= Solution
<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 representation>[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:
$$
\overline P=\frac1{|H|}\sum_{h\in H}\rho(h)P\rho(h)^{-1}.
$$
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 representation>[irreducible representations].
The isometry group of $\mathbb Z$ is the <infinite dihedral group>
$$
G=\langle t,s\mid s^2=1,\ sts=t^{-1}\rangle.
$$
Let $\rho$ 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 quotient
$$
\langle t,s\mid t^m=s^2=1,\ sts=t^{-1}\rangle.
$$
Maschke'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 gives
$$
t(sv)=st^{-1}v=\lambda^{-1}sv,
$$
so $\operatorname{span}\{v,sv\}$ 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:
$$
(Tp)(x)=p(x+1).
$$
In the basis $(1,x)$,
$$
T=\begin{pmatrix}1&1\\0&1\end{pmatrix}.
$$
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 $V=\mathbb C[x]_{\le2}$ define
$$
(Tp)(x)=p(x+1),
\qquad
(Sp)(x)=p(-x).
$$
Then $S^2=I$ 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 are
$$
T=
\begin{pmatrix}
1&1&1\\
0&1&2\\
0&0&1
\end{pmatrix},
\qquad
S=
\begin{pmatrix}
1&0&0\\
0&-1&0\\
0&0&1
\end{pmatrix}.
$$
Here $(T-I)^2\ne0$ 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.
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