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Given a prime factorization $n=p_1\cdots p_r$, repeatedly apply part (c) to the cyclic group of order $n$: it has a chain of subgroups of successive ordersRead in increasing order, this is a composition series with factors isomorphic to the cyclic groups of orders $p_1,\dots ,p_r$. Any second prime factorization of the same integer produces another composition series of the same group $C_n$.

The Jordan–Hölder theorem equates the two multisets of simple factors. Since $C_p\cong C_q$ for primes $p,q$ implies $p=q$ by equality of group orders, the two prime multisets agree. HenceFor $n=1$ both factor lists are empty, corresponding to the trivial group.

The exact orders printed in the question are inconsistent. The source PDF says that the orders of group elements $a,b,c$ are $4,12,24$, respectively, and also imposes $3a+2b=0$. Multiplying this relation by $6$ and using $12b=0$ gives $18a=0$. Since $4a=0$ also holds, subtracting $4(4a)=0$ gives $2a=0$, contradicting the stated order of a group element $4$. Thus

The natural abelian group presentation interpretation is to impose the relations $4a=12b=24c=0$ and $3a+2b=0$, with no further relations, without requiring the orders of the images of the generators to stay $4,12,24$. Under this interpretation the calculation is as follows. Since $3a=-a$ follows from $4a=0$, the mixed relation gives $a=2b$. Consequently $8b=0$, and together with $12b=0$ this gives $4b=0$. Conversely, $a=2b$, $4b=0$, and $24c=0$ imply every original relation. Eliminating $a$ leaves two independent cyclic groups:To verify that these relations force no further identifications, map $a,b,c$ to $(2,0),(1,0),(0,1)$ in ${\mathbb{Z}}_4\oplus {\mathbb{Z}}_{24}$. All defining relations hold, and the inverse group homomorphism sends the two standard generators to $b,c$. These group homomorphisms are mutually inverse. The divisibility condition $4\mid 24$ gives the requested invariant factors of a finitely generated module over $\mathbb{Z}$. In this presented group, $a,b,c$ actually have orders $2,4,24$.

Equivalently, the relation matrixhas nonzero Smith normal form entries $1,4,24$, with the trivial ${\mathbb{Z}}_1$ factor omitted. The displayed direct sum of modules is the answer to the presentation interpretation; the inconsistency of the literal exact-order formulation remains.

Write $q=|\mathbb{F}|$. The unit group $U(\mathbb{F})=\mathbb{F}\setminus \{0\}$ is a finite abelian group of group order $q-1$, since multiplication in a field satisfies commutativity. The structure theorem for finitely generated abelian groups, applied to this finite group, gives invariant factors of a finitely generated moduleafter omitting trivial factors. If there are no factors, the unit group is trivial and hence a cyclic group.

Otherwise put $D=d_r$. Every element $u\in U(\mathbb{F})$ satisfies $u^D=1$, because the order of a group element in each coordinate divides $D$. Hence all $q-1$ distinct nonzero elements of $\mathbb{F}$ are roots of a polynomial $X^D-1$. The root bound for a polynomial over a field givesFor completeness, that bound follows by induction from the factor theorem: if a nonzero polynomial $f$ vanishes at $a$, write $f=(X-a)g$; every other root of a polynomial is a root of $g$, because a field has no zero divisors. Each new distinct root therefore uses one unit of polynomial degree.

On the other hand, the group order in the decomposition isEquality is forced, so $d_1\cdots d_{r-1}=1$. As every retained factor is greater than $1$, there can be only one. ConsequentlyThe same proof establishes the more general result finite multiplicative subgroup of a field is cyclic. The square-free integer result in part (c)(i) alone would not suffice, since $q-1$ need not be a square-free integer; the root bound for a polynomial rules out multiple invariant factors of a finitely generated module even then.

Let $N=|G|$ be a square-free integer. If $N=1$, then $G$ is the trivial cyclic group. Otherwise, finiteness means that a direct sum of modules decomposition into cyclic groups has only finitely many nontrivial factors, each of finite group order. Omit trivial factors and writeThe integers $n_i,n_j$ are coprime integers whenever $i\ne j$. If a prime number $p$ divided both, then $p^2\mid N$, contradicting the square-free integer hypothesis.

Moreover, $n_1\cdots n_{j-1}$ and $n_j$ are coprime integers: any common prime divisor would divide one of the earlier factors as well as $n_j$. Thus part (b)(ii) can be applied repeatedly to combine factors, givingIn particular $G$ is a cyclic group. Under the displayed identifications, the element whose coordinate in every factor is the class of $1$ is a generator of a group.

