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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.

Suppose $\sigma \ne e$. Choose a symbol $i$ moved by $\sigma$ and write $j=\sigma (i)\ne i$. Since $n\ge 3$, choose a third symbol $k$ distinct from $i,j$. Let $\tau =(jk)$, a transposition that fixes $i$. ThenThus ${\tau }^{-1}\sigma \tau \ne \sigma$, proving that the conjugacy class of any nonidentity permutation has more than one element. By part (a)(iii), no such permutation lies in the center of a group. The identity always does, soThe restriction on $n$ is necessary: $S_2$ is an abelian group and equals its own center.

The hypothesis implies $x^{-1}=x$ for every $x\in G$, including $e$. For $a,b\in G$,Thus $G$ is an abelian group. This is the group of exponent two is abelian argument.

The converse is false: the cyclic group $C_3$ is an abelian group, but its two nonidentity elements have order of a group element three.

If $G=⟨g⟩$, any two elements are $g^a,g^b$, andThus every cyclic group is an abelian group.

For the converse, the Klein four-group $C_2\times C_2$ is an abelian group, with componentwise multiplication. Each nonidentity element has order of a group element two, so none generates its four elements. It is therefore not a cyclic group.

Fix $x\in G$. The integer powers of a group element satisfyFor nonnegative integers these rules follow by associativity and counting factors; negative integers are handled by $x^{-r}=(x^{-1}{)}^r$, and opposite signs by cancellation of adjacent inverse elements. All factors are powers of the same element, so no abelian group assumption is involved.

It follows that $z\mapsto \rho (z,x)=x^z$ is a group homomorphism from the additive group of integers to $G$, andThe homomorphism uses ordinary addition on $\mathbb{Z}$; composition of the power functions corresponds to ordinary multiplication of exponents.

Suppose the quotient group $G/A$ is a cyclic group, generated by $gA$. For any $x,y\in G$, there are integers $r,s$ and $a,b\in A$ such thatSince $a,b$ lie in the center of a group, they commute with every element, in particular with each other and with the integer powers of a group element $g$. ThereforeAs $x,y$ were arbitrary, $G$ is an abelian group. This is the cyclic quotient by the center.

A vector space over a field $\mathbb{F}$ is a set $V$ with vector addition $V\times V\to V$ and scalar multiplication $\mathbb{F}\times V\to V$ satisfying the following vector space axioms. Under vector addition, $V$ is an abelian group: addition is associative and commutative, there is a zero vector $0_V$, and each $v$ has an additive inverse $-v$. For all $a,b\in \mathbb{F}$ and $u,v\in V$,The operations are defined for every indicated pair and take their values in $V$.

If $G$ is an abelian group, its elements commute, so for all $x,y\in G$,Thus $\boxed{T\text{ is a group homomorphism}}$. Also $T(e)=e^3=e$, as required.

If $T$ is a surjective group homomorphism, then $G=T(G)$, which is an abelian group by the preceding part.

If $T$ is an injective group homomorphism, then for all $a,b\in G$,because $T(a),T(b)$ lie in the abelian group $T(G)$. Injectivity gives $ab=ba$. Hence $\boxed{G\text{ is abelian in either case}}$.

Yes: use the non-abelian group $H$ from part (a). Every element is $I_3+N$ for a strictly upper triangular matrix $N$ with $N^3=0$. In the finite field ${\mathbb{Z}}_3$, the binomial theorem for commuting matrices givesThus $\boxed{T(h)=I_3\text{ for all }h\in H}$. This constant identity-valued map is a group homomorphism, since $T(xy)=I_3=T(x)T(y)$. Part (a)(iv) already proves that $H$ is not an abelian group. Here $T$ is neither injective nor surjective, consistently with the preceding part.

Conjugacy classes. For $a\in G$, its conjugacy class isEvery element belongs to its own conjugacy class, by taking $g=e$. To prove uniqueness, suppose $x=ga{g}^{-1}=hb{h}^{-1}$ belongs to both $\mathrm{Cl}(a)$ and $\mathrm{Cl}(b)$. ThenFor every $t\in G$, this makes $tb{t}^{-1}$ a conjugate of $a$ as well. Hence $\mathrm{Cl}(b)\subseteq \mathrm{Cl}(a)$, and reversing the argument gives the other inclusion. Two conjugacy classes that meet are therefore equal, proving that the classes partition $G$.

Abelian case. If $G$ is an abelian group, then $ga{g}^{-1}=ag{g}^{-1}=a$ for every $g$, so $\boxed{\mathrm{Cl}(a)=\{a\}}$ for each $a$.

Centre. An element has a singleton conjugacy class precisely when $gz{g}^{-1}=z$ for every $g$, equivalently $gz=zg$ for every $g$. Thus the definition in the paper agrees with the usual centre of a group:The identity lies in $Z(G)$. If $z,w\in Z(G)$, then for any $g$, $gzw=zgw=zwg$, so $zw\in Z(G)$. Also $gz=zg$ implies $z^{-1}g=g{z}^{-1}$, so $z^{-1}\in Z(G)$. This proves it is a subgroup. Any two of its elements commute because either one commutes with every element of $G$. Consequently $\boxed{Z(G)\text{ is an abelian subgroup of }G}$.

