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I worked the proof for 3) in this link out, but I have problems with the last step:

Let $\sigma$ be a irred. representation of a normal subgroup $H=\langle z\rangle$ of $G$ and $\sigma'$ its dual representation. The proof comes to the conclusion that $$\sigma \text { is equivalent to } \sigma' \Leftrightarrow z^{-1}=z^{p^k}, $$ where $p$ is the charakteristic of the field $K$.

Where is the connection between saying that " $\sigma$ is equivalent to $\sigma'$ " and " $\sigma$ expands to a representation of $G$ "?

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Please consider registering yourself with the site so that SE can help maintain your questions. Regards, –  user21436 Aug 26 '12 at 17:39
    
I'll do it when I'm back at home. Now i haven't my password with me. –  Haruki S. Aug 26 '12 at 17:51
    
I've changed $<z>$ to $\langle z\rangle$. The code for the former is <z>; that for the latter is \langle z\rangle. –  Michael Hardy Aug 27 '12 at 3:19

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