Direct product of faithful representations into direct sum


by rkrsnan
Tags: faithful, product, representations
rkrsnan
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#1
Feb11-12, 11:27 AM
P: 54
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.
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morphism
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#2
Feb11-12, 01:49 PM
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Quote Quote by rkrsnan View Post
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?
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.

My second question is (if it makes sense), can we have a finite group in which none of the irreducible representations are faithful?
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.
rkrsnan
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#3
Feb11-12, 04:06 PM
P: 54
Thank you so much! That was totally what I wanted to know.

PS: Yes, I should have written 'tensor product' instead of 'direct product'.

morphism
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#4
Feb14-12, 05:14 PM
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Direct product of faithful representations into direct sum


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! ).
rkrsnan
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#5
Feb15-12, 07:48 AM
P: 54
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.
morphism
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#6
Feb15-12, 09:09 AM
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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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