Square symmetry. Use the paper's notation $D_4$ for the eight-element dihedral group of a square. Let $r$ be a quarter-turn and $s$ a reflection, so $r^4=s^2=e$ and $srs=r^{-1}$. A rotation $r^j$ commutes with $s$ exactly when $r^j=r^{-j}$, or $2j=0$ modulo four. This selects $e$ and $r^2$. No reflection $r^{j}s$ commutes with $r$, becauseand $r^{j-1}\ne r^{j+1}$. Therefore

A group is a nonempty set with a closed binary operation that is associative, has an identity element, and gives every element an inverse element. A subgroup is a subset that is itself a group under the inherited operation. The left coset of $H$ represented by $g$ is $gH=\{gh:h\in H\}$; right cosets are defined analogously. Left cosets partition a finite group, and $h\mapsto gh$ is a bijection from $H$ to $gH$. Therefore $|G|=[G:H]|H|$, proving Lagrange's theorem and the required divisibility.

For the given matrices, put $a=A$ and $c=AB=iI$. Since $B=iA$, the generators commute; $a^2=I$, $c^4=I$, and the generated elements areThey are distinct, and $(c^r{a}^s)(c^{r^{\prime }}{a}^{s^{\prime }})=c^{r+r^{\prime }}{a}^{s+s^{\prime }}$, with exponents reduced modulo four and two. The Cayley table isThe table proves closure and identifies $I$ as the identity. The inverse element of $c^r{a}^s$ is $c^{-r}{a}^s$, and associativity follows from matrix multiplication. Thus this is the direct product of cyclic groups of orders four and two, an abelian group of order eight.

Its subgroups are exactlyTo see the list is complete, Lagrange's theorem leaves orders one, two, four and eight. There are exactly three elements of order two, giving the three order-two subgroups. An order-four subgroup containing an element of order four must be one of the two cyclic groups displayed; without such an element it must be the four-element set of elements whose squares are $I$. This is the enumeration in subgroups of C4 times C2.

For example choose $H=\{I,c,c^2,c^3\}$. A complete set of cosets is $\boxed{H,aH=\{a,ca,c^{2}a,c^{3}a\}}$. Since $G$ is Abelian, the left and right cosets coincide.

Observe that $A=-I$, $B^2=I$ and $C=-B$. Thus $A^2=B^2=C^2=I$, while $AB=BA=C$, $AC=CA=B$ and $BC=CB=A$. The multiplication table, with the left factor labeling the row, isIt proves closure of the set. Associativity is inherited from matrix multiplication, $I$ is the identity element, and every element is its own inverse element. Therefore this set is a group. The table is symmetric, so it is an Abelian group; its three nonidentity elements have order two, identifying it as the Klein four-group $C_2\times C_2$.

Let $T$ be the set of torsion elements of the given Abelian group. It contains the identity element, which has order one. If $g,h\in T$ have orders $m,n$, set $\ell =\mathrm{lcm}(m,n)$. Commutativity permits powers of the product to split:Hence $g{h}^{-1}$ has finite order, dividing $\ell$, and lies in $T$. The nonempty set $T$ therefore satisfies the subgroup test and is a subgroup:This is the torsion subgroup. The proof uses commutativity exactly when rearranging the factors in $(g{h}^{-1}{)}^{\ell }$; one cannot use that step in an arbitrary noncommutative group. The subgroup need not itself be a finite set.

The hypothesis implies $g^2=e$ for every $g\in G$, including the identity, so $g^{-1}=g$. For arbitrary $a,b\in G$, their product also equals its own inverse. The inverse-of-a-product rule therefore givesThus $G$ is an abelian group. Finiteness is not needed for this implication; it holds for every group in which each element squares to the identity.

By Lagrange's theorem, the possible values of the order of a group element areThese are possibilities across groups of order eight; they need not all occur in one group.

If every element has order at most two, every element equals its own inverse. Thus for any $x,y$,so $G$ is an abelian group. Choose $a\ne e$, and then $b\notin ⟨a⟩$. The four elements $e,a,b,ab$ are distinct and form a subgroup $H$: commutativity and $a^2=b^2=e$ prove closure and inverses. Choose $c\notin H$. The elements of $H$ and $Hc$ are disjoint, since $hc\in H$ would imply $c\in H$. These two four-element sets therefore exhaust $G$.

Every element consequently has a unique expression $a^i{b}^j{c}^k$, with $i,j,k\in \{0,1\}$. The map from triples to these products is bijective and respects multiplication by addition of exponents modulo two. Hencea direct product of groups and an elementary abelian group.

The sum of two odd integers plus one is odd, so the operation is closed. Both associations of a triple give $a+b+c+2$, proving associativity. Its identity element is $-1$, and the inverse element of $a$ is $-a-2$, also odd, since $a+(-a-2)+1=-1$. Thus it is an abelian group.

More explicitly, shifted addition on odd integers is ordinary integer addition in different coordinates:This bijection is a group isomorphism to $(\mathbb{Z},+)$.

The requested two-dimensional irreducible representation does not exist for the Vierergruppe. The Klein four-group $V_4$ is an abelian group in which every nonidentity element has order two. Its complex irreducible representations are all one-dimensional. The same holds over the real numbers here: the commuting representing involutions are simultaneously diagonalizable with real eigenvalues $\pm 1$, so every two-dimensional representation has an invariant line. Each genuine irreducible character satisfies $\chi (g^2)=\chi (1)=1$, and hence has Frobenius-Schur indicator $\nu =1$.

The printed notation $C_4$ suggests a different possible interpretation, which must not be conflated with the Klein four-group. For the cyclic group of order four, a real two-dimensional irreducible representation is generated byIts character on $1,J,-I,-J$ is $(2,0,-2,0)$, giving $\nu =(2-2+2-2)/4=0$. Over the complex numbers this representation splits into two conjugate one-dimensional characters, so it is not a two-dimensional complex irreducible representation. These alternatives explain what can be computed without silently changing the printed group or the ground field.