# Direct product of faithful representations into direct sum

1. Feb 11, 2012

### rkrsnan

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. Feb 11, 2012

### morphism

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 $V^{\otimes n}$ 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 $\chi$ has $\chi(g)=\chi(1)$ for some $g\neq 1$ then necessarily $g \in \ker \chi$ and $\chi$ isn't faithful.

Thus there are lots of other examples, e.g. any noncyclic abelian group, and more generally any group with noncyclic center.

3. Feb 11, 2012

### rkrsnan

Thank you so much! That was totally what I wanted to know.

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

4. Feb 14, 2012

### morphism

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! ).

5. Feb 15, 2012

### rkrsnan

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.

6. Feb 15, 2012

### morphism

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