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Direct product of faithful representations into direct sum

  1. Feb 11, 2012 #1
    Direct product of two irreducible representations of a finite group can be decomposed into a direct sum of irreducible representations. So, starting from a single faithful irreducible representation, is it possible generate every other irreducible representation by successively taking direct products?

    My second question is (if it makes sense), can we have a finite group in which none of the irreducible representations are faithful?

    Thanks.
     
  2. jcsd
  3. Feb 11, 2012 #2

    morphism

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    Do you mean to say tensor products here instead of direct products? If so, then the answer to your question is in some sense affirmative. A result due to Molien (sometimes called the Burnside-Molien theorem) says that every irreducible representation of a finite group is contained inside some tensor power [itex]V^{\otimes n}[/itex] of a faithful irreducible representation V.

    Yes. For example Z/2Z x Z/2Z doesn't have any. You can spot this by looking at the character table: if the the column corresponding to [itex]\chi[/itex] has [itex]\chi(g)=\chi(1)[/itex] for some [itex]g\neq 1[/itex] then necessarily [itex]g \in \ker \chi[/itex] and [itex]\chi[/itex] isn't faithful.

    Thus there are lots of other examples, e.g. any noncyclic abelian group, and more generally any group with noncyclic center.
     
  4. Feb 11, 2012 #3
    Thank you so much! That was totally what I wanted to know.

    PS: Yes, I should have written 'tensor product' instead of 'direct product'.
     
  5. Feb 14, 2012 #4

    morphism

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    No problem. By the way, the end of my first paragraph above should of course read "of a faithful representation V" and not "of a faithful irreducible representation V" (as there might not be such a V! :smile:).
     
  6. Feb 15, 2012 #5
    Yep, understood.
    About the cyclic center and having faithful irreducible reps, does this result work if the center is identity? I can find examples of groups in which center is identity; in some cases faithful irreps exist and in some others it doesn't.
     
  7. Feb 15, 2012 #6

    morphism

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    Yes, you're right - the center being cyclic is a necessary but by no means sufficient condition!
     
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