Choose a generator $u$ of the cyclic group $U(R)$, so $u^p=1$. Part (a)(iii) gives a unital ring homomorphism ${\mathbb{F}}_2\to R$. Its image lies in the centre of $R$, so even if $R$ is not a commutative ring, evaluating a polynomial at $u$ defines a ring homomorphismBecause $\varphi (X^p-1)=0$, it factors through the quotient ring to giveIts image is the subring $S={\mathbb{F}}_2[u]$. Every unit of $S$ is a unit of $R$, because an inverse in $S$ also lies in $R$. Conversely, every member of $U(R)$ is $u^j$ for some $j$, and both $u^j$ and its inverse $u^{p-j}$ lie in $S$. HenceThe inverse argument matters: merely belonging to a subring would not by itself make an ambient unit invertible there.

A simple group is a nontrivial group whose only normal subgroups are itself and the trivial subgroup.

A composition series is a chain $G=G_0▹G_1▹\cdots ▹G_r=1$ in which each $G_{i+1}$ is a normal subgroup of $G_i$ and each quotient group $G_i/G_{i+1}$ is simple and nontrivial. Its length is $r$, and these quotients are its composition factors. The individual terms need not all be normal in $G$.

A solvable group is a group whose derived series reaches the trivial group: starting with $G^{(0)}=G$, define $G^{(i+1)}=[G^{(i)},G^{(i)}]$. For finite groups this is equivalent to having a composition series whose factors are all cyclic groups of prime order.

The Sylow theorems also give $n_5\mid 14$ and $n_5\equiv 1(\mathrm{mod}5)$. Of $1,2,7,14$, only $1$ is congruent to one modulo five. Thus $H$ is a normal subgroup of $G$ as well.

By Lagrange's theorem, $H\cap K=\{1\}$, since its group order divides both $5$ and $7$. For $h\in H$ and $k\in K$, the group commutator $hk{h}^{-1}{k}^{-1}$ lies in $H$ by the normal subgroup property of $H$, and in $K$ by the normal subgroup property of $K$. It is therefore the identity element, so all elements of $H$ commute with all elements of $K$. This is the fact that normal subgroups of coprime order commute.

The subgroup $HK$ consequently has $|H||K|=35$ elements: the expression $hk$ is unique because $H\cap K=\{1\}$. Since $H$ and $K$ have prime orders, they are cyclic groups. Write $H=⟨h⟩$ and $K=⟨k⟩$. The product of commuting elements of coprime order has order $35$, so $hk$ generates $HK$. Finally, $HK$ is a normal subgroup because both its factors are normal subgroups; equivalently, it has subgroup index two. Therefore

Let $|G|=p^2$. The allowed center of a group theorem gives $|Z(G)|>1$. By Lagrange's theorem its order is $p$ or $p^2$. In the first case $G/Z(G)$ has order $p$ and is a cyclic group. A cyclic quotient by the center forces $G$ to be Abelian: if $G/Z(G)=⟨gZ(G)⟩$, write arbitrary elements as $g^{a}z,g^{b}w$ with $z,w\in Z(G)$, and commute them directly. Hence in every caseFor two examples at prime-cube order, take the dihedral group of a square, $⟨r,s\mid r^4=s^2=1,srs=r^{-1}⟩$, and the quaternion group $Q_8=\{\pm 1,\pm i,\pm j,\pm k\}$. Both have order $8=2^3$. They are nonabelian because $sr\ne rs$ and $ij=k\ne -k=ji$, respectively. They are distinct examples even up to group isomorphism: the dihedral group has five elements of order two and the quaternion group has only one. The request asks for two examples; it does not require the prime to be arbitrary.

The Abelian group criterion for a product of two prime squares applies because $11<19$ and $11\nmid 19^2-1=360$. Here its Sylow theorems proof is especially short:The candidates for $n_{19}$ are $1,11,121$, of which only $1$ has the required remainder. The candidates for $n_{11}$ are $1,19,361$, again leaving only $1$. Thus both Sylow subgroups $P,Q$ are normal subgroups. Their group commutators lie in $P\cap Q=\{1\}$, and $PQ=G$, so $G\cong P\times Q$. Both factors are abelian groups of prime-square order by (a).

The Fundamental theorem of finitely generated abelian groups gives preciselyThese four groups are pairwise nonisomorphic: their characteristic Sylow subgroups differ in whether they contain an element of order $11^2$ or $19^2$. The first is also the cyclic group $C_{43681}$.

