## The Dual Group

**Point of post: **In this post we discuss how one can turn a subset of naturally into a group.

*Motivation*

We’ve done extensive work with the set of all equivalency classes of irreps on a finite group . A natural question that we have put off until now is if there is a natural way to make into a group itself. In this post we provide a partial affirmative answer to this question by showing that the set of all linear representations (representations of degree one) forms a group under a natural product.

*Linear Representations and Their Tensor Products*

* *

Let be a finite group an irrep of is said to be *linear *if . We call an equivalency class *linear* if (or equivalently if every one of its members is a linear irrep). If is a linear irrep then is called a *linear character *of . We denote the set of all linear by .

What we first show is that if is an irrep and a linear irrep then the tensor product is an irrep. Indeed:

**Theorem: ***Let be a finite group. Then, if is a linear irrep of and an irrep then is an irrep of with .*

**Proof: **To see that is irreducible we note that

(where we’ve used the simple fact that for any linear representation one has that for every ) from where the conclusion follows from our previous characterization of irreducibility. To see that we merely note that

.

Now, it’s clear from this that for any two linear irreps their tensor product is also a linear irrep, and since it’s evident that if and that we may unambiguously consider the *product *of , denoted to be the equivalency class for any and .

It’s evident that for any three representations that so that if are the equivalency classes of and respectively then so that the map is associative. Moreover, it’s clear that if denotes the equivalence class of the trivial character then is a two-sided identity for . Indeed, recall that two irreps are equivalent if and only if the admit the same character and for any and one has that

for every so that or, said equivalently, . Lastly, we claim that if then . Indeed,

Thus, is a finite set which has the associative binary operation with the two-sided identity and which satisfies the left and right cancellative property. There’s a name for such a structure: a group. Moreover, it’s clear that is commutative since for any any two representations and it’s clearly true that . Also, since for any and any we have that it’s clear that for any complex conjugate on representation space, and so . Summing this all up we have:

**Theorem: ***Let be a finite group, then is an abelian group with identity and .*

* *

We call the group the *dual group of .*

As a last note, we know that where is the abelianization of and is the commutator subgroup.

**References:**

1. Simon, Barry. *Representations of Finite and Compact Groups*. Providence, RI: American Math. Soc., 1996. Print.

[…] our last post we discussed how to naturally turn a subset of into a group, namely the subset of linear […]

Pingback by Representation Theory: Another Way of Viewing the Dual Group « Abstract Nonsense | April 12, 2011 |

[…] it’s difficult to deal directly with the dual group even with our alternate characterization of it. In this post we show that the lay of the land […]

Pingback by Representation Theory: Dual Group of a Cyclic Group « Abstract Nonsense | April 13, 2011 |

[…] We recall that the dual group has order equal the the abelianization . But, by inspection we see that and so and so […]

Pingback by University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) (August-2003) « Abstract Nonsense | May 1, 2011 |

[…] and to be honest I was always a little hazy on the end of it. A while ago when I was thinking about dual groups and trying to develop stuff independent of my books I realized that there was a ‘much […]

Pingback by Clearer Proof for the Number of Degree One Irrep Classes « Abstract Nonsense | May 6, 2011 |