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For any $x$, choose $z$ with $x\ast z=e$. By ii, $z\ast x=e$ as well. Associativity and the right identity element property now giveThus $e$ is also a left identity element. The given binary operation is closed on $X$ and associative, $e$ is a two-sided identity element, and each $x$ has the two-sided inverse element $z$. All group axioms hold, so $(X,\ast )$ is a group. This is the one-sided group axioms criterion.

Suppose $y\ast y=y$. Choose $z$ with $y\ast z=e$, using the given right inverse element property. Multiply the idempotent relation on the right by $z$. By associativity and the given right identity element property,Thus the only idempotent element of a semigroup is $e$. This argument uses no cancellation and does not assume a left identity element.

The shift $s:\mathbb{N}\to \mathbb{N}$, $s(n)=n+1$, is injective, so $s\in I(\mathbb{N})$. It is not surjective: the least natural number is missing from its image of a function, whether $\mathbb{N}$ starts at zero or one. For every $t:\mathbb{N}\to \mathbb{N}$, the image of a function of $s\circ t$ is contained in the image of a function of $s$, so $s\circ t\ne {\mathrm{id}}_{\mathbb{N}}$.

The identity function belongs to $I(\mathbb{N})$ and is its identity element under function composition; any group structure using that operation would have this same identity element. Since $s$ has no right inverse element, $I(\mathbb{N})$ is not a group under function composition. It is a monoid, since injectivity is preserved by function composition.

Suppose an arbitrary binary operation $\star$ makes $\mathbb{Z}$ a group and makes $\rho$ a left group action on $G$. For any group action, each acting element induces a bijection: the function induced by its inverse element is its inverse function.

In particular, the acting element whose underlying integer is $0$ must induce a bijection $G\to G$. But its power function isa constant function. This is injective only when all elements of $G$ equal $e$. ThereforeThis proves the power-map group action obstruction for every possible $\star$; it never assumes that $\star$ is ordinary integer multiplication, or that the integer $0$ is the acting identity element.

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.

Write $g\cdot x=gx{g}^{-1}$. If $e$ is the identity element of $G$, then $e\cdot x=ex{e}^{-1}=x$. Also, for $g,h,x\in G$,The middle equality uses associativity and the inverse element rule $(gh{)}^{-1}=h^{-1}{g}^{-1}$. Thus this is a left group action, the conjugation action of $G$ on itself.

Write $T_{\theta ,\mathbf{b}}(\mathbf{v})=R_{\theta }\mathbf{v}+\mathbf{b}$. The sine addition formula and cosine addition formula give $R_{\theta }{R}_{\psi }=R_{\theta +\psi }$. Therefore function composition satisfiesThe resulting angle is real and the new translation vector lies in ${\mathbb{R}}^2$, so $G$ is closed under function composition. Associativity holds because, for any functions $S,T,U$ and any $\mathbf{v}$, both $(S\circ T)\circ U$ and $S\circ (T\circ U)$ take $\mathbf{v}$ to $S(T(U(\mathbf{v})))$.

The identity element is $T_{0,0}$, since $R_0=I$. Every element has the two-sided inverse elementIndeed $R_{\theta }{R}_{-\theta }=R_{-\theta }{R}_{\theta }=I$, and substituting into the composition formula makes the translation vector zero in either order. These verify all the group axioms. Thus $G$ is a group under function composition, the special Euclidean group in two dimensions.

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.

A group homomorphism $h:G\to F$ preserves multiplication: $h(g_1{g}_2)=h(g_1)h(g_2)$. It sends the identity element to the identity and inverse elements to inverse elements. Its group kernel is $K=\{g\in G:h(g)=e_F\}$. This contains $e_G$ and is closed under multiplication and inverse elements, so it is a subgroup. For $k\in K$ and $g\in G$,Thus $gK{g}^{-1}\subseteq K$; applying the same argument with $g^{-1}$ gives equality. Hence the group kernel is a normal subgroup.

Let $S_2=\{e,\tau \}$ with $\tau =(12)$. The parity of a permutation defines the desired map from $S_3$:The sign of a permutation is multiplicative, so parity adds modulo two under composition, exactly matching ${\tau }^2=e$. Thus this is a group homomorphism; it is surjective because a transposition maps to $\tau$. Its group kernel isThis is the alternating group on three objects, a normal subgroup of index two in $S_3$.

Lagrange's theorem states that the group order of a finite group divides the group order; more precisely $|G|=[G:H]|H|$.

If every element squares to the identity element, every element is its own inverse element, and $ab=(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$. Thus a group of exponent two is abelian. For distinct nonidentity $a,b$, the four elements $1,a,b,ab$ are distinct: for example $ab=1$ would imply $a=b$, and $ab=a$ would imply $b=1$. Products remain in this set because $a,b$ commute and square to one. Thus they form a Klein four-group subgroup of order four.

Now let $|G|=2p$ with odd prime number $p$. By Lagrange's theorem, every element order is one of $1,2,p,2p$. If an element has order $2p$, its square has order $p$. If no element had order $p$, all nonidentity elements would therefore have order two. There are at least two distinct such elements, producing a subgroup of order four. This contradicts $4\nmid 2p$. Hence there is an element of order $p$.

