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Suppose $\hat{\Gamma }\cong {\mathbb{Z}}_p$. This completion is an abelian group, so every finite quotient of $\Gamma$ is abelian. Consequently the quotient map to the abelianization $A={\Gamma }^{\mathrm{ab}}$ inducesThe group $A$ is a finitely generated abelian group. If its free rank is zero, $\hat{A}=A$ is finite. If its free rank is positive, $A$ and hence $\hat{A}$ have a nontrivial quotient $\mathbb{Z}/q\mathbb{Z}$ for every sufficiently chosen prime $q$. But ${\mathbb{Z}}_p$ has no nontrivial finite quotient of order coprime to $p$. Both cases are impossible, so $\hat{\Gamma }\ncong {\mathbb{Z}}_p$.

A group $G$ is abelian when $xy=yx$ for every $x,y\in G$. It is cyclic when some generator of a group $g$ satisfiesIf $x=g^r$ and $y=g^s$ belong to a cyclic group, thenso every cyclic group is abelian.

The Klein four-group $C_2\times C_2$ is abelian, but every nonidentity element has order two, so no element generates all four elements. Thus an abelian group need not be cyclic.

The condition on proper subgroups does not force $G$ to be abelian. The quaternion groupis nonabelian because $ij=k$ whereas $ji=-k$. Its proper subgroups are the trivial subgroup, $\{\pm 1\}$, and the three cyclic subgroupseach of order four. They are all cyclic, so

The vertical tangent space of the trivial principal $\mathbb{R}$-bundle is spanned by ${\partial }_z$. Sinceprojects isomorphically onto the tangent space of $S^1\times \mathbb{R}$, it is always complementary to the vertical direction. It is the horizontal distribution of a principal connection precisely when it is invariant under the principal translations $z\mapsto z+a$. The horizontal lifts of ${\partial }_{\theta }$ and ${\partial }_t$ are unique, so this invariance is equivalent toThe functions must also be smooth and periodic in $\theta$, as is already required for them to be functions on the cylinder.

Under these conditions the connection form isIt sends ${\partial }_z$ to $1$, is translation-invariant, and has kernel $D$, proving sufficiency as well. Since the structure group $\mathbb{R}$ is abelian, the bracket term vanishes and

An $n$-qubit stabilizer state is the unique simultaneous $+1$ eigenstate of an abelian group $G$ of Pauli operators having $2^n$ elements and not containing $-I$. The group $G$ is the state's stabilizer group; equivalently, it is generated by $n$ independent commuting Hermitian Pauli operators.

Irreducible degrees divide the order of a finite $2$-group and satisfyThe number of linear characters is $|G/G^{\prime }|$. For a nontrivial finite $2$-group this is at least $4$ unless the group is abelian, and every nonlinear degree is at least $2$. A degree $4$ constituent is therefore impossible in the nonabelian case. If there are $\ell$ linear and $m$ degree-two characters, thenThus the only degree collections arewith respectively $r=16,10,7$.

For $1^{16}$ take the cyclic group $G=C_{16}=⟨g⟩$. Every conjugacy class is a singleton, and its character table iswhere ${\zeta }_{16}=e^{2\pi i/16}$.

For $1^8,2^2$, take $G=D_8\times C_2$, where
$D_8=⟨r,s:r^4=s^2=1,srs=r^{-1}⟩$ and
$C_2=⟨z⟩$. On the five classesthe four linear characters of $D_8$, indexed by
$\epsilon ,\delta \in \{\pm 1\}$, and its degree-two character areThe ten classes of $G$ are $(C,1)$ and $(C,z)$. Its full table consists ofwhere $\chi$ runs over the above five rows and
$\eta \in \{1,-1\}$ runs over the two characters of $C_2$. This explicitly gives eight linear and two degree-two rows.

For $1^4,2^3$, take the dihedral groupOrder its seven classes asThe four linear rows areThe three degree-two rows, for $k=1,2,3$, areThe row norms and mutual inner products, weighted by the displayed class sizes, verify irreducibility and completeness.

Embed $E$ as a smooth plane cubic with $O_E$ an inflection point. A line through $P,Q$, using the tangent when $P=Q$, has a third intersection $S$ counted with multiplicity. The chord-and-tangent group law defines $P+Q=-S$, where $-S$ is the third point on the line through $S$ and $O_E$.

The divisor cut out by the first line isand the line through $S$ and $O_E$ gives $(S)+(-S)-2(O_E)$. Their quotient therefore showsUnder the bijection from part (a), the chord-and-tangent operation is exactly addition in the abelian group ${\mathrm{Pic}}^0(E)$. It is consequently associative and commutative, has identity $O_E$, and has the geometrically defined point $-P$ as inverse. Thus it makes $E$ an abelian group.

