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Use the coordinate conventionIn this convention the Heisenberg group over a prime field has the productThe subset $H=\{A(t,b,t):t,b\in {\mathbb{F}}_p\}$ is an Abelian subgroup of order $p^2$: the middle coordinate in a product is $b+b^{\prime }+t{t}^{\prime }$, which is symmetric in the factors. Also $A(t,b,t{)}^{-1}=A(-t,t^2-b,-t)$ lies in $H$. The map $A(a,b,c)\mapsto a-c$ is a surjective group homomorphism to the additive group of ${\mathbb{F}}_p$, with kernel of a group homomorphism $H$. Hence $H$ is a normal subgroup of subgroup index $p$, and its left cosets are exactly the level sets of $a-c$.

For odd $p$, an explicit group isomorphism from $H$ to the additive group ${\mathbb{F}}_p^2$ isThe term $t{t}^{\prime }$ in matrix multiplication cancels the cross term in $(t+t^{\prime }{)}^2/2$. Thus $H$ is an elementary abelian group, isomorphic to $C_p\times C_p$.

In the Heisenberg group over a commutative ring with $R=\mathbb{Z}/4\mathbb{Z}$, the subset $H=\{A(t,y,t):t,y\in R\}$ is an Abelian subgroup of group order 16. Its product has central coordinate $y+y^{\prime }+t{t}^{\prime }$, symmetric in the two factors. Its inverse element is $A(-t,t^2-y,-t)$. The map $A(x,y,z)\mapsto x-z$ is a surjective group homomorphism to the additive group of $R$, with kernel of a group homomorphism $H$. Thus $H$ is a normal subgroup and has four cosets in the 64-element group.

Unlike the odd-prime case, division by two is unavailable. In fact take $u=A(1,0,1)$ and $v=A(2,0,2)$. Then $u^n=A(n,(\genfrac{}{}{0ex}{}{n}{2}),n)$, where the binomial coefficient is computed as an integer before reduction modulo four. Hence $u^4=A(0,2,0)$ and $u$ has element order eight. The element $v$ has element order two and is not a power of $u$: the powers with outer coordinates two have central coordinates one or three. Their generated subgroups intersect trivially, so the 16-element Abelian group is the direct product of groups $C_8\times C_2$.

A finite abelian subgroup of the Möbius group is cyclic or isomorphic to $C_2\times C_2$. If it contains an element of order greater than two, conjugate that element's two fixed points to $0,\infty$. Every commuting map must preserve these points individually: an interchange would conjugate the element to its inverse. All maps are then scalings, forming a cyclic finite subgroup of ${\mathbb{C}}^{\times }$. If every nonidentity element has order two, choose one with fixed points $0,\infty$; the subgroup acts on this pair with image of size at most two and kernel contained in $\{z,-z\}$. Its size is therefore at most four, giving the asserted possibilities.

Use the product from part (ii). Move each selected element of each small set $X_i$ to the left, conjugating every approximate-group factor that it crosses. For each tuple $\mathbf{x}\in X_1\times \cdots \times X_m$, the corresponding part of the product is therefore contained inwhere every $D_{j,\mathbf{x}}$ is a conjugate subset of some $C_j$. Conjugation preserves cardinality, the approximation parameter, and the property that the generated subgroup is abelian. The conjugating elements belong to $A^{O_s(1)}$, so $D_{j,\mathbf{x}}\subseteq A^{O_s(1)}$ after enlarging the implicit constant.

The number of tuples is at mostThese translated products cover $A$, so one has size at least the reciprocal fraction of $|A|$. For that tuple, put $D_j=D_{j,\mathbf{x}}$. Thenwhere $q=n\le O_s({\log }^{O_s(1)}(2K))$, and each $D_j$ is a $K^{O_s(1)}$-approximate group generating an abelian subgroup.

Cosets and Lagrange. A left coset is $gH=\{gh:h\in H\}$. The left cosets of a subgroup partition a group, each with $|H|$ elements. Lagrange's theorem states $\boxed{|G|=[G:H]|H|}$ for a finite group, so every subgroup order divides the group order.

The 64-element group. Work in the commutative ring $R=\mathbb{Z}/4\mathbb{Z}$, which has zero divisors. Denote the displayed lower triangular matrix by $A(a,b,c)$, where $a,b,c$ are its three entries below the diagonal. Direct matrix multiplication givesAll arithmetic is modulo four, so the set is closed. Associativity comes from matrix multiplication, the identity element is $A(0,0,0)$, andThe formula works in either multiplication order and requires no division. Hence these matrices form a group of order $\boxed{|G|=4^3=64}$. This is the lower triangular version of the Heisenberg group over a commutative ring.

The abelian subgroup. If $a=c$, write $h(a,b)=A(a,b,a)$. ThenThe identity is $h(0,0)$ and the inverse is $h(-a,a^2-b)$, proving an abelian subgroup. Its two independent entries give $\boxed{|H|=16}$ and $\boxed{[G:H]=4}$, in accordance with Lagrange's theorem. This is the abelian plane subgroup of the Heisenberg group modulo four.

