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If $H$ is a subgroup of a finite abelian group, $a\notin H$, and $k$ is the least positive integer with $ka\in H$, then each character of a finite abelian group on $H$ has exactly $k$ extensions to $H+⟨a⟩$. The displayed root choice makes the extension well-defined. Starting with the trivial subgroup and adjoining generators proves that the number of characters equals the group order without using a decomposition into cyclic groups.

Let $H\le G$ be finite abelian and let $\chi :H\to \mathbb{T}$ be a character of a finite abelian group. To adjoin $x\in G$, let $d$ be the least positive integer with $dx\in H$ and choose a complex number $z$ of modulus one with $z^d=\chi (dx)$. Then $\tilde{\chi }(h+jx)=\chi (h)z^j$ is well-defined and extends $\chi$ to $H+⟨x⟩$. There are exactly $d$ choices, corresponding to the $d$ roots. Iterating proves character extension to $G$, proves that $G$ has exactly $|G|$ characters, and proves separation of points by extending a primitive character of the cyclic subgroup generated by a nonzero point. No structure theorem is needed for this argument.

We prove the Roth theorem on three-term arithmetic progressions by a density increment on an ordinary integer arithmetic progression. The required increment will be small, but depend only on the current subset density.

First establish the increment step. Suppose $A\subseteq [1,L]$ has subset density $\rho =|A|/L>0$ and no nonconstant three-term arithmetic progression. Put $M=4L+1$, regard $[1,L]$ as a subset $I$ of the cyclic group ${\mathbb{Z}}_M$, and set $f=1_A-\rho 1_I$. Then $\mathbb{E}f=0$ and $|f|\le 1$. Use the normalized Fourier coefficient on a finite abelian groupThe character orthogonality geometric-series identity ${\mathbb{E}}_x{e}^{2\pi irx/M}=1$ for $r=0$ and zero otherwise gives Fourier inversion on a finite group, Parseval identity on a finite group, andSince $M$ is odd, multiplication by two permutes its frequencies. If one argument is $f$ and the other two have absolute value at most one, Cauchy-Schwarz and the Parseval identity on a finite group give $|T|\le {\max }_r|\hat{f}(r)|$.

A modular arithmetic progression whose three entries lie in $I$ satisfies the corresponding integer equation: the absolute value of $x+z-2y$ is less than $2L<M$, so congruence to zero forces equality. Thus $T(1_A,1_A,1_A)=|A|/M^2\le 1/(16L)$, since only constant arithmetic progressions remain. Meanwhile, the choices $1\le x\le \lfloor L/2\rfloor$ and $0\le h\le \lfloor L/4\rfloor$ give at least $L^2/16$ arithmetic progressions in $I$ for $L\ge 8$. As $M\le 5L$, this gives $T(1_I,1_I,1_I)\ge 1/400$. Expand $1_A=\rho 1_I+f$. The seven terms other than the main term are each bounded by $\max |\hat{f}|$. For $L\ge 50{\rho }^{-3}$,Consequently a nonzero frequency $r$ obeys $|\hat{f}(r)|\ge \gamma$, where $\gamma ={\rho }^3/6000$. It is nonzero because $\hat{f}(0)=0$.

We turn this coefficient into a local increase. Let $Q=\lfloor \sqrt{L}\rfloor$. Partition the unit circle into $Q$ equal intervals and consider the $Q+1$ points $jr/M$, $0\le j\le Q$, modulo one. Two occupy the same interval, yielding an integer $1\le q\le Q$ with $\Vert qr/M{\Vert }_{\mathbb{R}/\mathbb{Z}}\le 1/Q$. Set $\ell =\lfloor \gamma Q/200\rfloor$. If $L$ exceeds a sufficiently large absolute multiple of ${\rho }^{-6}$, then $\ell \ge 1$ and each residue-class arithmetic progression of step $q$ in $[1,L]$ has at least $\ell$ terms. Partition each into pieces of lengths between $\ell$ and $2\ell$, distributing the final remainder among the pieces.

On each such arithmetic progression $P$, the character of a finite abelian group $\chi (x)=e^{-2\pi irx/M}$ differs from its value at the first point by at most $4\pi \ell /Q<\gamma /4$. Write $s_P={\sum }_{x\in P}f(x)$. Thereforeso ${\sum }_P|s_P|\ge 3\gamma L/4$. But ${\sum }_P{s}_P={\sum }_{I}f=0$, and hence the sum of the positive $s_P$ is at least $3\gamma L/8$. Some piece must consequently satisfy $s_P/|P|\ge \gamma /4$. On that piece,Also, once $\gamma Q\ge 400$, $|P|\ge \ell \ge \gamma \sqrt{L}/800$. This proves the one-frequency density increment on an integer interval: for absolute constants $c,C>0$, if $L\ge C{\rho }^{-6}$, there is a arithmetic progression $P\subseteq [1,L]$ of length at least $c{\rho }^3\sqrt{L}$ on which the subset density rises by at least $c{\rho }^3$. For example one may decrease $c$ to $10^{-7}$ and choose $C$ large enough for the displayed elementary inequalities.

Now fix $0<\delta \le 1$ and suppose the required conclusion fails. On each new arithmetic progression, identify its points in order with $[1,L^{\prime }]$. This affine identification preserves all integer arithmetic progressions, so its copy of $A\cap P$ is still progression-free. All densities remain at least $\delta$. Put $\epsilon =c{\delta }^3$, $H=C{\delta }^{-6}$, and choose an integer $t>1/\epsilon$. Every increment raises the subset density by at least $\epsilon$ and changes the length by $L_{j+1}\ge \epsilon \sqrt{L_j}$. Iterating the latter inequality givesChoose $N>(H/{\epsilon }^2{)}^{2^t}$. Then every length up to step $t$ exceeds $H$, so every increment is available. After $t$ increments the subset density exceeds one, a contradiction. ThusFor $\delta >1$ there are no sets satisfying the subset density assumption. The argument above proves the substantive case without invoking the theorem being asked for.

