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The Noncommutative Ruzsa triangle inequality says that nonempty finite subsets $B,C,D$ of a group satisfyThe Ruzsa covering lemma says that if $|CD|\le K|D|$, then some $X\subseteq C$ with $|X|\le K$ satisfies

Now let $A$ be a symmetric subset of a group, so $A^{-1}=A$ and $1\in A$. Apply the triangle inequality with the middle set $A$ to obtain, for $m\ge 4$,The hypothesis therefore gives $|A^m|\le K|A^{m-1}|$. Starting from $|A^3|\le K|A|$ proves

In particular $|A^5|\le K^3|A|$. Apply the covering lemma to $C=A^4$ and $D=A$. There is $X\subseteq A^4$ with $|X|\le K^3$ such thatThe set $A^2$ is symmetric and contains the identity element, so this inclusion is exactly the covering condition showing that

Small doubling alone is insufficient in a noncommutative group. Let $H$ be a finite group, let $G=H\ast ⟨x⟩$ be its free product with an infinite cyclic group, and putThen $A$ is symmetric and $|A^2|\le 5|H|+4=O(|A|)$, while $A^3$ contains the double coset $HxH$. Distinct pairs $(h_1,h_2)\in H^2$ give distinct reduced words $h_{1}x{h}_2$, so $|HxH|=|H{|}^2$. Letting $|H|\to \infty$ proves the small doubling does not control tripling in a noncommutative group phenomenon.

A Lie group is a group that is also a smooth manifold, with differentiable multiplication and inversion. A Lie algebra is a vector space with a bilinear alternating Lie bracket satisfying the Jacobi identity.

For a Matrix Lie group $G$, put $\mathfrak{g}=T_{I}G$. If $X,Y\in \mathfrak{g}$, the matrix commutator $[X,Y]=XY-YX$ again lies in $\mathfrak{g}$: the group commutator $e^{tX}{e}^{sY}{e}^{-tX}{e}^{-sY}$ lies in $G$, and the coefficient of $ts$ in its matrix logarithm is $[X,Y]$. Bilinearity and antisymmetry are immediate, while associativity of matrix multiplication gives the Jacobi identity. Thus $T_{I}G$, with the commutator bracket, is the Lie algebra of a matrix Lie group $\mathcal{L}(G)$.

The special linear group $SL(2,\mathbb{R})$ is the inverse image of the regular value $1$ under the smooth determinant map, and multiplication and inversion are smooth. Differentiating $\det (I+tA)=1+t\mathrm{tr}A+O(t^2)$ shows thatThe Cayley-Hamilton theorem applied to a trace-zero two-by-two matrix gives

The exponential map of a matrix Lie group is the matrix exponentialSince $\det (e^A)=e^{\mathrm{tr}A}=1$, its image lies in $SL(2,\mathbb{R})$. Put $d=\det A$. The identity $A^2=-dI$ sums the series explicitly. If $d>0$, with $r=\sqrt{d}$,If $d=0$, the trace is $2$, while if $d<0$, with $r=\sqrt{-d}$,HenceBut $\mathrm{diag}(-2,-1/2)\in SL(2,\mathbb{R})$ has trace $-5/2$. It is therefore outside the image, so the exponential map is not surjective.

The hidden subgroup problem is to recover a subgroup $K\le G$, usually by giving generators of a group, from an oracle machine providing $f:G\to Y$ with the promiseThus $f$ is constant on each left coset and distinguishes different left cosets. The input includes an effective representation of elements of the group and access to the group operations; the desired efficiency is measured against the number of bits describing an element, rather than necessarily against $|G|$.

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 set $S(\mathbb{N})$ consists of all bijections $\mathbb{N}\to \mathbb{N}$. The composition of two bijections is a bijection, composition is associative, the identity map is an identity, and every bijection has an inverse, so $S(\mathbb{N})$ is a group.

An element of $S_{\mathrm{fin}}(\mathbb{N})$ has finite support. The support of a composition is contained in the union of the two supports, and a permutation and its inverse have the same support. The identity has empty support. Hence $S_{\mathrm{fin}}(\mathbb{N})$ is a subgroup.

Let $\sigma$ be a cycle and choose $n$ that it moves. Because $\sigma$ has finite support, the sequencemust repeat. Since $\sigma$ is invertible, its first repetition returns to $n$; let the least positive return time be $l$. The cycle condition says that every moved point occurs in this orbit, so ${\sigma }^l$ fixes every point. No smaller positive power fixes $n$. Thereforewhich is finite.