More explicitly, $A(a,b,c)\mapsto a-c$ is a surjective group homomorphism to the additive group of $R$, with kernel $H$. The four left cosets are exactly the four sets with fixed $a-c=t$, $t=0,1,2,3$; representatives are $A(t,0,0)$. Thus $H$ is also a normal subgroup.

Matrices squaring to the identity. The product formula givesThe first and third coordinates vanish only when $a,c\in \{0,2\}$. Then $ca=0$ modulo four, so the middle coordinate vanishes exactly when $b\in \{0,2\}$. Therefore the complete set isIt is a subgroup: multiplying two such elements keeps all three entries even, and their cross term $cd$ vanishes modulo four, so the product simply adds the three even coordinates. The identity is included and each element is its own inverse. In fact $K$ is isomorphic to the direct product of groups $C_2\times C_2\times C_2$. The seven nonidentity elements are involutions.

Finally, the last subset in the PDF is defined by $a\ne 0\ne b$, with no restriction on $c$. It has $3\cdot 3\cdot 4=36$ elements, which cannot be a subgroup because $36$ does not divide $64$. It also excludes the identity element. Thus

Cosets and counting. A left coset of a subgroup $H$ in a group $G$ is $gH=\{gh:h\in H\}$. The left cosets partition $G$ and each has $|H|$ elements. Lagrange's theorem states that for a finite group, $\boxed{|G|=[G:H]|H|}$; in particular, the subgroup order divides the group order.

The matrix group. Work throughout in the finite field ${\mathbb{F}}_5$, so every nonzero entry has a multiplicative inverse. Denote a matrix by $A(a,b,c)$, with $a,c\ne 0$. Ordinary matrix multiplication, interpreted using modular arithmetic, givesIts diagonal entries remain nonzero, proving closure. Associativity is inherited from matrix multiplication, $A(1,0,1)$ is the identity element, andThis lies in the same set and multiplication checks the inverse in either order. Thus these matrices form a group. There are four choices each for $a$ and $c$ and five for $b$, giving $\boxed{|G|=4\cdot 5\cdot 4=80}$.

The subgroup. Write $h(a,c)=A(a,a-c,c)$. ThenIt contains the identity $h(1,1)$, is closed, and has inverse $h(a^{-1},c^{-1})$. Interchanging the two factors leaves the product unchanged because multiplication in ${\mathbb{F}}_5$ is commutative. Therefore $H$ is an abelian subgroup, withEquivalently $H$ is isomorphic to ${\mathbb{F}}_5^{\times }\times {\mathbb{F}}_5^{\times }$ by $h(a,c)\mapsto (a,c)$, which also explains its abelian property. Its order divides $80$, verifying Lagrange's theorem.

There are five distinct left cosets. Representatives can be taken as $U_t=A(1,t,1)$, $t\in {\mathbb{F}}_5$: indeedEvery matrix lies in exactly the coset specified by this unique $t$.

Elements whose square is the identity. Squaring givesThus $a,c\in \{1,-1\}=\{1,4\}$ and $b(a+c)=0$. For equal signs, $a+c\ne 0$ in ${\mathbb{F}}_5$, so $b=0$. For opposite signs, $a+c=0$ and all five $b$ values work. The complete set is thereforeIt contains $2+5+5=12$ elements, including the identity; the other eleven are involutions. Since $12$ does not divide $80$, Lagrange's theorem proves that this set is not a subgroup.

Two elements are conjugate group elements if $h=xg{x}^{-1}$ for some $x\in G$. The conjugacy class of $g$ is $\{xg{x}^{-1}:x\in G\}$. Conjugacy is an equivalence relation: it is reflexive using $x=e$, symmetric by replacing $x$ with $x^{-1}$, and transitive becauseThus the classes partition the group: every element belongs to its own class, and any two classes that intersect are equal. This proves that every element belongs to exactly one class.

If $G$ is Abelian, then $xg{x}^{-1}=g$, so every class is a singleton. In general the elements whose classes are singletons are precisely those commuting with every group element. Thus the center of a group isIt contains $e$. If $a,b\in Z(G)$, then $a{b}^{-1}$ commutes with every $g$, since $b^{-1}$ does and $a$ does. The subgroup test therefore proves that $Z(G)$ is a subgroup. Any two of its elements commute with each other, so it is an abelian subgroup.

For the square centred at the origin with vertices $(\pm 1,\pm 1)$, putThe dihedral group $D_4$ consists of $R^j$ and $R^{j}S$, $j=0,1,2,3$. Its eight actions and their matrices areThese exhaust the symmetries: the image of one vertex has four choices and the image of an adjacent vertex has two choices, determining the entire square. For an explicit pair of nonconjugate elements take $R^2$ and $S$. Both have order two, but their determinants are $+1$ and $-1$. A similarity transformation preserves the determinant, so $R^2$ and $S$ are not conjugate.

