## Mackey Irreducibility Criterion

**Point of Post: **In this post we prove the Mackey Irreducibility Criterion.

*Motivation*

In this post we put to rest probably one of the most niggling questions about induced representations. Namely, we have so far figured out how to take representations on and create a representation on but have yet to give a condition for which the a property we are so fond of us is true. To be direct, we have yet to decide when is actually an irrep of . The idea is simple, namely we know that being an irrep is equivalent to having but from Frobenius Reciprocity theorem we know that this is equivalent to showing the slightly more scary looking . But, thanks to Mackey we have information on this right hand term.

*Mackey’s Irreducibility Criterion*

Before we get to the theorem we need a few definitions. Namely, let be any group and to representations on . We call and *disjoint *if they have no equivalent subrepresentations and we denote this property by . This notation is supposed to be provocative since it’s evident that if and only if , in other words if the characters of and are orthogonal. Also, as was mentioned in our last post given a subgroup of and a representation we can define a representation on for any given by by . We denote the character of by . We note that if then and for any we have that and so . With these the irreducibility criterion becomes simple. Namely:

**Theorem (Mackey Irreducibility): ***Let be a finite group, , and a representation of with character . Then, the induced representation is an irreducible -representation if and only if is irreducible and for every one has that .*

**Proof: **As stated in the motivation the proof is simple, namely we know that is irreducible if and only if . But, using Frobenius Reciprocity and the our knowledge of we know that, picking some transversal for the set of double cosets we see that

Then, making our previous observation that for the we have by the above observation that we see that the above is equal to

And since we see from the above that if and only if and

for . But, unraveling this it says precisely that is irreducible if and only if is irreducible and using our disjoint notation as desired. But, since the possible choices for a transversal with is precisely the conclusion follows.

Of course, if the above greatly simplifies greatly since and evidently .

**References:**

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

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