LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-sol)audited 2026-09-27

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Kernel of an induced representation is the intersection of the conjugates of the kernel of the inducing representation

Statement

Let $G$ be a finite group, let $H \leq G$ , and let $W \neq 0$ be a finite-dimensional complex representation of $H$ with kernel $\ker W = {h \in H : h \cdot w = w for all w \in W}$ . Then the kernel of the induced representation ${Ind}_{H}^{G} W$ is

$\ker ({Ind}_{H}^{G} W) = ⋂_{g \in G} g (\ker W) g^{- 1} \subseteq {Core}_{G} (H) = ⋂_{g \in G} g H g^{- 1} \leq H .$

In particular, if $W = 1_{H}$ is the trivial representation, then $\ker ({Ind}_{H}^{G} 1_{H}) = {Core}_{G} (H)$ , which is the kernel of the permutation action of $G$ on the left cosets $G / H$ . By The kernel of a complex character agrees with the kernel of any representation affording it the same formula computes the kernel $\ker χ$ of the induced character $χ = {Ind}_{H}^{G} χ_{W}$ .

Facts & Assumptions

Given: A finite group $G$ , a subgroup $H \leq G$ , a nonzero finite-dimensional complex $H$ -representation $W$ with kernel $\ker W$ and character $χ_{W}$ , a left transversal $T$ for the left cosets $G / H$ , and $V = {Ind}_{H}^{G} W$ .

[F1]

$V = {Ind}_{H}^{G} W = {f : G \to W : f (g h) = h^{- 1} \cdot f (g) for all g \in G, h \in H}$ , with $(x \cdot f) (g) = f (x^{- 1} g)$ . (The induced $R$ -linear $G$ -module ${Ind}_{H}^{G} W$ as $H$ -covariant functions on $G$ ).

[F2]

Evaluation on $T$ is a linear isomorphism ${ev}_{T} : V \to ⨁_{t \in T} W$ , $f \mapsto (f (t))_{t \in T}$ ; hence the tuples $(f (t))_{t \in T}$ run through all of $⨁_{t \in T} W$ , independently in each coordinate. (A left transversal identifies ${Ind}_{H}^{G} W$ with a direct sum of $[G : H]$ copies of $W$ ).

[F3]

$\ker W$ is the kernel of the homomorphism $H \to GL (W)$ defining $W$ , hence a normal subgroup of $H$ ; and the kernel $\ker χ_{W}$ of the character equals $\ker W$ . (The kernel and image of a group homomorphism, The image of a group homomorphism is a subgroup and its kernel is a normal subgroup, The kernel of a complex character agrees with the kernel of any representation affording it, The kernel of a complex character).

[F4]

Left multiplication on $G / H$ is a permutation action whose kernel is ${Core}_{G} (H) = ⋂_{g \in G} g H g^{- 1}$ . (Left multiplication on $G / H$ is transitive, has stabiliser $H$ at $H$ , and has kernel ${Core}_{G} (H)$ , The core ${Core}_{G} (H) = ⋂_{g \in G} g H g^{- 1}$ of a subgroup).

[F5]

Inducing the trivial representation $1_{H}$ gives the permutation representation of $G$ on $G / H$ . (Inducing the trivial representation gives the permutation representation on $G / H$ ).

Proof

1.1

Conjugation by any $x \in G$ permutes the set ${g (\ker W) g^{- 1} : g \in G}$ , so $N : = ⋂_{g \in G} g (\ker W) g^{- 1}$ satisfies $x N x^{- 1} = N$ ; hence $N ⊴ G$ . Since $\ker W \leq H$ by [F3], each conjugate $g (\ker W) g^{- 1}$ lies in $g H g^{- 1}$ , so $N \subseteq ⋂_{g \in G} g H g^{- 1} = {Core}_{G} (H) \leq H$ .

F3F4given

1.2

If $t^{'} = t h$ with $t \in T$ and $h \in H$ , then $t^{'} (\ker W) t^{' - 1} = t (h (\ker W) h^{- 1}) t^{- 1} = t (\ker W) t^{- 1}$ by normality of $\ker W$ in $H$ , so the conjugate depends only on the coset. As $T$ meets every left coset exactly once, $⋂_{t \in T} t (\ker W) t^{- 1} = ⋂_{g \in G} g (\ker W) g^{- 1} = N$ .

F3given

1.3

By [F2] the map ${ev}_{T}$ is injective, so $x \in G$ acts as the identity on $V$ exactly when $(x \cdot f) (t) = f (t)$ for all $f \in V$ and all $t \in T$ . Fix $t \in T$ and write $x^{- 1} t = t_{*} k$ with $t_{*} \in T$ and $k \in H$ . Then $f (x^{- 1} t) = f (t_{*} k) = k^{- 1} \cdot f (t_{*})$ by the covariance rule of [F1], so $x$ acts as the identity exactly when $k^{- 1} \cdot f (t_{*}) = f (t)$ for every $f \in V$ and every $t \in T$ .

F1F2given

2.1

If $t_{*} \neq t$ for some $t \in T$ , choose by [F2] an element $f \in V$ with $f (t_{*}) = w \neq 0$ and $f (t) = 0$ ; then $k^{- 1} \cdot w \neq 0 = f (t)$ , so the condition of step 1.3 fails. Hence an element $x \in G$ acts as the identity on $V$ only if $x^{- 1} t \in t H$ for every $t \in T$ , that is, $t_{*} = t$ for every $t \in T$ .

F2step 1.3

3.1

Suppose then that $x^{- 1} t = t k_{t}$ with $k_{t} \in H$ for every $t \in T$ . By step 1.3 the element $x$ acts as the identity if and only if $k_{t}^{- 1} \cdot w = w$ for all $w \in W$ , that is, if and only if $k_{t} \in \ker W$ for every $t \in T$ ; equivalently $t^{- 1} x^{- 1} t \in \ker W$ for every $t \in T$ , which says $x \in t (\ker W) t^{- 1}$ for every $t \in T$ .

F3step 1.3step 2.1

4.1

Steps 2.1 and 3.1 together characterise the kernel: $\ker ({Ind}_{H}^{G} W) = ⋂_{t \in T} t (\ker W) t^{- 1} = N$ , which by step 1.1 lies in ${Core}_{G} (H) \leq H$ .

step 1.1step 1.2step 3.1

5.1

For $W = 1_{H}$ one has $\ker W = H$ , so step 4.1 gives $\ker ({Ind}_{H}^{G} 1_{H}) = ⋂_{g \in G} g H g^{- 1} = {Core}_{G} (H)$ , which by [F4] and [F5] is the kernel of the permutation action on $G / H$ ; and by [F3] the same formula computes the kernel of the induced character ${Ind}_{H}^{G} χ_{W}$ in general. ∎

F3F4F5step 4.1

Depends on

Used by

M-groups are solvable (Taketa) Corollary

Dependency tree · two levels

28 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

SLMath, Character Theory of Finite Groups, Chapter 9 — slides 387–390 (induced-character kernel lemma $\ker(\alpha^G)\subseteq H$ with proof) (standard reference, not scraped)
I. M. Isaacs, Character Theory of Finite Groups — Chapter 5, kernel of an induced character, p. 67 (standard sharp form, located through the book's index) (standard reference, not scraped)