For subgroups and an -representation , Mackey restriction formula is
Frobenius reciprocity states
Applying these with gives
The identity double coset contributes one exactly when is irreducible, and every summand is a nonnegative integer. This proves Mackey irreducibility criterion: the induced representation is irreducible exactly when is irreducible and every nonidentity double-coset summand vanishes.
Now take , , and . Then
the subgroup fixing both and . Conjugation by fixes every element of this subgroup, so the two representations appearing in the corresponding Mackey summand are both . Their inner product is positive because this restriction is nonzero. Hence the Mackey irreducibility criterion fails whenever is irreducible. If is reducible, induction preserves its direct-sum decomposition, so the induced representation is again reducible. Therefore is never irreducible for .
The Mackey restriction formula says, for an -representation ,
Frobenius reciprocity says
Applying both with an irreducible character gives
The identity double coset contributes one, and all terms are nonnegative integers. Therefore Mackey irreducibility criterion says that is irreducible exactly when every term belonging to a nonidentity double coset is zero.
Now take . Write
so . For , commutators of with generate , because conjugation sends to and one can choose . Every degree-one character of is therefore trivial on and has the form
where is a multiplicative character.
The Bruhat decomposition of SL2 over a finite field has two double cosets,
and . Conjugation by inverts , so
The only nonidentity Mackey term vanishes exactly when these two one-dimensional characters differ. Consequently
Thus the trivial character is excluded, as is the unique quadratic character when is odd. All other degree-one characters of induce irreducibly; the induced representations have degree . This is the irreducible principal series of finite SL2.