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For an $H$-representation $V$ and $g\in G$, the conjugate representation ${}^{g}V$ of $gH{g}^{-1}$ is defined by $(gh{g}^{-1})v=hv$. This transports the acting subgroup; it is distinct from conjugating complex matrix entries. It occurs in the Mackey restriction formula.

For a finite group $G$, subgroups $H,K$, and a complex group representation $V$ of $H$, the Mackey restriction formula isChoose one representative $g$ of each double coset $KgH$. The conjugate subgroup representation ${}^{g}V$ is a representation of $gH{g}^{-1}$ on the same vector space, with $(gh{g}^{-1})v=hv$. Restriction of a representation keeps the vector space and restricts the acting subgroup. induced representation is $\mathbb{C}G{\otimes }_{\mathbb{C}H}V$, and analogously for each subgroup of $K$. The displayed isomorphism is an isomorphism of $K$-representations. These definitions specify every summand and its subgroup.

There is an index error in the displayed identity of the original PDF. With its definition of the signed Young permutation module, the tensor identity for induced representations givesBoth superscripts must be $\alpha$. In particular ${\tilde{M}}^{{\lambda }^{\prime }}\cong \mathrm{sgn}\otimes M^{{\lambda }^{\prime }}$, rather than the printed sign twist of $M^{\lambda }$. For $\lambda =(n)$, $n\ge 2$, the printed left side has dimension $n!$, while the printed right side has dimension one. Thus that identity cannot hold as written.

The requested character inner product between $M^{\lambda }$ and ${\tilde{M}}^{{\lambda }^{\prime }}$ nevertheless has a well-defined answer. Let $R$ be the row Young subgroup of a tableau of shape $\lambda$, and let $C$ be its column subgroup, conjugate to $S_{{\lambda }^{\prime }}$. By Frobenius reciprocity and the Mackey restriction formula, the inner product is a sum over $R\backslash S_n/C$ ofThese double cosets are encoded by nonnegative integer matrices $A=(a_{ij})$ with row sums ${\lambda }_i$ and column sums ${\lambda }_j^{\prime }$, recording intersection sizes of row and column blocks. The intersection subgroup is a product of $S_{a_{ij}}$. Its sign representation is trivial exactly when every $a_{ij}\le 1$. Each zero-one matrix contributes one, and every other matrix contributes zero.

There is exactly one zero-one matrix with these margins. Its first row has length ${\lambda }_1$, equal to the number of nonzero column sums, so that row must consist entirely of ones. Delete it, subtract one from every column margin and discard zero columns. The remaining margins are those of $({\lambda }_2,{\lambda }_3,\dots )$ and its conjugate partition; induction forces the rest. The unique matrix is $a_{ij}=1$ precisely for $j\le {\lambda }_i$. ThereforeThis calculation uses the actual two modules requested in the PDF and does not depend on its erroneous displayed isomorphism.

Let $P$ be a Sylow p-subgroup. Since $[G:P]$ is invertible in $k$, every $kG$-module is relatively $P$-projective.

Suppose first that $kP$ has only finitely many indecomposable modules $U_1,\dots ,U_t$. For every indecomposable $kG$-module $M$, decompose ${\mathrm{Res}}_P^{G}M$ into the $U_i$. Relative projectivity makes $M$ a summand of the corresponding finite direct sum of the ${\mathrm{Ind}}_P^G{U}_i$. The Krull–Schmidt theorem leaves only finitely many possible indecomposable summands, so $kG$ has finite representation type.

Conversely, suppose $kG$ has finitely many indecomposables. For an indecomposable $kP$-module $U$, the identity double coset in the Mackey restriction formula shows that $U$ is a direct summand ofDecomposing the induced module into the finitely many $kG$-indecomposables and restricting them shows, again by Krull–Schmidt, that only finitely many $U$ can occur. Thus

If $P$ is cyclic, the indecomposable modules of a cyclic p-group in characteristic p form a finite list. If $P$ is noncyclic, its Frattini quotient has rank at least two and therefore has a quotient $C_p\times C_p$. Inflation preserves indecomposability and nonisomorphism, while $k(C_p\times C_p)$ has infinitely many indecomposable modules. The Higman criterion for finite representation type of a group algebra now gives

Existence of a minimal subgroup follows because $G$ has only finitely many subgroups and every module is relatively $G$-projective. Suppose an indecomposable module $M$ is relatively projective for both $H$ and $K$. Then $M$ is a summand of ${\mathrm{Ind}}_H^{G}U$ and of ${\mathrm{Ind}}_K^{G}V$ for suitable modules $U,V$. Applying the Mackey restriction formula and the Krull–Schmidt theorem shows that $M$ is relatively projective for some subgroupIf $H$ and $K$ are minimal, this forces $H\subseteq {}^{g}K$. Reversing their roles gives the reverse containment after conjugacy; since the groups are finite, $H$ and $K$ are conjugate. Thus vertices form a unique conjugacy class.

Let $Q$ be a vertex and let $P$ be a Sylow p-subgroup of $Q$. Since $[Q:P]$ is invertible in $k$, every $kQ$-module is relatively $P$-projective. Transitivity of relative projectivity makes $M$ relatively $P$-projective, so minimality forces $Q=P$. Hence every vertex is a p-group.

For the trivial module $k$, every $kH$-endomorphism is scalar, and its relative trace to $G$ is multiplication by $[G:H]$. The D. Higman criterion says that $k$ is relatively $H$-projective exactly when $p\nmid [G:H]$. The minimal such subgroups are precisely the Sylow p-subgroups. Therefore the vertex of an indecomposable module gives

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 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.

The Mackey restriction formula says that for subgroups $H,K\le G$ and a $K$-representation $W$,where ${}^{x}W(xk{x}^{-1})=W(k)$.

For $H=K=B$, the Bruhat decomposition of SL2 over a finite field givesThus $B\backslash G/B$ has representatives $1,w$. The identity double coset contributes $\theta$, whilecontributes an induced representation from the diagonal subgroup. ThereforewhereFor $t(a)=\mathrm{diag}(a,a^{-1})$, one has $w^{-1}t(a)w=t(a^{-1})$.

Over the complex numbers, the signed Young permutation module isThe isomorphism is the tensor identity for induced representations. The partition index is unchanged by this twist; conjugation instead changes the labels of its irreducible constituents. For the conjugate partition ${\lambda }^{\prime }$, the character inner product of $M^{\lambda }$ and ${\tilde{M}}^{{\lambda }^{\prime }}$ is one: Mackey restriction formula reduces it to the unique zero-one matrix with row margins $\lambda$ and column margins ${\lambda }^{\prime }$.