The multiplicative group ${\mathbb{F}}_p^{\times }$ is a cyclic group of order $p-1$, so choose a primitive root $g$. Every nonzero residue is uniquely $g^j$, with $j$ taken modulo $p-1$. The condition becomes $(p-1)\mid kj$. Writing $d=\gcd (k,p-1)$, its solutions are $j=0,(p-1)/d,\dots ,(d-1)(p-1)/d$. Zero is not a solution. Therefore the number isThis includes $p=2$. The cyclic group argument gives the exact count, whereas the bound of $k$ roots for a nonzero degree-$k$ polynomial over a field would only give an upper bound.

There are six functions: the four power maps $x\mapsto x^k$ for $k=1,2,3,4$, and the two constant maps $0,1$. All calculations in this part are in the finite field ${\mathbb{Z}}_5$.

Put $e=f(1)$. Multiplicativity gives $e^2=e$, so the absence of zero divisors forces $e=0$ or $e=1$. If $e=0$, then $f(x)=f(x)f(1)=0$ for every $x$, giving only the constant-zero map.

Suppose $e=1$. For nonzero $x$, the equality $f(x)f(x^{-1})=1$ shows $f(x)\ne 0$. The nonzero elements form a cyclic group generated by $2$, in the order $1,2,4,3$. Choosing $a=f(2)$ determines every value on this group by $f(2^j)=a^j$. Each $a\in \{1,2,3,4\}$ satisfies $a^4=1$, so all four choices give consistent group homomorphisms.

Finally, $z=f(0)$ satisfies $z^2=z$ and $z=zf(x)$ for every $x$. The choice $z=0$ works for all four group maps. The choice $z=1$ forces every nonzero value to be $1$, giving one extra map, the constant-one map. Thus the count is $1+4+1=6$. The four zero-preserving possibilities are exactly the four displayed power maps. This is the special case with five elements of multiplicative self-maps of a finite field.

A group is poly-(cyclic or finite) when it has a finite subnormal series $1=G_0◃G_1◃\cdots ◃G_r=G$ in which each quotient $G_i/G_{i-1}$ is a cyclic group or a finite group. Each term need only be normal in the next term. The polycyclic group condition instead requires every factor to be cyclic.

The scalar sign action of a cyclic group of order two on ${\mathbb{C}}^2$ has invariant ring $\mathbb{C}[X^2,XY,Y^2]\cong \mathbb{C}[a,b,c]/(ac-b^2)$. It is not factorial: $a,b,c$ are pairwise nonassociate irreducibles, since every nonconstant invariant has degree at least two, while $ac=b^2$ gives two distinct factorizations.

Let $J$ be the plane rotation matrix through $\pi /2$, generating a cyclic group of order four. A quadratic form $a{x}^2+b{y}^2$ has its two coefficients exchanged by a quarter-turn and unchanged by a half-turn. Thus the pullback action on this two-dimensional space gives the displayed group representation: $D(J^k)=D(J{)}^k$, with $D(J{)}^2=I$. It is not a faithful representation, because its kernel of a group homomorphism is $\{I,J^2\}$. Its two invariant coefficient lines correspond to $x^2+y^2$ and $x^2-y^2$, on which a quarter-turn acts by $+1$ and $-1$, respectively. Restricting to this diagonal space is valid for these signed coordinate permutations; a general linear transformation can introduce an $xy$ term.

For a group $G$ and a normal subgroup $N$, the quotient group $G/N$ is the set of cosets $gN$, with operationIts identity element is the coset $N$, and the inverse element of $gN$ is $g^{-1}N$. A quotient group treats all elements within one coset as the same input while retaining a consistent operation on the classes. The normal subgroup condition is what makes this operation independent of which representatives are chosen. ^[ref].

Take the additive group $G=\mathbb{Z}/6\mathbb{Z}$ and $N=\{0,3\}$. Since $G$ is an abelian group, $N$ is a normal subgroup. There are three elements in the quotient group:Adding two copies of $2+N$ gives $4+N=1+N$; adding $1+N$ and $2+N$ gives $3+N=N$. Hence the quotient group has the addition table of $\mathbb{Z}/3\mathbb{Z}$. It does not select one representative and discard the others: its elements are the entire cosets.

This is useful whenever distinctions within a normal subgroup should be ignored. For example, the sign of a permutation retains only even or odd parity of a permutation in $S_3$. Its group kernel is the alternating group $A_3$, and the quotient group $S_3/A_3$ has two elements. The first isomorphism theorem explains why a group homomorphism's image is a quotient group of its domain.

