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No, it does not. As a simple example, let $G\cong S_3$, $\Omega = G/C_2 \sqcup G/C_2 \sqcup G/C_3$. You can easily check that the corresponding permutation character is isomorphic to the regular character plus two copies of the trivial. In other words, $\mathbb{C}[G/1] \oplus \mathbb{C}[G/G]^{\oplus 2} \cong \mathbb{C}[G/C_3] \oplus \mathbb{C}[G/C_2]^{\oplus 2}$.

One systematic source of such counterexamples are Brauer relations in which the regular set $G/1$ enters with non-zero coefficient, of which both Tim's and my answer are particular examples (for more on Brauer relations see e.g. this MO question, as well as google).

No, it does not. As a simple example, let $G\cong S_3$, $\Omega = G/C_2 \sqcup G/C_2 \sqcup G/C_3$. You can easily check that the corresponding permutation character is isomorphic to the regular character plus two copies of the trivial. In other words, $\mathbb{C}[G/1] \oplus \mathbb{C}[G/G]^{\oplus 2} \cong \mathbb{C}[G/C_3] \oplus \mathbb{C}[G/C_2]^{\oplus 2}$.

One systematic source of such counterexamples are Brauer relations in which the regular set $G/1$ enters with non-zero coefficient, of which both Tim's and my answer are particular examples (for more on Brauer relations see e.g. this MO question, as well as google).

No, it does not. As a simple example, let $G\cong S_3$, $\Omega = G/C_2 \sqcup G/C_2 \sqcup G/C_3$. You can easily check that the corresponding permutation character is isomorphic to the regular character plus two copies of the trivial. In other words, $\mathbb{C}[G/1] \oplus \mathbb{C}[G/G]^{\oplus 2} \cong \mathbb{C}[G/C_3] \oplus \mathbb{C}[G/C_2]^{\oplus 2}$.

Your question can be rephrased as "does there exist a Brauer relation for $G$ in which the regular set $G/1$ enters with non-zero coefficient" (for more on Brauer relations see e.g. this MO question, as well as google). My hunch is that such a Brauer relation exists in any non-cyclic finite group, but I would have to think about it for a bit. Let me know if you need to know this.

No, it does not. As a simple example, let $G\cong S_3$, $\Omega = G/C_2 \sqcup G/C_2 \sqcup G/C_3$. You can easily check that the corresponding permutation character is isomorphic to the regular character plus two copies of the trivial. In other words, $\mathbb{C}[G/1] \oplus \mathbb{C}[G/G]^{\oplus 2} \cong \mathbb{C}[G/C_3] \oplus \mathbb{C}[G/C_2]^{\oplus 2}$.

One systematic source of such counterexamples are Brauer relations in which the regular set $G/1$ enters with non-zero coefficient, of which both Tim's and my answer are particular examples (for more on Brauer relations see e.g. this MO question, as well as google).

No, it does not. As a simple example, let $G\cong S_3$, $\Omega = G/C_2 \sqcup G/C_2 \sqcup G/C_3$. You can easily check that the corresponding permutation character is isomorphic to the regular character plus two copies of the trivial. In other words, $\mathbb{C}[G/1] \oplus \mathbb{C}[G/G]^{\oplus 2} \cong \mathbb{C}[G/C_3] \oplus \mathbb{C}[G/C_2]^{\oplus 2}$.

Your question can be rephrased as "does there exist a Brauer relation for $G$ in which the regular set $G/1$ enters with non-zero coefficient" (for more on Brauer relations see e.g. this MO question, as well as google). My hunch is that such a Brauer relation exists in any non-cyclic finite group, but I would have to think about it for a bit. Let me know if you need to know this.

No, it does not. As a simple example, let $G\cong S_3$, $\Omega = G/C_2 \sqcup G/C_2 \sqcup G/C_3$. You can easily check that the corresponding permutation character is isomorphic to the regular character plus two copies of the trivial. In other words, $\mathbb{C}[G/1] \oplus \mathbb{C}[G/G]^{\oplus 2} \cong \mathbb{C}[G/C_3] \oplus \mathbb{C}[G/C_2]^{\oplus 2}$.

Your question can be rephrased as "does there exist a Brauer relation for $G$ in which the regular set $G/1$ enters with non-zero coefficient" (for more on Brauer relations see e.g. this MO question, as well as google). My hunch is that such a Brauer relation exists in any non-cyclic finite group, but I would have to think about it for a bit. Let me know if you need to know this.