For $\tau \in S_{\mathrm{fin}}(\mathbb{N})$, partition its finite support into the orbits of the cyclic group $⟨\tau ⟩$. On each orbit, let ${\sigma }_i$ agree with $\tau$ and fix every point outside that orbit. Then each ${\sigma }_i$ is a permutation cycle, their supports are pairwise disjoint, andDisjoint cycles commute because at every point at most one of them acts nontrivially.

Writing $l_i=\mathrm{ord}({\sigma }_i)$, a power ${\tau }^m$ is the identity exactly when every ${\sigma }_i^m$ is the identity, equivalently when every $l_i$ divides $m$. Thus

For $N=1$, writeThe given evolution of the discrete scattering data givesSolving the resulting two scalar linear equations for the components of ${\psi }_1$ and substituting in the reconstruction formula givesChoosing the additive multiple of $2\pi$ so that $u\to 0$ as $X\to +\infty$ and integrating in $X$ yields the One-soliton solution of the sine-Gordon equation in light-cone coordinatesIt depends only on $X+T$ exactly when the two positive coefficients in the exponent agree:The unique positive solution is $l=1/2$, and then

The transformations satisfy $g^s{g}^r=g^{s+r}$, $g^0$ is the identity map, and $(g^s{)}^{-1}=g^{-s}$, so they form a one-parameter group. Ifthen the chain rule givesThus $g^s$ is a Lie point symmetry. Applied to the one-soliton family, it replaces $l$ by $l^{\prime }=e^{-s}l$. Taking $e^s=2l$ gives $l^{\prime }=1/2$, so every member is transformed to the function $F(X+T)$ found above.

For the stated $N=2$ solution, set $x=X+T$ and $t=T-X$. When $l^2+m^2=1/4$, its two arguments reduce toAt fixed $x$, only $\sin (2mt)$ varies, so the Sine-Gordon breather has fundamental period

Finally put $r^2=l^2+m^2$. Under the same symmetry, the parameters becomeand hence $(l^{\prime }{)}^2+(m^{\prime }{)}^2=e^{-2s}{r}^2$. Choosing $e^s=2r$ makes this sum $1/4$, proving that every solution in the family is symmetry-equivalent to the normalized breather.

True. The group has order six. A Sylow $3$-subgroup $P$ has order three, while the given normal subgroup $N$ has order two. Their intersection is trivial and $NP=G$. Conjugation by $P$ on $N$ is trivial because $\mathrm{Aut}(C_2)$ is trivial, so $G=N\times P\cong C_2\times C_3\cong C_6$.

No. In any group choose a nonidentity $x$ and put $y=x^{-1}$. They commute, but $xy=e$ has order one rather than $o(x)o(y)$.

The Lagrange theorem states that $|H|$ divides $|G|$ for every subgroup $H$ of a finite group $G$, with $|G|=|G:H||H|$. The left cosets partition $G$, and multiplication by a representative bijects $H$ with each coset, proving the formula.

The intersection $H\cap K$ contains the identity and is closed under $x{y}^{-1}$, so it is a subgroup. Its order divides both $|H|$ and $|K|$; if these are coprime, $H\cap K=\{e\}$.

The order $o(x)$ is the least positive $m$ with $x^m=e$. Division $k=qm+r$, $0\le r<m$, shows that $x^k=e$ exactly when $r=0$, so exactly when $o(x)\mid k$.

If $x,y$ commute and have coprime orders $m,n$, then $(xy{)}^{mn}=e$. Conversely $(xy{)}^r=e$ implies $x^r=y^{-r}$ lies in $⟨x⟩\cap ⟨y⟩=\{e\}$, so $m,n\mid r$ and $mn\mid r$. Hence $o(xy)=mn$.

Cauchy theorem for groups says that every prime divisor $p$ of $|G|$ occurs as the order of an element of $G$. For $|G|=26$, the Sylow $13$-subgroup $P\cong C_{13}$ is unique and normal, and Cauchy's theorem supplies a complement $C_2$. Thus $G=C_{13}⋊C_2$. The homomorphism $C_2\to \mathrm{Aut}(C_{13})\cong C_{12}$ is either trivial or has the unique image of order two, inversion. These give exactly $C_{26}$ and the dihedral group of order $26$.

The first isomorphism theorem says that for a homomorphism $\varphi :G\to H$,The map is well-defined because equal cosets differ by a kernel element; it is a surjective homomorphism onto the image, and injectivity follows because its kernel is the identity coset.

