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Every group representation of a finite group over a field whose characteristic does not divide the number of elements of the group is semisimple. Given an invariant subspace, averaging any projection onto it over the group produces an equivariant projection, whose kernel is an invariant complement.

Restriction of an irreducible representation to a subgroup is multiplicity-free when its semisimple representation decomposition has no repeated irreducible summands. Over the complex numbers, for finite groups, this is equivalent to the restricted module's endomorphism algebra being commutative. This follows by writing that algebra as a product of matrix algebras of sizes equal to the multiplicities.

A two-dimensional Lie algebra representation already provides a counterexample. In the basis $e_1,e_2$ of $V={\mathbb{C}}^2$, defineThe commutator $[\rho (x),\rho (y)]$ is zero, agreeing with $\rho (z)$; the other defining Lie brackets also map to zero. Thus $\rho$ is a Lie algebra homomorphism, although it is not a Faithful Lie algebra representation.

The line $L=\mathbb{C}{e}_1$ is an invariant subspace. Any complementary line has the form $\mathbb{C}(e_2+a{e}_1)$ for some $a\in \mathbb{C}$. Butso no complementary line is an invariant subspace. Equivalently, every invariant line is killed by $\rho (x)$: its restriction to a line is a scalar, and its square is zero, forcing that scalar to be zero. A decomposition into two invariant lines would then force $\rho (x)=0$, a contradiction. This representation is therefore not a semisimple representation. Thus

Maschke's theorem says that if $H$ is a finite group and the characteristic of a field $F$ does not divide $|H|$, then every finite-dimensional $F$-representation of $H$ is semisimple.

To prove it, let $W$ be an invariant subspace of a representation $V$, and choose any linear projection $P:V\to W$. Average its conjugates:The scalar $|H{|}^{-1}$ exists by the characteristic assumption. Reindexing the sum shows that $\overline{P}$ commutes with every $\rho (g)$. Since each summand restricts to the identity on $W$, so does $\overline{P}$; hence $\overline{P}$ is an equivariant projection onto $W$. Its kernel is an invariant complement, and induction on $\dim V$ decomposes $V$ into irreducible representations.

The isometry group of $\mathbb{Z}$ is the infinite dihedral groupLet $\rho$ be a nonfaithful finite-dimensional complex representation. Its nontrivial normal kernel contains $t^m$ for some $m>0$, by the stated hint, so $\rho$ factors through the finite quotientMaschke's theorem decomposes this quotient representation into irreducibles. To bound their dimensions, choose a $t$-eigenvector $v$ with eigenvalue $\lambda$ in an irreducible constituent. The relation givesso $\mathrm{span}\{v,sv\}$ is invariant under both generators. Irreducibility makes it the whole constituent, whose dimension is therefore at most two. This proves the result on nonfaithful representations of the infinite dihedral group.

For the requested representation of $(\mathbb{Z},+)$, let its generator act on the polynomials of degree at most one by translation:In the basis $(1,x)$,This is a nontrivial Jordan block, so it is not diagonalizable. A direct sum of one-dimensional subrepresentations would make $T$ diagonalizable, proving that this nonsemisimple translation representation of the infinite cyclic group has the required property.

Finally, on $V=\mathbb{C}[x{]}_{\le 2}$ defineThen $S^2=I$ and $STS=T^{-1}$, so these operators define a quadratic-polynomial representation of the infinite dihedral group. In the basis $(1,x,x^2)$ they areHere $(T-I{)}^2\ne 0$ but $(T-I{)}^3=0$, so $T$ consists of one size-three Jordan block. If $V$ were a direct sum of subrepresentations of dimensions at most two, then $T$ would be block diagonal with Jordan blocks of sizes at most two, a contradiction. Thus this representation has the required failure of decomposition.

A representation is completely reducible, or semisimple, when it is a direct sum of irreducible representation. Equivalently, every invariant subspace has an invariant complement.

Maschke's theorem says that every finite-dimensional representation of a finite group $G$ over a field of characteristic zero is completely reducible. More generally, it is enough that the field characteristic does not divide $|G|$.

Schur lemma states the following.

Indeed, $\ker T\subseteq V$ and $\mathrm{im}T\subseteq W$ are invariant subspaces. If $T\ne 0$, irreducibility forcesso $T$ is an isomorphism. For an endomorphism of a finite-dimensional complex representation, choose an eigenvalue $\lambda$ of $T$. The intertwiner $T-\lambda I$ has nonzero kernel, so the first part forcesThis proves the Proof of Schur lemma.

Now let $\rho :G\to \mathrm{GL}(V)$ be faithful and irreducible over $\mathbb{C}$. For every $z\in Z(G)$, the operator $\rho (z)$ commutes with every $\rho (g)$. Schur's lemma therefore givesfor some ${\lambda }_z\in {\mathbb{C}}^{\times }$. Faithfulness makes $z\mapsto {\lambda }_z$ an embeddingSince every finite subgroup of the multiplicative complex numbers is cyclic,This is the cyclic-center obstruction to a faithful irreducible representation.