Let $E$ be a smooth plane cubic whose identity $O$ is an inflection point. A line through $P$ and $Q$, using the tangent when $P=Q$, has a third intersection $R$ counted with multiplicity. The chord-and-tangent group law defines $P+Q$ by drawing the line through $R$ and $O$ and taking its third intersection.

The clean verification of the group axioms uses divisors. The line at infinity meets a Weierstrass cubic in $3(O)$, so three collinear points $P,Q,R$ satisfyConsequently the mapsends the chord-and-tangent construction to addition of divisor classes. The principal divisor criterion on an elliptic curve shows that this map is bijective. Associativity and commutativity therefore follow from the abelian group law on ${\mathrm{Pic}}^0(E)$. The tangent convention handles repeated intersections, $O$ represents the zero class, and the third point on the line through $P$ and $O$ represents the inverse of $P$. Hence all group axioms hold.

The Plünnecke-Ruzsa inequality says that if $A,B$ are finite nonempty subsets of an abelian group andthen for all nonnegative integers $m,n$,

Choose a nonempty $X\subseteq A$ minimizingthen $K^{\prime }\le K$. We first prove the Petridis minimal-growth lemmafor every finite $C$, by induction on $|C|$. Remove $c\in C$, write $C^{\prime }=C\setminus \{c\}$, and letThe new points contributed to $X+C$ by $X+c$ are exactly $|X\setminus X^{\prime }|$. Moreover, $X^{\prime }+B+c\subseteq (X+B+c)\cap (X+B+C^{\prime })$, soThe induction hypothesis, the identity $|X+B|=K^{\prime }|X|$, and minimality, which gives $|X^{\prime }+B|\ge K^{\prime }|X^{\prime }|$, yieldIteration with $C=(r-1)B$ gives

Finally, the Ruzsa triangle inequality givesas required.

The Schur multiplier of a group $G$ isthe second group homology group with trivial integral coefficients. If $G\cong F/R$ is a free presentation, Hopf's formula states that

Write $I_F=\ker (\mathbb{Z}F\to \mathbb{Z})$ for the augmentation ideal. The presentation relation sequence iswhere $r[R,R]\mapsto 1\otimes (r-1)$. If $F$ is free on a set $S$, then $I_F$ is free as a left $\mathbb{Z}F$-module on the elements $s-1$, so the two modules immediately preceding $\mathbb{Z}$ are free $\mathbb{Z}G$-modules. Resolving the relation module $R/[R,R]$ by free modules and splicing produces a free resolution of $\mathbb{Z}$.

Apply the right-exact functor $\mathbb{Z}{\otimes }_{\mathbb{Z}G}-$ to this partial resolution. Its degree-two homology is the kernel ofThe coinvariant module on the left is $R/[F,R]$. On the right, the map $f-1\mapsto f[F,F]$ identifies the coinvariants with the abelianization $F/[F,F]$. The displayed map is induced by the inclusion $R\hookrightarrow F$, so its kernel isThis proves Hopf's formula.

For an abelian group $A$, the Schur multiplier of an abelian group is $M(A)\cong {\bigwedge }^{2}A$. One way to see the direct-sum rule is the degree-two Künneth theorem:A cyclic group has zero second integral group homology, whileConsequently

The integers are closed under addition, addition is associative, zero is an identity, and $-n$ is the additive inverse of $n$. Thus $(\mathbb{Z},+)$ is an abelian group.

The statement that limits of shape $\mathcal{I}$ commute with colimits of shape $\mathcal{J}$ means that for every $D:\mathcal{I}\times \mathcal{J}\to \mathcal{C}$, the canonical commutation of limits and colimits mapis an isomorphism whenever the iterated limits and colimits exist.

A filtered category is a nonempty category in which every finite diagram has a cocone. Equivalently, any two objects map to a common object and any parallel pair becomes equal after postcomposition. A weakly filtered category requires cocones only for finite connected diagrams; equivalently, each connected component is filtered.

Write a weakly filtered category as the disjoint union ${\coprod }_s{\mathcal{J}}_s$ of its filtered connected components. Its colimit is the coproduct of the filtered colimits over the ${\mathcal{J}}_s$. In sets, a connected finite limit commutes with coproducts: connectedness forces all coordinates of a compatible tuple to lie in the same coproduct summand. By the assumed theorem, the filtered colimit over each ${\mathcal{J}}_s$ commutes with every finite limit. Applying these two facts successively proves that weakly filtered colimits commute with connected finite limits in $\mathbf{Set}$.

