I am trying to understand proof of Prop. 5.1(page 64) from Fulton and Harris representation theory. I am unable to prove one statement in the proof. Which is

Let $V$ be an irreducible representation of a finite group $G$. Let $W$ be restriction of $V$ to a normal subgroup $H$ with index $2$. Now $W = W' \oplus W''$, where $W', W''$ are irreducible and conjugate but not isomorphic.

I am unable to understand why they are not self conjugate? The proof says,

Since $W$ is self conjugate and if $W', W''$ were self conjugate $V$ wouldn't be irreducible.

I am trying to assume $W',W''$ are self conjugate and produce a decomposition of $V$, but I couldn't. Please help me in filling details.

UPDATE: My idea is We know that $W', W''$ are different irreducible representations(if not, $<W,W>=4$, a contradiction as it's value is $2$) and as $W', W''$ are self conjugate by an element of $G\backslash H$. For any $g \in G, g.W' \subset W'$, if not the map $g$ gives an isomorphism between $W'$ and $W''$ but we started with two different representations.


I am currently reading the same text and also stumbled upon this question. It seems to me that there should be an easy argument, but I cannot seem to find it.

However, Clifford theory gives an immediate answer. It says that when you restrict an irreducible representation to a normal subgroup, it becomes some multiple of the sum of all its conjugates. Since $W'$ and $W''$ are distinct (so the multiple is $1$), they must be conjugate.

Clifford theory is mentioned after the proof of prop. 5.1 in Fullton and Harris' text, so it is definitely overkill to utilize it. Nonetheless, it provides an answer.

  • $\begingroup$ "Since $W′$ and $W′′$ are distinct" -- why? Why is it a direct sum of two irreps to begin with? $\endgroup$ – darij grinberg Feb 9 at 23:44
  • $\begingroup$ Let $\chi$ be the character afforded by the irreducible representation $V$ of $G$. Then $|G|$=$\sum_{g\in G}|\chi(g)|^2$=$\sum_{h\in H}|\chi(h)|^2$+$\sum_{t\notin H}|\chi(t)|^2$. Since the first sum on the right side is some integral multiple of $|H|$ and since $H$ has index 2 in $G$, it follows that $V$ restricted to $H$ either is irreducible or is a sum of two distinct irreducible representations. We have simply assumed the latter case. $\endgroup$ – Joakim Færgeman Feb 10 at 15:30

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