The expectations use normalized counting measure on a finite abelian group $G$. This is the finite-group convention implicit in the averaging notation; an arbitrary infinite group has no normalized counting probability measure on all its points. For $h\in G$, define ${\Delta }_{h}f(x)=f(x)\overline{f(x+h)}$, and let $\mathcal{C}$ denote complex conjugation. The Gowers U3 norm isAveraging first over $h_3$ gives the Derivative identity for the Gowers U3 normIndeed, if $u(x)={\Delta }_{h_1}{\Delta }_{h_2}f(x)$, the average over $(x,h_3)$ is ${\mathbb{E}}_{x,h_3}u(x)\overline{u(x+h_3)}=|{\mathbb{E}}_{x}u(x){|}^2$. The eighth root is consequently well-defined. Homogeneity follows because the cube has four unconjugated and four conjugated factors, so $\Vert af{\Vert }_{U^3}=|a|\Vert f{\Vert }_{U^3}$. If the norm is zero, every squared term in the finite average vanishes. Taking $h_1=h_2=0$ gives ${\mathbb{E}}_x|f(x){|}^4=0$, so $f=0$ pointwise. This proves positive definiteness.

For the triangle inequality, we prove the needed Gowers-Cauchy-Schwarz inequality. For eight functions $F_{\omega }$ on a product of finite sets $X_1\times X_2\times X_3$, form the box norm multilinear averageFix a coordinate $j$. After averaging its two variables separately, this is $\mathbb{E}A\overline{B}$, where $A$ uses all vertices with ${\omega }_j=0$ and $B$ those with ${\omega }_j=1$, with the remaining parity conjugations. The Cauchy-Schwarz inequality giveswhere $S_j^b$ replaces the function at each vertex by the original function with its $j$th index set to $b$. Each right-hand average is a squared absolute value averaged over the other coordinates, hence nonnegative. Apply this operation successively for $j=1,2,3$, to every resulting factor. The eight final families each use one original $F_{\omega }$ at all vertices. Taking roots gives the box Cauchy-Schwarz inequalityNow take $X_j=G$ and $F_{\omega }(a,b,c)=f_{\omega }(a+b+c)$. With $x=a_0+b_0+c_0$ and $h_j=x_j^1-x_j^0$, these independent variables induce uniform $(x,h_1,h_2,h_3)$: each such tuple has exactly $|G{|}^2$ preimages. Therefore each box norm is the corresponding $U^3$ norm, and the mixed cube obeysExpand the eight factors in the cube expression for $f+g$. Its $2^8$ mixed terms are bounded by this inequality, and their bounds sum to $(\Vert f{\Vert }_{U^3}+\Vert g{\Vert }_{U^3}{)}^8$. Thus $\boxed{\Vert f+g{\Vert }_{U^3}\le \Vert f{\Vert }_{U^3}+\Vert g{\Vert }_{U^3}}$, completing the norm proof.

Two consequences will be useful below. Multiplication by a character of a finite abelian group leaves the norm unchanged, because the alternating character product around a three-dimensional cube equals one. Alsowhere the last inequality uses the $h=0$ term and $\Vert u{\Vert }_{U^2}^4={\sum }_{\chi }|\hat{u}(\chi ){|}^4\ge |\mathbb{E}u{|}^4$. Hence $\Vert f{\Vert }_{U^3}\ge |G{|}^{-1/8}\Vert f{\Vert }_2$.

Write the finite abelian group additively. A character of a finite abelian group is a group homomorphism $\chi :G\to \{z:|z|=1\}$. We first prove that there are exactly $|G|$ such characters of a finite abelian group, without assuming a structure theorem.

Use extension of a character across a cyclic quotient. Given a subgroup $H$ and $a\notin H$, let $k$ be the least positive integer with $ka\in H$. The subgroup $K=H+⟨a⟩$ has $k$ cosets of $H$. If $\varphi$ is a character of a finite abelian group on $H$, choose any of the $k$ roots ${\lambda }^k=\varphi (ka)$ and defineThis is well-defined: two representations differ by an integer multiple of $ka$, and the root equation exactly cancels that difference. It is a character of a finite abelian group, and every extension arises from one of the $k$ choices of $\lambda$. Build a chain from the trivial subgroup to $G$ by adjoining elements. The character of a finite abelian group count multiplies by the same factor as the subgroup order at every step, so $|\hat{G}|=|G|$.

For a nontrivial character of a finite abelian group $\eta$, choose $a$ with $\eta (a)\ne 1$. Translating the group sum showsso the sum is zero. Applied to $\chi \overline{\psi }$, this provesThe $|G|$ orthogonal nonzero characters of a finite abelian group therefore form a basis of all complex functions on $G$, a vector space of dimension $|G|$.

With the unnormalized Fourier coefficients $\hat{f}(\chi )={\sum }_{x\in G}f(x)\overline{\chi (x)}$, expansion in that basis gives the Fourier inversion on a finite groupUnder a normalized forward-transform convention the prefactor would instead be one. Thus the unspecified constant is determined by the convention, and the inversion itself follows directly from character of a finite abelian group counting and orthogonality.