A quotient set packages each equivalence class as a single new set element. It remembers which class an original set element belongs to and forgets which member of that class it was. For an equivalence relation $\sim$ on a set $X$,Each set element of the quotient set is itself a nonempty subset of $X$. The canonical projection of a quotient set sends an original set element to this entire equivalence class. ^[ref]

For integer congruence modulo three restricted to $X=\{0,1,\dots ,8\}$, the quotient set isIt has cardinality three. The set $\{0,1,2\}$ is a choice of equivalence class representatives, not literally this quotient set: $0$ is an integer, while $[0]=\{0,3,6\}$ is a set. The function sending each chosen equivalence class representative to its equivalence class is a bijection between these two different sets.

A quotient set alone specifies no arithmetic. To obtain the familiar modular arithmetic example with addition, use integer congruence on the whole set of integers. Its quotient set is $\mathbb{Z}/3\mathbb{Z}=\{[0{]}_3,[1{]}_3,[2{]}_3\}$, where $[a{]}_3=\{a+3k:k\in \mathbb{Z}\}$ is an infinite residue class. Define addition byThe answer must be independent of the equivalence class representatives. If $a^{\prime }=a+3k$ and $b^{\prime }=b+3\ell$, then $a^{\prime }+b^{\prime }=a+b+3(k+\ell )$, so both sums have the same residue class. This is what it means for the group operation to be well-defined. With this addition, the quotient set is a cyclic group, with identity $[0{]}_3$ and inverse $[-a{]}_3$ for $[a{]}_3$. ^[ref]

For example, $[2{]}_3+[2{]}_3=[4{]}_3=[1{]}_3$. Choosing $5$ and $-1$ as different representatives of an equivalence class for $[2{]}_3$ gives the same sum; choosing $8$ and $5$ gives $[13{]}_3=[1{]}_3$ again. These calculations use residue classes of all integers. Raw addition on the finite $X=\{0,\dots ,8\}$ does not stay in $X$: $8+8=16\notin X$. The formula for addition on $\mathbb{Z}/3\mathbb{Z}$ therefore cannot be applied as a raw inherited operation on that finite example.

Representation theory describes abstract symmetries by linear maps on vector spaces. A group representation assigns an invertible linear map to each group element, respecting multiplication. The payoff is that a symmetry problem becomes a problem about matrices, invariant subspaces, and eigenvalues. Decomposing the action can reveal combinations of variables that a symmetric system cannot mix.

Take the cyclic group $C_3=\{e,g,g^2\}$, with $g^3=e$. On three real coordinates, let $g$ send $(a,b,c)$ to $(c,a,b)$:This permutation representation does more than relabel a list: it acts linearly on every possible assignment of values to the three positions. The vector $(1,1,1)$ stays fixed. The vector subspace $W=\{(a,b,c):a+b+c=0\}$ is also preserved, since a permutation does not change the sum.

Every vector separates uniquely into these two types:For example, $(4,1,1)=2(1,1,1)+(2,-1,-1)$. The first term belongs to a trivial representation; the second is a contrast with zero coordinate sum. This is a direct sum decomposition ${\mathbb{R}}^3=\mathbb{R}(1,1,1)\oplus W$. The symmetry preserves the mean and rotates the contrasts.

The rotation is literal after choosing the orthonormal basis $u=(2,-1,-1)/\sqrt{6}$, $v=(0,1,-1)/\sqrt{2}$ of $W$. In this basis,It is a rotation through $120$ degrees, and three applications return to the starting point. There is no nonzero invariant real line, because this rotation has no real eigenvector. Thus the two-dimensional action is an irreducible representation over the real numbers. Over the complex numbers it instead splits into two eigenlines with eigenvalues $e^{2\pi i/3}$ and $e^{-2\pi i/3}$. The coefficient field is part of an irreducibility claim.

Why would this decomposition matter for a physical calculation? Consider three equal masses with scalar displacements $x_i$ and the symmetric quadratic coupling energyThe equations of motion are $m\ddot{x}=-kLx$, whereis a graph Laplacian and $J$ is the all-ones matrix. The constant vector has eigenvalue zero: translating all coordinates equally costs no coupling energy. On $W$, $J$ vanishes and $L=3I$, so both independent contrast directions have angular frequency $\sqrt{3k/m}$. A three-coordinate coupled problem reduces to a free mean and two identical harmonic oscillators.

For initial displacement $x(0)=(4,1,1)$ and initial velocity zero, the exact solution is $x(t)=2(1,1,1)+\cos (\sqrt{3k/m}t)(2,-1,-1)$. The fixed mean and oscillating contrast can be read directly from the decomposition. More generally, a linear operator commuting with the group action carries its fixed-vector subspace into itself; with an invariant inner product, it preserves the corresponding orthogonal complement too.

