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in the answer to this post there was the statement that a representation $\vartheta$ of a subgroup $\langle z\rangle$ can extend to a representation of the whole group $D_{2n}$.

If I start the other way round, this would mean that if the induced representation $\operatorname{Ind}(\vartheta)$ decomposes into two irreducibles $\vartheta_1$ and $\vartheta_2$,

  1. $\vartheta_1$ and $\vartheta_2$ have the same degree (Why can't $\operatorname{Ind}(\vartheta)$ decompose into $2$ representation with different degrees?)
  2. $\vartheta_1$ is the same representations as $\vartheta_2$. ( Why can't they be different?)
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  • Thanks for the nice job formatting! Just a little FYI: you can obtain the brackets shown in the question statement by using \langle z \rangle to get $\langle z \rangle$. – amWhy Apr 20 '13 at 18:59
  • Ok thanks, I will use it next time – user73227 Apr 20 '13 at 19:01

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