# Self conjugate representations

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.

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.
• "Since $W′$ and $W′′$ are distinct" -- why? Why is it a direct sum of two irreps to begin with? – darij grinberg Feb 9 at 23:44
• 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. – Joakim Færgeman Feb 10 at 15:30