To find the centre, use $SR=R^{-1}S$. A rotation $R^j$ commutes with $S$ exactly when $R^j=R^{-j}$, so $j$ is even. No reflection $R^{j}S$ commutes with $R$, since $(R^{j}S)R=R^{j-1}S$ while $R(R^{j}S)=R^{j+1}S$, and $R^2\ne I$. Therefore

The identity commutes with every group element, so it lies in the center of a group. If $z,w$ are central and $g\in G$, then $gw=wg$ implies $g{w}^{-1}=w^{-1}g$. ConsequentlyThus $z{w}^{-1}$ is central. The subgroup test provesEquivalently, the centre is nonempty, closed under products, and closed under inverses. This proof does not require finiteness. In fact the centre is an Abelian subgroup, and it is normal because $gz{g}^{-1}=z$ for each central element.

Perform the product and inverse formulas from part (b) in the commutative ring $\mathbb{Z}/4\mathbb{Z}$. They involve only addition, multiplication and negatives, so closure and inverses remain valid; no division by a nonunit is needed. Associativity again comes from matrix multiplication, and $A(0,0,0)$ remains the identity. The Heisenberg group over a commutative ring therefore hasThese are residues modulo four, not arithmetic in the finite field of four elements.

For the additional subgroup task, write an element of $H$ as $A(t,y,t)$. Its product and inverse areBoth remain in $H$, and the product is unchanged on exchanging the factors because $ts=st$. The identity is included, so $H$ is an Abelian subgroup. There are four independent choices each for $t,y$, givingThus there are four distinct left cosets. Explicitly the surjective group homomorphism $A(x,y,z)\mapsto x-z(\mathrm{mod}4)$ has kernel $H$; the four cosets are the level sets of that difference, with representatives $A(d,0,0)$, $d=0,1,2,3$. This is the Abelian plane subgroup of the Heisenberg group modulo four.

A member of $H$ is $A(t,b,t)$. The multiplication law from part (a) becomesIt stays in $H$ and is unchanged when the two factors are exchanged, since $ts=st$. The identity lies in $H$, and $A(t,b,t{)}^{-1}=A(-t,t^2-b,-t)$ also lies in $H$. Thus $H$ is an Abelian subgroup, namely the Abelian plane subgroup of a finite Heisenberg group in this coordinate convention.

There are five choices each for $t,b$, soThis verifies Lagrange's theorem: the subgroup order divides the group order, and the number of distinct left cosets is five.

Explicitly, the map $\phi (A(a,b,c))=a-c$ is a surjective group homomorphism to the additive group of ${\mathbb{F}}_5$, by the product formula. Its kernel of a group homomorphism is $H$. Thus $H$ is even a normal subgroup; the five left cosets are the level sets $a-c=d$ for $d=0,1,2,3,4$, with representatives $A(d,0,0)$. Each contains 25 elements, providing a direct count as well.

The center of a group contains $e$. If $z,w$ lie in the center, then for every $g\in G$,It is therefore closed under the group operation and inverse elements, and inherits associativity, proving it is a subgroup. Each of its elements commutes with every element of $G$, in particular with every other central element. HenceThus the center is an abelian subgroup.

A stabilizer code is the common $+1$ eigenspace of a stabilizer group $S$, an abelian subgroup of the Pauli group excluding $-I$. Choose $r$ independent stabilizer generators $g_1,\dots ,g_r$ on $n$ physical qubits. They are commuting Hermitian operators, each squaring to $I$. The stabilizer subspace and its orthogonal projector areThe stabilizer-projector formula gives $\dim \mathcal{C}=2^{n-r}$, so $k=n-r$ logical qubits remain. Independent here means that no nonempty product of the chosen stabilizer generators is $I$. Excluding $-I$ is necessary: a vector cannot simultaneously be fixed by $I$ and $-I$ unless it is zero.

For the three-qubit bit-flip repetition code, take $g_1=Z_1{Z}_2$ and $g_2=Z_2{Z}_3$, with $Z_j$ the Pauli Z gate on site $j$ and identities elsewhere. The stabilizer group is $S=\{I,Z_1{Z}_2,Z_2{Z}_3,Z_1{Z}_3\}$. Both stabilizer generators have eigenvalue $+1$ exactly on the span of $|000⟩$ and $|111⟩$, giving $2^{3-2}=2$ dimensions for one logical qubit. A projective measurement of these checks distinguishes four orthogonal two-dimensional sectors rather than eight individual computational basis states.

A Pauli error that anticommutes with a stabilizer generator changes that check's sign, producing an error syndrome. A Pauli operator commuting with every check can still act on the logical qubits; the centralizer of a stabilizer group describes these possibilities. Thus the number of checks determines the encoded dimension, while the quantum code distance determines protection against arbitrary local errors. The five-qubit error correcting code, for instance, has four stabilizer generators and quantum code distance three; the three-qubit bit-flip repetition code has two checks but quantum code distance one because a single phase flip is unflagged. ^[ref].