U_I$ has the unique complement '$U_{I^{\prime}},$ $U_{I^{\prime}},$ where $I^{\prime} = \{ 1,2,\ldots ,n \} \backslash I.$'
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1
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answered Jan 15 2012 at 11:24 |
This happens if and only if $V$ is expressible as a sum of pairwise non-isomorphic irreducible $G$-modules. If, for example $V \equiv U \cong U$ for an irreducible $G$-module $U,$ then the natural invariant submodule $ {(u,0): u \in U }$ has at least two complements: one is the natural choice ${(0,u): u \in U }.$ Another is ${(u,u): u \in U }.$ On the other hand, if $V \equiv \bigoplus_{i=1}^{n} U_i,$ where $U_i \not \equiv U_j$ for $i \neq j,$ then the only $G$-submodules of $V$ are of the form $U_I = \bigoplus_{i \in I} U_i$ for a subset $I$ of ${1,2,\ldots ,n},$ and each such $G$-submodule $U_I$ has the unique complement '$U_{I^{\prime}},$ where $I^{\prime} = { 1,2,\ldots ,n } \backslash I.$'
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