The forgetful functor from abelian group to sets creates finite limits and filtered colimits and reflects isomorphisms. The comparison map for a filtered colimit and a finite limit therefore becomes the corresponding isomorphism of sets, so filtered colimits commute with finite limits in $\mathbf{AbGp}$.

The dual claim fails because inverse limit need not preserve epimorphisms. Take the inverse systemswith identity bonding maps on $A_n$, reduction maps on $B_n$, and levelwise epimorphisms $q_n:A_n\to B_n$. Thenand the induced map $\mathbb{Z}\to {\mathbb{Z}}_2$ is not surjective. Since an epimorphism in abelian groups is a finite-colimit cokernel, cofiltered limits do not commute with finite colimits in $\mathbf{AbGp}$.

The Entropic Balog-Szemerédi-Gowers theorem states that for finitely supported random variables $A,B$ in an abelian group,where the left-hand side is the Simultaneous conditional entropic Ruzsa distance.

Put $S=A+B$ and take two copies $(A_1,B_1)$ and $(A_2,B_2)$ that are conditionally independent given $S$. Thus both sums equal $S$, and, given $S$, $A_1$ and $B_2$ are independent with the required conditional marginals. By conditioning reduces entropy,The two conditional entropies are equal, since either $A$ or $B$ together with $S$ determines the other, and the chain rule for information entropy gives

Set $D=A_1-B_2$. By entropy submodularity,The first term is $H(A_1,B_2)\le H(A)+H(B)$ by subadditivity of information entropy. Since $A_1+B_1=A_2+B_2$, we also have $D=A_2-B_1$, so the second term is at most $H(A)+H(B)$. Finally $(D,A_1,B_1)$ determines all four copied variables, and conditional independence givesAs $H(A,B)=H(A)+H(B)-I(A;B)$, these estimates implySubtracting the common conditional-entropy term proves the stated Entropic Balog-Szemerédi-Gowers theorem.

The Dold–Kan correspondence says thatis an equivalence from simplicial abelian groups to nonnegatively graded chain complexes of abelian groups. The restriction of the right adjoint $K$ to ${\mathbf{Ch}}_{\ge 0}(\mathbf{Ab})$ is a quasi-inverse: both the unit $C_{\ast }\to N_{\ast }K(C_{\ast })$ and counit $K(N_{\ast }A)\to A$ are natural isomorphisms.

For nonempty finite subsets $A,B$ of an abelian group, $|A+B|\le K|A|$ implies that some nonempty $X\subseteq A$ satisfies this bound simultaneously for every integer $m\ge 0$, with $0B=\{0\}$. The Petridis minimal-growth lemma proves this by choosing $X$ to minimize $|X+B|/|X|$. The Ruzsa triangle inequality then gives the Plünnecke-Ruzsa inequality $|kB-\ell B|\le K^{k+\ell }|A|$.

For a finite nonempty subset of an abelian group with doubling constant at most $K$, the Plünnecke-Ruzsa inequality gives $|3A-2A|\le K^5|A|$. Applying the Ruzsa covering lemma to $2(A-A)$ using $A$ gives $2T\subseteq D+T$ for $T=A-A$ and $|D|=q\le K^5$. Thus $\ell T\subseteq (\ell -1)D+T$, and counting multiplicities in the finite set $D$ proves the displayed polynomial bound. For fixed $K>1$, it is eventually at most $K^{ϵ\ell }|A|$ for every $ϵ>0$.

A category is preadditive if every hom-set is an abelian group and composition is additive in each variable. Zero morphisms and differences of parallel morphisms are therefore defined. A zero object or finite biproducts need not exist; these are extra requirements in an additive category.

If $S^n\to N$ has prime degree $p$ and $N$ is a closed connected oriented $n$-manifold, every integral homology group of $N$ in degrees strictly between zero and $n$ is a finite $p$-primary abelian group. Consequently one power of $p$ annihilates all these groups. The proof combines Poincare duality over each field ${\mathbb{F}}_q$, $q\ne p$, with the universal coefficient theorem for homology and finite generation.

If $K/{\mathbb{Q}}_p$ is finite of degree $d$, then as a topological abelian groupwhere ${\mu }_{p^a}(K)$ is the finite group of all roots of unity in $K$ whose orders are prime powers dividing a power of $p$.

A profinite abelian group is an inverse limit of finite abelian groups. Its topology is compact and totally disconnected. An abelian pro-p group acquires scalar multiplication by p-adic integers by taking limits of integer multiples.