Here $G{L}_n(\mathbb{R})$ consists of invertible real $n\times n$ matrices and $S{L}_n(\mathbb{R})$ consists of those with determinant one. The latter is the kernel of the surjective determinant map to ${\mathbb{R}}^{\ast }$, so it is normal and

The given integral matrices form a group because products and inverses remain integral. The inverse formula shows that an integral inverse exists exactly when the determinant divides every cofactor; taking determinants shows more directly that $\det A\det A^{-1}=1$, hence $\det A=\pm 1$, and the adjugate formula proves the converse. The matrices $\left(\begin{array}{cc}1& m\\ 0& 1\end{array}\right)$ show infinitude.

Reduction modulo two is a homomorphism $G\to G{L}_2({\mathbb{F}}_2)$. Its kernel is exactly $H$, so $H$ is normal; its index is finite because the target has only six elements.

Suppose the additive group of the integral domain $R$ is isomorphic to ${\mathbb{Z}}^n$. Every ideal $I$ is then an additive subgroup of a finitely generated free abelian group, so it has finitely many additive generators $x_1,\dots ,x_m$. These also generate $I$ as an $R$-ideal: integer coefficients are coefficients from the canonical copy of $\mathbb{Z}$ in $R$. Thus every ideal is finitely generated, proving that $R$ is Noetherian by noetherianity from finite additive rank.

For an example that fails (i)--(iv), takeIt is an integral domain and has additive group ${\mathbb{Z}}^2$. The idealhas index two: modulo $I$, one has $2=0$ and $\sqrt{-5}=-1=1$, and the resulting quotient is ${\mathbb{F}}_2$. If $I=(a+b\sqrt{-5})$, multiplication by its generator would have determinant and absolute indexwhich has no integer solution. Hence $I$ is not principal, as detailed in nonprincipal ideal in the integers adjoined a square root of minus five. Condition (i), and therefore all four equivalent conditions, fails.

Count pairs $(x,P)$ where $P$ is a Sylow $2$-subgroup fixing $x$. Each of the seven point stabilizers contains three such subgroups, giving 21 pairs. Conversely, a $2$-group acting on seven points has a fixed point, and a Sylow $2$-subgroup cannot fix two points because a two-point stabilizer has order four. Hence every Sylow $2$-subgroup occurs in exactly one pair, so

Similarly, each point stabilizer contains four Sylow $3$-subgroups, giving 28 pairs. A group of order three acting on seven points has a fixed point, and it cannot fix two because a two-point stabilizer is a $2$-group. Thus

The Sylow congruence and divisibility conditions give $n_7\in \{1,8\}$. Regard the faithful action as an embedding $G\le S_7$. If $n_7=1$, its Sylow $7$-subgroup $P$ is normal, so $G\le N_{S_7}(P)$. The stated fact gives $|N_{S_7}(P)|=7\cdot 6=42$, impossible for a subgroup $G$ of order 168. HenceThese are the Sylow counts in a faithful degree-seven action with S4 point stabilizers.

Choose an angle $2\pi \alpha$ with $\alpha \notin \mathbb{Q}$. Rotation through that angle has infinite order, so the subgroup it generates is isomorphic to $\mathbb{Z}$.

The group also contains a copy of ${\mathbb{Z}}^2$. Choose real numbers $\alpha ,\beta$ such that $1,\alpha ,\beta$ are linearly independent over $\mathbb{Q}$, for example $\alpha =\sqrt{2}$ and $\beta =\sqrt{3}$. DefineThis is a homomorphism. If $\varphi (m,n)=I$, then $m\alpha +n\beta \in \mathbb{Z}$, and the stated rational independence forces $m=n=0$. Thus $\varphi$ is injective and its image is isomorphic to ${\mathbb{Z}}^2$.

Let $\varphi :A_5\times A_5\to \{\pm 1\}$ be a homomorphism. Its restrictions to $A_5\times \{1\}$ and $\{1\}\times A_5$ cannot be nontrivial, because any nontrivial subgroup of $\{\pm 1\}$ is the whole group and part (iii) rules out such a surjection. Both restrictions are therefore trivial. Since$\varphi$ itself is trivial and in particular is not surjective.

Five examples arewhere $D_{12}$ denotes the symmetry group of a hexagon and ${\mathrm{Dic}}_3$ is the $3$-dicyclic group.

The first two are abelian, while the last three are not. Of the first two, only $C_{12}$ is cyclic. Among the nonabelian groups, $A_4$ has no element of order $6$; both of the others do. The dihedral group $D_{12}$ has seven involutions, whereas ${\mathrm{Dic}}_3$ has the unique involution $a^3$ and also has elements outside $⟨a⟩$ of order four. These invariants prove that all five groups are pairwise non-isomorphic.