[proofplan] We reduce the statement to the aperiodic case by passing to the [quotient group](/page/Quotient%20Group) $G/H$, where $H=\operatorname{Stab}(A+B)$. In that quotient, the images of $A$, $B$, and $A+B$ under the canonical quotient map have aperiodic sumset, so the aperiodic form of Kneser's theorem gives the lower bound for the quotient sumset. We first verify that $H$ is finite, and then multiply the quotient inequality by $|H|$ to convert it exactly into the desired estimate for $A+B$. [/proofplan] [step:Pass from $G$ to the quotient by the stabilizer] Let $\pi:G\to G/H$ denote the quotient homomorphism from $G$ to the abelian [quotient group](/theorems/790) $G/H$. Define finite nonempty subsets $\bar A,\bar B\subset G/H$ by \begin{align*} \bar A&:=\pi(A), & \bar B&:=\pi(B). \end{align*} Since $A$ and $B$ are finite and nonempty, their images $\bar A$ and $\bar B$ are finite and nonempty. Moreover \begin{align*} \bar A+\bar B=\pi(A)+\pi(B)=\pi(A+B). \end{align*} We claim that $\operatorname{Stab}_{G/H}(\bar A+\bar B)=\{0_{G/H}\}$. Let $g+H\in G/H$ stabilize $\bar A+\bar B$. Then \begin{align*} (g+H)+\pi(A+B)=\pi(A+B), \end{align*} which means \begin{align*} g+A+B+H=A+B+H. \end{align*} Because $H=\operatorname{Stab}(A+B)$, the set $A+B$ is $H$-invariant, so $A+B+H=A+B$. Hence $g+A+B=A+B$, and therefore $g\in \operatorname{Stab}(A+B)=H$. Thus $g+H=0_{G/H}$, proving the claim. We also record that $H$ is finite. Choose an element $x_0\in A+B$, which exists because $A$ and $B$ are nonempty. Define \begin{align*} \iota:H&\to A+B\\ h&\mapsto x_0+h. \end{align*} This map is well-defined because $H=\operatorname{Stab}(A+B)$ implies $x_0+h\in A+B$ for every $h\in H$. If $\iota(h_1)=\iota(h_2)$, then $x_0+h_1=x_0+h_2$, hence $h_1=h_2$ by cancellation in the group $G$. Therefore $\iota$ is injective. Since $A+B$ is finite, $H$ is finite. [guided] The purpose of introducing the quotient is to remove exactly the periodicity that prevents the elementary Cauchy-Davenport-type lower bound from being true. Let \begin{align*} \pi:G&\to G/H\\ g&\mapsto g+H \end{align*} be the quotient homomorphism, and define \begin{align*} \bar A&:=\pi(A), & \bar B&:=\pi(B). \end{align*} The sets $\bar A$ and $\bar B$ are finite and nonempty because $A$ and $B$ are finite and nonempty and $\pi$ is a function. Since $\pi$ is a group homomorphism, it respects sums of subsets: \begin{align*} \bar A+\bar B &=\{\pi(a)+\pi(b):a\in A,\ b\in B\}\\ &=\{\pi(a+b):a\in A,\ b\in B\}\\ &=\pi(A+B). \end{align*} Now we verify that the quotient sumset has no nonzero period. Suppose $g+H\in G/H$ satisfies \begin{align*} (g+H)+(\bar A+\bar B)=\bar A+\bar B. \end{align*} Using $\bar A+\bar B=\pi(A+B)$, this becomes \begin{align*} (g+H)+\pi(A+B)=\pi(A+B), \end{align*} or equivalently \begin{align*} g+A+B+H=A+B+H. \end{align*} The definition $H=\operatorname{Stab}(A+B)$ implies $A+B+H=A+B$: adding any element of $H$ to $A+B$ leaves $A+B$ unchanged. Therefore the preceding equality reduces to \begin{align*} g+A+B=A+B. \end{align*} This says precisely that $g\in\operatorname{Stab}(A+B)=H$. Hence $g+H=0_{G/H}$. So the only stabilizer element of $\bar A+\bar B$ in $G/H$ is the identity coset. We will also need $H$ to be finite when translating quotient cardinalities back to $G$. Because $A$ and $B$ are nonempty, choose an element $x_0\in A+B$. Define \begin{align*} \iota:H&\to A+B\\ h&\mapsto x_0+h. \end{align*} The map $\iota$ is well-defined because every $h\in H$ stabilizes $A+B$, so $x_0+h\in A+B$. It is injective: if $\iota(h_1)=\iota(h_2)$, then $x_0+h_1=x_0+h_2$, and cancellation in the group $G$ gives $h_1=h_2$. Thus $H$ injects into the finite set $A+B$, so $H$ is finite. [/guided] [/step] [step:Apply the aperiodic form of Kneser's theorem in the quotient] We use the previously established aperiodic form of Kneser's theorem: if $K$ is an abelian group and $C,D\subset K$ are finite nonempty subsets with $\operatorname{Stab}_K(C+D)=\{0_K\}$, then \begin{align*} |C+D|\ge |C|+|D|-1. \end{align*} Apply