The group order of a finite group is its number of elements, $|G|$. The order of a group element $g$ is the least positive integer $r$ with $g^r=e$. Such an $r$ exists in a finite group because two powers must coincide, and cancellation then gives a positive power equal to the identity element.

For the direct product of groups, the componentwise operation is closed because each factor is closed. Associativity holds in each coordinate:The identity element is $(e_1,e_2)$, and the inverse element of $(g,h)$ is $(g^{-1},h^{-1})$. Thus all group axioms hold. For finite factors, the multiplication principle gives

For ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$, write $E=(e,e)$, $u=(w,e)$, $v=(e,w)$, $z=(w,w)$, with $w^2=e$. Its Cayley table, with the row element multiplying the column element, isThis is the Klein four-group: every nonidentity element has order two.

In the cyclic group ${\mathbb{Z}}_4=⟨a⟩$, the set $\{e,a^2\}$ is closed because $(a^2{)}^2=e$, and both elements have inverses in the set. It is therefore a proper subgroup of order two. By Lagrange's theorem, a proper subgroup has order one or two. An order-two subgroup must be generated by an element of order two. Neither $a$ nor $a^3$ has order two, whereas $a^2$ does. HenceThe PDF's claim needs “nontrivial”: $\{e,a^2\}$ is the only nontrivial proper subgroup, but $\{e\}$ is also a proper subgroup under the usual definition.

A group isomorphism preserves element orders: $\varphi (g{)}^r=e$ if and only if $g^r=e$, using the homomorphism property and injectivity. Since ${\mathbb{Z}}_4$ has elements of order four and the Klein four-group does not,

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 $C_2\times C_2$.

A group is a set $G$ equipped with a binary operation satisfying the group axioms: closure; associativity, $(gh)k=g(hk)$; an identity element $e$ with $eg=ge=g$; and an inverse element $g^{-1}$ for every $g$, with $g{g}^{-1}=g^{-1}g=e$.

A subgroup $H$ is a subset that is a group under the restricted operation. Equivalently, it is nonempty and contains $x{y}^{-1}$ whenever $x,y\in H$. A generating set of a group is a set whose elements and inverses can be multiplied in finite words to produce every group element.

The group order is its number of elements, $|G|$. The order of a group element $g$ is the least positive integer $r$ with $g^r=e$, or infinity if there is none. Lagrange's theorem states that for a finite group and subgroup $H$, $|H|$ divides $|G|$ and $|G|=[G:H]|H|$. In particular, every finite element order divides the group order.

A cyclic group is generated by one element, so every element is a power of that generator. A group isomorphism is a bijective group homomorphism: a bijection $\varphi :G\to G^{\prime }$ satisfying $\varphi (gh)=\varphi (g)\varphi (h)$ for all $g,h$.

The substitution $F(a)=e^a$, with $F(-\infty )=0$, is a bijection onto $[0,\infty )$ and obeysThus the log-sum-exp function transfers nonnegative addition to the given set. It is closed, associative and commutative, with identity element $-\infty$. But a finite $a$ has $e^a>0$, and adding a nonnegative number cannot give zero, so no finite $a$ has an inverse element. Therefore it is a commutative monoid but not a 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 $(\mathbb{Z},+)$.

A group homomorphism preserves multiplication:It automatically sends the identity element to the identity and sends inverse elements to inverses. A group isomorphism is a bijective group homomorphism. For finite groups this requires equal group orders, and, once those orders agree, either injectivity or surjectivity of the group homomorphism is sufficient. Equal orders alone do not suffice.

Closure of the componentwise operation follows from closure in each factor. For three pairs, associativity holds coordinatewise:The identity element is $(e_1,e_2)$ and the inverse element of $(g_1,g_2)$ is $(g_1^{-1},g_2^{-1})$. These verify all group axioms for the direct product of groups. Counting independent choices givesThe subset $N=G_1\times \{e_2\}$ contains the identity and is closed under products and inverses, so it is a subgroup. For any $x=(a,b)$ in the product and $(g,e_2)\in N$,Conjugation by $x^{-1}$ gives the reverse inclusion, and henceThis is the definition of a normal subgroup.

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.

Use group axioms from two-sided local inverse identities. Choose any $a$ in the nonempty semigroup and put $e_R=a_L^{-1}a$, $e_L=a{a}_R^{-1}$. The given identities say $b{e}_R=b$ and $e_{L}b=b$ for every $b$. Therefore $e_L{e}_R=e_L$ by the right-identity property, and $e_L{e}_R=e_R$ by the left-identity property. Thus $e_L=e_R=e$ is a two-sided identity element.

For an arbitrary $a$, set $b=a_L^{-1}$ and $c=a_R^{-1}$. Inserting $e$ into the original identities gives $ba=e$ and $ac=e$. By associativity,Thus every element has a two-sided inverse element. Closure and associativity were assumed, so all the group axioms hold. No cancellation or identity was used before it had been established.

The map is continuous if it is continuous at every point of its domain. Equivalently the inverse image of every open subset of $Y$ is open in $X$. These equivalent formulations use the given metrics' induced topologies.