= Solution
We use left multiplication by $\Gamma=SL_2(\mathbb Z)$. For an integral matrix $A$ of determinant $n>0$, the <Bezout identity> gives a determinant-one row operation sending its first column to $(a,0)^t$, where $a$ is the positive gcd of that column. The resulting matrix is $\begin{pmatrix}a&b\\0&d\end{pmatrix}$ with $d=n/a>0$. Adding a multiple of the second row to the first makes $0\leq b<d$.
These representatives are unique. Left multiplication by a <unimodular matrix> preserves the gcd of the first column, so two representatives in one orbit have the same $a$ and hence $d$. A matrix taking $(a,0)^t$ to itself has the form $\begin{pmatrix}1&t\\0&1\end{pmatrix}$, so it changes $b$ by $td$; the prescribed range makes $b$ unique. Thus the <determinant-n matrix representatives for Hecke operators> are exactly $\Pi_n$. In the printed set, $0\leq b<d$ already forces $d>0$, and $ad=n>0$ forces $a>0$.
For positive determinant extend the <slash operator for modular forms> by
$$
(f|_k A)(z)=\det(A)^{k/2}(cz+d)^{-k}f(Az).
$$
The determinant factor makes this a right action: $(f|_k A)|_k B=f|_k(AB)$. Right multiplication by $\delta\in\Gamma$ permutes the left cosets represented by $\Pi_n$, since it preserves the set of determinant-$n$ integral matrices. Hence $T_nf$ satisfies the modular transformation law. Its summands are <holomorphic> on the upper half-plane. For an upper-triangular representative,
$$
T_nf=n^{k-1}\sum_{ad=n\atop a,d>0}d^{-k}
\sum_{b=0}^{d-1}f\left(\frac{az+b}{d}\right).
$$
Substitute the <Fourier expansion of a modular form>. The sum over $b$ is zero unless the original index is divisible by $d$, in which case it equals $d$. Writing that index as $dr$ gives
$$
T_nf=\sum_{ad=n\atop a,d>0}a^{k-1}\sum_{r\geq0}a_{dr}(f)q^{ar}.
$$
Only nonnegative powers occur, so $T_nf$ is <holomorphic> at infinity, and all level-one cusps are equivalent to infinity. We have proved $T_nf\in M_k(\Gamma)$ and the <Fourier coefficients of a composite-index Hecke operator> formula
$$
\boxed{a_r(T_nf)=\sum_{a\mid\gcd(n,r)}a^{k-1}a_{nr/a^2}(f)
\quad(r\geq0),\qquad a_1(T_nf)=a_n(f).}
$$
Here $\gcd(n,0)=n$, giving $a_0(T_nf)=\sum_{a\mid n}a^{k-1}a_0(f)$. In particular <cusp forms> stay <cusp forms>. For nonzero level-one <cusp forms> the weight is an even integer at least twelve, so the powers $a^{k-1}$ are integers; the zero cusp space causes no exception. Thus every $T_n$ preserves integral <Fourier coefficients>, and closure under addition, multiplication and integer scalars proves
$$
\boxed{\mathbb T\,S_k(\Gamma,\mathbb Z)\subseteq S_k(\Gamma,\mathbb Z).}
$$
This is the <integral Hecke algebra of level-one cusp forms>.
We will also need commutativity. The coefficient formula directly gives
$$
T_uT_v=T_{uv}\quad(\gcd(u,v)=1),\qquad
T_pT_{p^r}=T_{p^{r+1}}+p^{k-1}T_{p^{r-1}}\quad(r\geq1).
$$
For the first identity, the divisors of $u$ and $v$ in the iterated coefficient sum combine uniquely into a divisor of $uv$. For the second, write a coefficient index as $p^t r_0$ with $p\nmid r_0$; the formula is
$$
a_{p^t r_0}(T_{p^r}f)=\sum_{j=0}^{\min(r,t)}p^{j(k-1)}a_{p^{r+t-2j}r_0}(f).
$$
Applying $T_p$ splits this into the two sums with shifts $t+1$ and $t-1$; their overlapping terms give exactly the stated recurrence. Since $T_1$ is the identity, induction expresses each $T_{p^r}$ as a polynomial in $T_p$, and coprime multiplicativity expresses every $T_n$ as a product of these polynomials. Distinct prime operators commute by the coprime identity. Hence all <Hecke operators> commute, and so do all elements of $\mathbb T$.
Let $\Lambda=S_k(\Gamma,\mathbb Z)$ and $m=\dim S_k(\Gamma)$. In the permitted <basis> $F_j=\Delta^jE_4^{a_j}E_6^{b_j}$, the <modular discriminant> begins $q+O(q^2)$ and the normalized <Eisenstein series> begin $1+O(q)$, so
$$
F_j=q^j+O(q^{j+1})\quad(1\leq j\leq m).
$$
The matrix of $a_1,\ldots,a_m$ on this integral <basis> is unitriangular and has determinant one. Integer elimination therefore gives an integral <basis> $f_1,\ldots,f_m$ with $a_i(f_j)=\delta_{ij}$ for $1\leq i,j\leq m$. Thus $a_1,\ldots,a_m$ are a $\mathbb Z$-basis of the <dual module> $\operatorname{Hom}_{\mathbb Z}(\Lambda,\mathbb Z)$.
The map $\alpha$ is well-defined by lattice preservation and is $\mathbb Z$-linear. Because $\alpha(T_j)=a_j$, it is surjective. For injectivity suppose $\alpha(T)=0$. For every $f\in\Lambda$ and every $n\geq1$, commutativity gives
$$
a_n(Tf)=a_1(T_nTf)=a_1(TT_nf)=\alpha(T)(T_nf)=0.
$$
The constant term is also zero because $Tf$ is a <cusp form>. The <identity theorem> applied to its <Fourier expansion of a modular form> gives $Tf=0$. The integral <basis> spans the complex cusp space, so $T=0$ as an endomorphism. This proves the <perfect integral Hecke pairing> and
$$
\boxed{\mathbb T\cong\operatorname{Hom}_{\mathbb Z}(\Lambda,\mathbb Z),
\qquad\mathbb T=\bigoplus_{j=1}^m\mathbb ZT_j.}
$$
If $m=0$, both modules are zero and the asserted <basis> is empty.
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