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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

A Wirtinger presentation from a connected knot diagram has one generator per arc and one relator per crossing, with one relator redundant. Its presentation complex is a finite two-dimensional CW complex $X$ with one zero-cell, $n$ one-cells, and $n-1$ two-cells, and the usual diagrammatic construction gives a homotopy equivalence $X\simeq E_K$.

The abelianization of the knot group is $H_1(E_K;\mathbb{Z})\cong \mathbb{Z}$, generated by a meridian of a knot. Every homomorphism to the cyclic group $\mathbb{Z}/2$ factors through this abelianization, and reduction modulo two is its unique surjection. Thus the requested map $\alpha$ is unique.

Its kernel determines a two-sheeted covering space $\hat{X}$. The nontrivial deck transformation acts on cellular chains and homology, giving them module structures over the group ringLift one copy of every cell of $X$; its two deck translates form a free $\hat{R}$-basis. Hence $C_0\cong \hat{R}$, $C_1\cong {\hat{R}}^n$, and $C_2\cong {\hat{R}}^{n-1}$. If $\tilde{X}$ is the infinite cyclic cover, its cellular chains are free over $R=\mathbb{Z}[t^{\pm 1}]$, and imposing $t^2=1$ gives

For an odd prime $p$, the two idempotents $e_{+}=(1+t)/2$ and $e_{-}=(1-t)/2$ split the group algebraThe plus summand is the cellular chain complex of $X$ with ${\mathbb{F}}_p$ coefficients, while the minus summand is $C_{-}=C_{\ast }^{\mathrm{cell}}(\tilde{X}){\otimes }_R{R}_{-}$. Therefore

On the minus summand, the boundary $t-1:C_1\to C_0$ becomes multiplication by $-2$, which is invertible in ${\mathbb{F}}_p$, so $H_0(C_{-})=0$. After the corresponding cancellation, the remaining square boundary matrix is an Alexander matrix specialized at $t=-1$. It is singular over ${\mathbb{F}}_p$ exactly whenor equivalently when $p$ divides the knot determinant $\det K$. Thus $H_{\ast }(C_{-})$ is nonzero exactly in that case.

Finally, $\det K=|{\Delta }_K(-1)|$ is a nonzero odd integer, so the minus complex is acyclic over $\mathbb{Q}$. Since a knot exterior has the rational homology of a circle,

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.

False. For a prime number $p$, the Prüfer p-group $C_{p^{\infty }}$ is an Artinian $\mathbb{Z}$-module: every proper subgroup is a finite cyclic group, so no infinite strictly descending chain of subgroups exists. It is not Noetherian because its cyclic subgroups form the strict ascending chain

The Lagrange root bound over a field says that a nonzero polynomial of degree $r$ over a field has at most $r$ roots. In particular, a polynomial modular congruence of degree $r$ modulo a prime number has at most $r$ incongruent solutions unless all its coefficients vanish modulo that prime.

Suppose that $d$ is good and that the positive integer divisor $e$ divides $d$. The $d$ roots of $X^d-1$ form a finite multiplicative subgroup $H$ of ${\mathbb{F}}_p^{\times }$. By the fact that every finite multiplicative subgroup of a field is cyclic, $H=⟨g⟩$ is a cyclic group of order $d$. The solutions in $H$ of $x^e=1$ areso there are at least $e$ of them. The Lagrange root bound over a field gives at most $e$ roots in all of ${\mathbb{F}}_p$, hence exactly $e$. Thus every divisor of a good number is good.

Now put $n=pq$. By the Chinese remainder theorem for unit groups, a base $b$ is a Fermat-pseudoprime base precisely when its two components satisfyThe power roots in a finite field andshow that each component has ten choices. Therefore there areFermat-pseudoprime bases.