this result with $K=G/H$, $C=\bar A$, and $D=\bar B$. The hypotheses hold because $G/H$ is abelian, $\bar A$ and $\bar B$ are finite and nonempty, and the previous step proved $\operatorname{Stab}_{G/H}(\bar A+\bar B)=\{0_{G/H}\}$. Therefore \begin{align*} |\bar A+\bar B|\ge |\bar A|+|\bar B|-1. \end{align*} [guided] The quotient step was designed exactly so that the aperiodic estimate applies. The previously established aperiodic form of Kneser's theorem states that for an abelian group $K$ and finite nonempty subsets $C,D\subset K$, if the stabilizer of $C+D$ is trivial, then \begin{align*} |C+D|\ge |C|+|D|-1. \end{align*} Here we take \begin{align*} K&:=G/H, & C&:=\bar A, & D&:=\bar B. \end{align*} We verify the hypotheses one by one. Since $G$ is abelian and $H\le G$, the quotient $G/H$ is abelian. The sets $\bar A$ and $\bar B$ are finite and nonempty because they are images of the finite nonempty sets $A$ and $B$ under $\pi$. Finally, the previous step proved \begin{align*} \operatorname{Stab}_{G/H}(\bar A+\bar B)=\{0_{G/H}\}. \end{align*} Thus the aperiodic lower bound gives \begin{align*} |\bar A+\bar B|\ge |\bar A|+|\bar B|-1. \end{align*} This is the desired inequality at the quotient level; the remaining task is only to translate each quotient cardinality back into the original group. [/guided] [/step] [step:Convert quotient cardinalities into $H$-periodic cardinalities in $G$] For any finite subset $S\subset G$, the $H$-periodic enlargement $S+H$ is the disjoint union of the cosets $s+H$ with $s+H\in\pi(S)$. Since $H$ is finite by the previous step, each such coset has cardinality $|H|$, and hence \begin{align*} |S+H|=|\pi(S)|\,|H|. \end{align*} Applying this identity to $S=A$, $S=B$, and $S=A+B$ gives \begin{align*} |A+H|&=|\bar A|\,|H|,\\ |B+H|&=|\bar B|\,|H|,\\ |A+B|&=|\pi(A+B)|\,|H|=|\bar A+\bar B|\,|H|. \end{align*} The last equality uses $A+B+H=A+B$, which follows from $H=\operatorname{Stab}(A+B)$. [guided] We now relate sizes in the quotient to sizes in $G$. Let $S\subset G$ be finite. The set $S+H$ is a union of $H$-cosets. Its distinct cosets are exactly the cosets appearing in $\pi(S)$, because \begin{align*} \pi(S)=\{s+H:s\in S\}. \end{align*} Distinct cosets of a subgroup are disjoint, and every coset of $H$ has cardinality $|H|$. The previous step proved that $H$ is finite by injecting it into the finite set $A+B$, so this cardinality is finite. Therefore the disjoint union of the $|\pi(S)|$ cosets in $S+H$ gives \begin{align*} |S+H|=|\pi(S)|\,|H|. \end{align*} Taking $S=A$ gives \begin{align*} |A+H|=|\pi(A)|\,|H|=|\bar A|\,|H|. \end{align*} Taking $S=B$ gives \begin{align*} |B+H|=|\pi(B)|\,|H|=|\bar B|\,|H|. \end{align*} Finally, because $H=\operatorname{Stab}(A+B)$, every element of $H$ stabilizes $A+B$, so \begin{align*} A+B+H=A+B. \end{align*} Therefore \begin{align*} |A+B|=|A+B+H|=|\pi(A+B)|\,|H|=|\bar A+\bar B|\,|H|. \end{align*} [/guided] [/step] [step:Multiply the quotient inequality by $|H|$] Multiplying \begin{align*} |\bar A+\bar B|\ge |\bar A|+|\bar B|-1 \end{align*} by $|H|$ and using the cardinality identities from the previous step yields \begin{align*} |A+B| &=|\bar A+\bar B|\,|H|\\ &\ge (|\bar A|+|\bar B|-1)|H|\\ &=|A+H|+|B+H|-|H|. \end{align*} This is the asserted Kneser inequality. [guided] We start from the quotient-level inequality already proved: \begin{align*} |\bar A+\bar B|\ge |\bar A|+|\bar B|-1. \end{align*} Because the previous step proved that $H$ is finite, multiplying both sides by the nonnegative integer $|H|$ preserves the inequality: \begin{align*} |\bar A+\bar B|\,|H|\ge (|\bar A|+|\bar B|-1)|H|. \end{align*} Now we substitute the cardinality identities obtained from the coset decomposition: \begin{align*} |A+B|&=|\bar A+\bar B|\,|H|,\\ |A+H|&=|\bar A|\,|H|,\\ |B+H|&=|\bar B|\,|H|. \end{align*} This gives \begin{align*} |A+B| &=|\bar A+\bar B|\,|H|\\ &\ge (|\bar A|+|\bar B|-1)|H|\\ &=|A+H|+|B+H|-|H|. \end{align*} This is exactly the inequality stated in the theorem. [/guided] [/step]