= Solution
In this part identify $\mathbb T$ with the <complex unit circle> by $t\mapsto e^{it}$. Write the finite <abelian group> $G$ additively and let $\widehat G=\operatorname{Hom}(G,\mathbb T)$, with pointwise multiplication, be the <character group of a finite abelian group> of $G$. A <linear character> has modulus one, since its values have finite order. We first prove the extension and counting facts needed for <Fourier analysis on a finite abelian group>.
If $H\le G$, $\chi\in\widehat H$, and $x\notin H$, let $d$ be the least positive integer with $dx\in H$. Choose $z\in\mathbb T$ with $z^d=\chi(dx)$ and define
$$
\widetilde\chi(h+jx)=\chi(h)z^j.
$$
This is well-defined: two representations differ by $j-j'=qd$ and $h-h'=-qdx$, whose contributions cancel. It is a <group homomorphism> and extends $\chi$. Conversely every extension has this form, and the equation for $z$ has exactly $d$ distinct roots. Since $[H+\langle x\rangle:H]=d$, successive adjunctions prove that each <linear character> of $H$ has exactly $[G:H]$ extensions. Taking $H=\{0\}$ gives $\boxed{|\widehat G|=|G|}$. This is the <extension of a character across a cyclic quotient>, proved here without a structure theorem. Extending the <linear character> sending a nonzero element $g$ of order $d$ to $e^{2\pi i/d}$ also proves that <linear characters> separate points of $G$.
Use the normalized <inner product> $\langle f,g\rangle=|G|^{-1}\sum_{x\in G}f(x)\overline{g(x)}$. For a nontrivial <linear character> $\chi$, choose $a$ with $\chi(a)\ne1$. Translation of the sum gives $\sum_x\chi(x)=\chi(a)\sum_x\chi(x)$, so that sum is zero. Applying this to $\chi\overline\psi$ gives
$$
\boxed{\langle\chi,\psi\rangle=\mathbf1_{\{\chi=\psi\}}.}
$$
There are $|G|$ such orthonormal vectors in the $|G|$-dimensional space $\mathbb C^G$, so they form an <orthonormal basis>. Define the <Fourier coefficients> by $\widehat f(\chi)=|G|^{-1}\sum_x f(x)\overline{\chi(x)}$. Expanding in that basis proves <Fourier inversion on a finite group> and the <Parseval identity on a finite group>:
$$
\boxed{f(x)=\sum_{\chi\in\widehat G}\widehat f(\chi)\chi(x),\qquad
\langle f,g\rangle=\sum_{\chi\in\widehat G}\widehat f(\chi)\overline{\widehat g(\chi)}.}
$$
For normalized <convolution> $(f*g)(x)=|G|^{-1}\sum_y f(y)g(x-y)$, changing the finite summation variables yields $\widehat{f*g}(\chi)=\widehat f(\chi)\widehat g(\chi)$. Translation $f(x)\mapsto f(x+a)$ multiplies its <Fourier coefficient> by $\chi(a)$. These formulas fix all normalization and conjugation conventions.
The PDF's double-hat identification is canonical <Pontryagin duality>: define $e:G\to\widehat{\widehat G}$ by $e_x(\chi)=\chi(x)$. This is a <group homomorphism> and is injective because <linear characters> separate points. Applying the already proved character-counting result to $\widehat G$ shows that both groups have size $|G|$, so
$$
\boxed{G\cong\widehat{\widehat G}\quad\text{canonically by evaluation}.}
$$
The TeX loses one hat. For completeness, the single-dual isomorphism $\widehat G\cong G$ also holds, but is generally noncanonical. Here is a proof including the needed finite cyclic decomposition. Choose an element $a$ of maximal order $d$. Every element order divides $d$: for two elements, extract their prime-power components by integer multiples: if an element has order $d$ and $p^\alpha$ is the exact power of $p$ in $d$, multiplying it by $d/p^\alpha$ gives order $p^\alpha$. Then combine the components with the larger order for each prime. Components of coprime orders have a sum of product order, because their <cyclic subgroups> have trivial intersection. This constructs an element of order the <least common multiple> of the two original orders, which cannot exceed maximal $d$; hence each order divides $d$.
Extend the primitive <linear character> of $\langle a\rangle$ to $\chi:G\to\mathbb T$. All its values are $d$th roots because all element orders divide $d$. Define $\rho(g)=ja$ when $\chi(g)=e^{2\pi ij/d}$. This is a <group homomorphism> fixing $\langle a\rangle$, and $g=\rho(g)+(g-\rho(g))$ shows $G=\langle a\rangle\oplus\ker\rho$. Induction on group size decomposes $G$ into finite <cyclic groups>. A <cyclic group> of order $d_j$ has precisely the <linear characters> $j\mapsto e^{2\pi imj/d_j}$ and hence a <character group of a finite abelian group> isomorphic to itself. The dual of a finite <direct product of groups> is the product of its factor duals, since a <linear character> is determined by its restrictions. This proves $\widehat G\cong G$ after choices of cyclic factors and generators, without using an unproved finite-abelian-group structure theorem.
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