To impose the strong pseudoprime condition, write $n-1=2^{s}m$ with $m$ odd. Since $10\mid n-1$, we have $5\mid m$. In each cyclic group of ten Fermat components, raising to the $m$th power sends five elements to $1$ and five elements to $-1$. A pair of components passes the strong test exactly when their signs agree: the pair $(1,1)$ satisfies the first alternative, and $(-1,-1)$ satisfies the second at $j=0$. Components of opposite sign become $(1,1)$ after squaring and can never jointly equal $-1$. Hence the number of strong-pseudoprime bases is

Write $\bar{a}=f(a)$, $\bar{t}=f(t)$, and $\bar{b}={\bar{a}}^{\bar{t}}={\bar{t}}^{-1}\bar{a}\bar{t}$. The elements $\bar{a}$ and $\bar{b}$ are conjugate group elements, so they have the same order of a group element, say $n$. The defining relation becomesIterating conjugation gives ${\bar{b}}^j\bar{a}{\bar{b}}^{-j}={\bar{a}}^{2^j}$. Since ${\bar{b}}^n=1$, taking $j=n$ yields $\bar{a}={\bar{a}}^{2^n}$, and henceThe stated consequence of Fermat's little theorem implies $n=1$. Thus $\bar{a}=1$, and the surjective group homomorphism $f$ shows that $Q$ is generated by $\bar{t}$. Therefore $Q$ is a cyclic group.

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

Yes. Choose the principal ultrafilter at any $j\ge 2$. Evaluation at the $j$th coordinate giveswhich is a finite cyclic group of order $j$.

A group homomorphism from an additive cyclic group ${\mathbb{Z}}_n$ is determined by the image $a$ of $1$, subject exactly to $a^n=e$ in multiplicative notation, or $na=0$ in an additive target.

For ${\mathbb{Z}}_{18}\to {\mathbb{Z}}_{15}$, the relation is $18a\equiv 0(\mathrm{mod}15)$, equivalently $3a\equiv 0(\mathrm{mod}15)$. The possibilities are $a=0,5,10$, and each defines a homomorphism between cyclic groups. Thus there are three such maps, an instance of the number of homomorphisms between finite cyclic groups.

For ${\mathbb{Z}}_9\to S_4$, the image of $1$ must be a permutation whose order of a group element divides $9$. A permutation on four symbols can have order $1,2,3$ or $4$, so the allowed images are the identity and the three-cycles. There are $\left(\genfrac{}{}{0ex}{}{4}{3}\right)(3-1)!=8$ three-cycles: choose their three symbols and one of the two cyclic orientations. Hence there are nine maps.

Finally, each transposition in $S_5$ has order $2$, so its image in ${\mathbb{Z}}_5$ must have order dividing both $2$ and $5$, and is therefore the identity. Every permutation is a product of transpositions, so every element must map to the identity. There is one map. In the order asked, the counts are

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.

Let $|G|=p$, with $p$ a prime number, and choose $g\ne e$. By Lagrange's theorem, the order of a group element $g$ divides $p$ and is greater than one. It is therefore $p$, and $G=⟨g⟩$. Hence a group of prime cardinality is a cyclic group.

The converse is false: the cyclic group $C_4$ has four elements, and $4$ is not a prime number.

Write $C_p=⟨g⟩$. A group homomorphism from this cyclic group is determined by the image of $g$, because $\varphi (g^j)=\varphi (g{)}^j$. Each choice $\varphi (g)=g^a$, with $a=0,\dots ,p-1$, gives a well-defined group homomorphism. It is an group automorphism exactly when $g^a$ generates $C_p$.

For a prime number $p$, every $a=1,\dots ,p-1$ is coprime to $p$, so the powers of $g^a$ run through all of $C_p$. The choice $a=0$ gives the trivial group homomorphism and fails to be bijective. Consequently

Choose a normal subgroup $N$ of prime order $p\in \{5,7\}$ as in ii, and choose a subgroup $H$ of the other prime order $q$. Their intersection is trivial by Lagrange's theorem. Both are cyclic groups, and ii gives $|\mathrm{Aut}(N)|=p-1$.

The image of a group homomorphism $\pi :H\to \mathrm{Aut}(N)$ from iii has order dividing $q$, because its kernel of a group homomorphism partitions $H$ into equally sized cosets. Its order also divides $p-1$ by Lagrange's theorem. Butso $\pi$ is trivial in either case. By iii, $⟨N,H⟩\cong C_p\times C_q$ has order $pq=35$ and therefore equals $G$.

If $n$ and $h$ generate $N$ and $H$, the element $(n,h)$ has order $\mathrm{lcm}(p,q)=pq=35$, since its $k$th power is the identity exactly when both $p$ and $q$ divide $k$. Thus this direct product of groups is cyclic, and

A smallest example is the Klein four-groupwith coordinatewise addition modulo $2$. Every nonidentity element has order of a group element $2$, so none generates all four elements; hence this group is not a cyclic group.

