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For subgroups $H,K\le G$ and an $H$-representation $W$, Mackey restriction formula isFrobenius reciprocity statesApplying these with $K=H$ givesThe identity double coset contributes one exactly when $W$ is irreducible, and every summand is a nonnegative integer. This proves Mackey irreducibility criterion: the induced representation is irreducible exactly when $W$ is irreducible and every nonidentity double-coset summand vanishes.

Now take $G=S_n$, $H=S_{n-1}$, and $x=(n-1n)\notin H$. Thenthe subgroup fixing both $n-1$ and $n$. Conjugation by $x$ fixes every element of this subgroup, so the two representations appearing in the corresponding Mackey summand are both ${\mathrm{Res}}_{S_{n-2}}^{S_{n-1}}W$. Their inner product is positive because this restriction is nonzero. Hence the Mackey irreducibility criterion fails whenever $W$ is irreducible. If $W$ is reducible, induction preserves its direct-sum decomposition, so the induced representation is again reducible. Therefore ${\mathrm{Ind}}_{S_{n-1}}^{S_n}W$ is never irreducible for $n\ge 2$.

The point-permutation character isBy the tensor identity for an induced character and Frobenius reciprocity,The restriction branching rule for a symmetric group is multiplicity-free with one constituent for each member of ${\lambda }^{-}$, so the right side is $|{\lambda }^{-}|$. Also $⟨{\chi }^{\lambda }{\chi }^{\lambda },{\chi }^{(n)}⟩=1$ because every symmetric-group character is real and irreducible. Subtracting the trivial constituent proves the standard-character multiplicity in a Specht self-product formula

The Mackey restriction formula says, for an $H$-representation $V$,Frobenius reciprocity saysApplying both with an irreducible character $\chi$ givesThe identity double coset contributes one, and all terms are nonnegative integers. Therefore Mackey irreducibility criterion says that ${\mathrm{Ind}}_H^G\chi$ is irreducible exactly when every term belonging to a nonidentity double coset is zero.

Now take $G=S{L}_2(k)$. Writeso $B=T⋉U$. For $q\ge 4$, commutators of $T$ with $U$ generate $U$, because conjugation sends $u(b)$ to $u(a^{2}b)$ and one can choose $a^2\ne 1$. Every degree-one character of $B$ is therefore trivial on $U$ and has the formwhere $\theta :k^{\times }\to {\mathbb{C}}^{\times }$ is a multiplicative character.

The Bruhat decomposition of SL2 over a finite field has two double cosets,and $B\cap wB{w}^{-1}=T$. Conjugation by $w$ inverts $T$, soThe only nonidentity Mackey term vanishes exactly when these two one-dimensional characters differ. ConsequentlyThus the trivial character is excluded, as is the unique quadratic character when $q$ is odd. All other degree-one characters of $B$ induce irreducibly; the induced representations have degree $[G:B]=q+1$. This is the irreducible principal series of finite SL2.

Apply Frobenius reciprocity and then the decomposition from part (i):Since all the representations in the last line are one-dimensional,It is therefore $2$ when both equalities hold, $1$ when exactly one holds, and $0$ when neither holds. This is the mackey inner product for the finite SL2 principal series.

For $\lambda ⊢n$ with $n\ge 2$,Indeed, adjoining the trivial character to the standard character gives the point-permutation character, and Frobenius reciprocity turns its multiplicity in the self-product into the norm of the multiplicity-free restriction.