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I am trying to understand why

1) all finite-dimensional complex representations $V$ of $G$ are self dual, and

2) How the derived subgroup $[G,G]$ is a union of particular conjugacy classes.

My thoughts: on

1) by definition the dual representation of $V$ is the dual representation $\rho ^*$ on the dual vector space $V^*:Hom(V,\mathbb{C})$ of linear maps defined by $\rho(g)^*L = L \circ \rho(g)^{-1}$. Now according to the source: Def: 5.7 this preserves the duality pairing of $V$ and $V^*$ by $L(\mathbf{v}) = \rho^*(g)(L)(\rho(g)(\mathbf{v}))$.

I don't see how this works though as $\rho(g)^*L = L \circ \rho(g)^{-1}$ means that the dual representation of a group element $g$ is really the inverse representation of an element of $g$ which is then mapped to a complex number. I am a bit lost on what how this machinery works and what $\rho(g)^{-1}$ really is.

2) If we take some $x \in [G,G]$ we have that $xg = gx$ for some element $g \in G$ then of course $g^{-1}xg = x$ and {$g^{-1}xg \forall g \in G$} is the conjugacy class. There are several things rolling around in my mind, the first being that perhaps I can take the union of the representations, but I am not sure how to go about writing it out.

Any thoughts on either question is appreciated.

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    $\begingroup$ It isn't true (for a general finite group $G$) that all finite dimensional representations are self-dual. What is true is that the dual of a representation is another (not necessarily equivalent) representation. $\endgroup$ Sep 15 '14 at 19:14
  • $\begingroup$ For an example, consider any non-trivial one-dimensional complex representations of a cyclic group of order three. $\endgroup$ Sep 15 '14 at 21:37
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Any normal subgroup, not only the commutator subgroup, Is the union of (disjoint) conjugacy classes.

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  • $\begingroup$ (More general: Any set/subset closed under the action of a group is a union of orbits.) $\endgroup$
    – whacka
    Sep 15 '14 at 21:50

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