A group of group order $1$ is a cyclic group. A group of group order $2$ or $3$ is also a cyclic group: choose a nonidentity element; by Lagrange's theorem, its order of a group element is a nontrivial divisor of that prime number, so equals the whole group order. Thus no smaller example exists, and the smallest possible group order is $4$.

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.

If $G$ is not Abelian, the contrapositive of ii says that $G/A$ is not a cyclic group. The quotient group is finite because $G$ is finite. By iii, every group with fewer than four elements is a cyclic group. HenceEquivalently, the center of a group satisfies $|A|\le |G|/4$, by the coset count in iii.

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.

Let $r$ be a quarter-turn and $s$ a reflection of the square. The dihedral group is generated by $r,s$ with $r^4=s^2=1$ and $srs=r^{-1}$, and its elements are $1,r,r^2,r^3,s,sr,s{r}^2,s{r}^3$. Two generators are minimal because this group is not cyclic. Its subgroups of order two are $⟨r^2⟩$ and $⟨s{r}^j⟩$ for $j=0,1,2,3$. Its subgroups of order four are $⟨r⟩$, $⟨r^2,s⟩$ and $⟨r^2,sr⟩$.

A group homomorphism obeys $\varphi (gh)=\varphi (g)\varphi (h)$ for all elements; a group isomorphism is a bijective group homomorphism. Take the rotation subgroup $⟨r⟩$ and defineThe quaternion element $i$ has order four, so the map is well defined modulo four, preserves products and is injective onto $\{1,i,-1,-i\}$. Surjectivity onto all eight quaternion elements is not required of a homomorphism.

A cyclic group is generated by the powers of one element. The final claim needs the usual qualification no nontrivial proper subgroup, and $|G|>1$: the trivial subgroup is proper in every nontrivial group, even one of prime order. With that interpretation, if $G=⟨g⟩$ has prime order $p$, Lagrange's theorem permits only subgroup orders one and $p$. Conversely, if its order $n>1$ is composite, choose a divisor $d$ with $1<d<n$. The subgroup $⟨g^d⟩$ has order $n/d$, strictly between one and $n$. This proves the assertion described by groups with no proper nontrivial subgroups.

A group homomorphism preserves multiplication, $\varphi (ab)=\varphi (a)\varphi (b)$. A group isomorphism is a bijective homomorphism. An isomorphism from a group to itself is a group automorphism.

The automorphism group contains the identity map. Composition is associative, and if $\varphi ,\psi$ are automorphisms, $(\varphi \circ \psi )(ab)=\varphi (\psi (a)\psi (b))=(\varphi \circ \psi )(a)(\varphi \circ \psi )(b)$; composition is also bijective. For $u=\varphi (a)$ and $v=\varphi (b)$, ${\varphi }^{-1}(uv)={\varphi }^{-1}(\varphi (ab))=ab={\varphi }^{-1}(u){\varphi }^{-1}(v)$, proving the inverse bijection is a homomorphism. These facts establish all group axioms.

A finite cyclic group is $G=⟨g⟩=\{e,g,\dots ,g^{n-1}\}$ with $g$ of order $n$. The map $g^k\mapsto e^{2\pi ik/n}$ is well defined modulo $n$, bijective onto the roots of unity, and preserves multiplication. It identifies $G$ with the multiplicative group of $n$th roots of unity.

For prime $n$, Lagrange's theorem leaves only subgroup orders one and $n$: there is no nontrivial proper subgroup. The trivial subgroup $\{e\}$ is nevertheless proper when $n>1$, so that qualification is necessary in the printed claim. The characterization of groups with no proper nontrivial subgroups makes the distinction explicit.

For composite $n$, choose a prime divisor $p$ and take $\boxed{H_0=\{e\},H_1=⟨g^p⟩}$, of orders $1$ and $n/p$. They are distinct proper subgroups, with $H_1$ nontrivial. If the question intended two nontrivial proper subgroups, that demand is not always possible: a cyclic group of order $p^2$ has only the one of order $p$. The subgroups and quotients of a cyclic group are controlled by its divisors. The order-one group is a separate trivial case, with no proper subgroups.