Trivial and one-dimensional monomial cases

Evaluation at the identity, $\mathrm{\Phi }:{\mathrm{Ind}}_G^{G}L\to L$, $\mathrm{\Phi }(f):=f(1)$, is a $\mathbb{C}$-linear map of $G$-modules. It is $\mathbb{C}$-linear because the module operations on ${\mathrm{Ind}}_G^{G}L$ are pointwise by [F1]; and for $f\in {\mathrm{Ind}}_G^{G}L$ and $x\in G$ one has $\mathrm{\Phi }(x\cdot f)=(x\cdot f)(1)=f(x^{-1})=x\cdot f(1)=x\cdot \mathrm{\Phi }(f)$, where the covariance law of [F1] with $g=1$ and $h=x^{-1}$ gives $f(x^{-1})=f(1\cdot x^{-1})=(x^{-1}{)}^{-1}\cdot f(1)=x\cdot f(1)$.

The map $\mathrm{\Phi }$ is bijective. For $v\in L$ define $f_v:G\to L$ by $f_v(g):=g^{-1}\cdot v$; then $f_v(gh)=(gh{)}^{-1}\cdot v=h^{-1}\cdot (g^{-1}\cdot v)=h^{-1}\cdot f_v(g)$ for all $g\in G$ and $h\in H=G$, so $f_v\in {\mathrm{Ind}}_G^{G}L$ by [F1], and $\mathrm{\Psi }(v):=f_v$ is $\mathbb{C}$-linear and $G$-equivariant because $f_{x\cdot v}(g)=g^{-1}\cdot (x\cdot v)=(x\cdot f_v)(g)$. Moreover $\mathrm{\Phi }(\mathrm{\Psi }(v))=f_v(1)=v$, and for $f\in {\mathrm{Ind}}_G^{G}L$ the same covariance law with $g=1$, $h=g$ gives $f(g)=f(1\cdot g)=g^{-1}\cdot f(1)=f_{f(1)}(g)$, that is $\mathrm{\Psi }(\mathrm{\Phi }(f))=f$. Hence $\mathrm{\Psi }={\mathrm{\Phi }}^{-1}$ and ${\mathrm{Ind}}_G^{G}L\cong L$ as $G$-modules, so their characters agree: ${\mathrm{Ind}}_G^G\lambda =\lambda$. The degrees match, since ${\dim }_{\mathbb{C}}{\mathrm{Ind}}_G^{G}L=[G:G]{\dim }_{\mathbb{C}}L=1$ by [F6].

By step 2.1 every one-dimensional complex character $\lambda$ of $G$ is monomial in the sense of [F2], with $H=G$ and ${\mathrm{Ind}}_G^G\lambda =\lambda$. In particular the trivial character $1_G$, the character of the trivial representation $\mathbb{C}$ of $G$, is one-dimensional and hence monomial; the trivial representation is irreducible because a one-dimensional space has no nonzero proper subspace, so $0$ and $\mathbb{C}$ are its only invariant subspaces by [F3].

For the trivial group $G=\{1\}$ one has $\mathrm{\mid }G\mathrm{\mid }=1$, so ${\sum }_i{\chi }_i(1{)}^2=1$ over the irreducible complex characters by [F5]; each term ${\chi }_i(1{)}^2$ is a positive integer, so the sum has exactly one term and ${\chi }_1(1)=1$. Hence $\{1\}$ has exactly one irreducible complex character, of degree one, and it is monomial by step 3.1; by [F2] the trivial group is an $M$-group.

For a finite abelian group $A$, the field $\mathbb{C}$ is a splitting field for $A$ by [F4], so every irreducible complex representation of $A$ has degree $1$ by [F4]; hence every irreducible complex character of $A$ is a one-dimensional character and is monomial by step 3.1, so $A$ is an $M$-group by [F2]. Moreover [F5] now evaluates to $\mathrm{\mid }A\mathrm{\mid }={\sum }_{i}1=\mathrm{\mid }\mathrm{Irr}(A)\mathrm{\mid }$, so a finite abelian group has exactly $\mathrm{\mid }A\mathrm{\mid }$ irreducible characters, all of them linear and monomial; the cases $\mathrm{\mid }A\mathrm{\mid }=1$ and $\mathrm{\mid }A\mathrm{\mid }=2$ are the extremes, the former being step 4.1.

All three assertions hold: every one-dimensional character of a finite group is induced from the group itself and is monomial, the trivial group is an $M$-group, and every finite abelian group is an $M$-group. The construction involves no proper subgroup and no choice: for $H=G$ the module ${\mathrm{Ind}}_G^{G}L$ is explicitly identified with $L$ by evaluation at $1$, with inverse $v\mapsto f_v$, and the only groups used have a specified single irreducible character or are handled by the degree count ${\sum }_i{\chi }_i(1{)}^2=\mathrm{\mid }A\mathrm{\mid }$ of [F5].