For a finite group, averaging over the action constructs an invariant inner product over the real numbers or complex numbers. The resulting invariant complements explain Maschke's theorem: a finite-dimensional group representation decomposes into irreducible representations. Characters compress the action into traces and help identify those pieces. The assumptions matter; in modular representation theory the coefficient field can have characteristic dividing the group order, and such decompositions can fail. Chapter 4 of Pavel Etingof and coauthors' notes develops these mechanisms. ^[ref].

These roots are taken in an algebraic closure, not merely in ${\mathbb{F}}_q$. If the characteristic of a field is $p$ and $n=p^{a}m$ with $p\nmid m$, then $X^n-1=(X^m-1{)}^{p^a}$; hence there are $m$ distinct roots. They form a cyclic group by the finite multiplicative subgroup of a field is cyclic theorem. In particular they have order $n$ when $\gcd (n,q)=1$.

A group action on a nonempty set $X$ is simply transitive when exactly one group element sends any chosen point to any other:Equivalently it is both a transitive group action and a free group action; this is also a regular group action. Transitivity supplies a transport, and freeness makes it unique. Conversely, putting $y=x$ in the uniqueness condition shows that every stabilizer subgroup is $\{e\}$, since the identity element already fixes $x$. In a transitive group action, checking that just one stabilizer subgroup is trivial suffices: all the others are conjugate subgroups.

The cyclic group $C_3$ in the diagram under transitive group action is simply transitive. More generally, the left regular action of any group $G$ on itself is simply transitive: the equation $gx=y$ has the unique solution $g=y{x}^{-1}$, using the inverse element of $x$. This shows why the property is useful: every position has a unique label relative to a chosen starting position.

Choose a point $x_0\in X$. Its orbit mapis a bijection and an equivariant map from the left regular action: ${\varphi }_{x_0}(hg)=h\cdot {\varphi }_{x_0}(g)$. This identifies $X$ with a copy of $G$, but $X$ does not come with a preferred identity element. If $x_1=a\cdot x_0$, then ${\varphi }_{x_1}(g)={\varphi }_{x_0}(ga)$: changing the chosen point composes this identification with right multiplication by $a$. An old label $g$ therefore becomes the new label $g{a}^{-1}$, which also specifies the side of multiplication for a noncommutative group.

The transport from $x$ to $y$ remains independent of the chosen point. Writing $x=b\cdot x_0$ and $y=c\cdot x_0$, it is $c{b}^{-1}$. With $x_1=a\cdot x_0$, the new labels are $b{a}^{-1}$ and $c{a}^{-1}$, andFor a finite group, simple transitivity implies $|X|=|G|$. Conversely, a transitive group action with $|X|=|G|<\infty$ is simply transitive by the orbit-stabilizer theorem. Equality of cardinalities alone is insufficient without transitivity: the cyclic group $C_2$ can fix both points of a two-point set, giving two group orbits and two possible self-transports at each point.

For a group action of a group $G$ on a set $X$, the stabilizer subgroup of $x\in X$ consists of the transformations that leave $x$ fixed:It is a subgroup: the identity element fixes $x$, products of transformations fixing $x$ fix $x$, and the inverse element of a transformation fixing $x$ also fixes $x$. A stabilizer subgroup is a subset of $G$, whereas the orbit of a group action is a subset of $X$. See Ackerman's algebra notes, definition 0.4.1.2.

For the square group actions defined under group action, direct calculation givesThe reflection $s$ fixes $A$ and $C$ individually. The reflection $rs$ preserves the unoriented side of a polygon $AB$ by exchanging its two vertices of a polygon. The half-turn rotation $r^2$ exchanges the two ends of the unoriented diagonal of a polygon $AC$ while preserving that diagonal of a polygon. Thus fixing a side of a polygon or diagonal of a polygon as an object does not mean fixing every point on it; the latter condition uses a pointwise stabilizer. The four stabilizer subgroups have group orders $2,2,4,8$, matching group orbit sizes $4,4,2,1$ through the orbit-stabilizer theorem.

A stabilizer subgroup records the transformations still available after imposing a constraint. Fixing a vertex of a polygon of the square leaves two choices; fixing its centroid leaves all eight. In a faithful group action, a stabilizer subgroup can still contain a nonidentity transformation, as $s\in (D_4{)}_A$ shows: a faithful group action requires that only the identity element fix every point of $X$ simultaneously, not that every individual stabilizer subgroup be trivial. The two-point cyclic group example under group action has both stabilizer subgroups equal to the group kernel $\{0,2\}$ of its associated group homomorphism.