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Let $p<q$ be distinct prime numbers. Every group of group order $p^2{q}^2$ is Abelian if and only if $p$ does not divide $q^2-1$.

For sufficiency, the Sylow theorems make the number of Sylow p-subgroups one of $1,q,q^2$, congruent to one modulo $p$. The divisibility hypothesis excludes both nontrivial choices, so that subgroup is normal. The number of Sylow subgroups for $q$ is one of $1,p,p^2$, congruent to one modulo $q$. The choice $p$ is impossible because $p<q$. If $q$ divided $p^2-1=(p-1)(p+1)$, primality and $q>p$ would force $q=p+1$, hence $(p,q)=(2,3)$; this pair violates the hypothesis. Thus the other Sylow subgroup is normal too. Their intersection is trivial, and their group commutators lie in that intersection, so the group is their direct product of groups. A group of prime-square order is abelian: its nontrivial center of a group is either the whole group or has cyclic quotient, and a cyclic central quotient also forces commutativity. Both factors are therefore abelian.

For necessity, suppose $p\mid q^2-1$. The multiplicative group of a finite field is cyclic, so ${\mathbb{F}}_{q^2}^{\times }$ contains $\alpha$ of order $p$. Let the cyclic group $C_{p^2}$ act on the additive group of ${\mathbb{F}}_{q^2}$ by mapping its generator to multiplication by $\alpha$. This is a well-defined nontrivial group action, since ${\alpha }^{p^2}=1$. The resulting semidirect product has order $p^2{q}^2$ and is nonabelian, because the generator does not commute with a nonzero field element. Thus a nonabelian group of this order exists exactly when $p\mid q^2-1$.

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 $C_8\times C_2$.

For vertices $(\pm 1,\pm 1,\pm 1)$, the symmetry group of a cube consists of signed permutation matrices. A matrix preserving the unoriented line through $(1,1,1)$ must have all three signs equal, so it has the form $\epsilon P$, where $P$ is a permutation matrix and $\epsilon \in \{1,-1\}$. Central inversion commutes with every $P$ and is not itself a permutation matrix. Consequently $(P,\epsilon )\mapsto \epsilon P$ identifies this twelve-element stabilizer subgroup with the direct product of groups $S_3\times C_2$. A body diagonal is different from a face-normal coordinate axis, whose line stabilizer has order sixteen.

Let $W$ be the subgroup fixed by the Coleman norm operator. The Coleman norm-fixed lift gives the unique factorization $f=w(\overline{f})v$, with $v\in 1+pR$. Multiplication consequently gives the displayed direct product of groups, while reduction identifies $W$ with ${\mathbb{F}}_p[[T]{]}^{\times }$. This is a decomposition of unit groups, not an identification of a power-series ring with a product of groups.

For an integer $n>1$, the Carmichael function $\lambda (n)$ is the exponent of a finite group for the unit group $U(n)=(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$: the least positive integer $m$ for whichHere the greatest common divisor condition selects the units. Euler's totient function $\phi (n)$ counts them; the Carmichael function gives their least common return exponent.^[ref]

Modulo eight, the units are $1,3,5,7$, and their squares are all congruent to one. Their multiplicative orders are $1,2,2,2$, so $\lambda (8)=2$ although $\phi (8)=4$. Modulo fifteen, the units $1,2,4,7,8,11,13,14$ have multiplicative orders $1,4,2,4,4,2,4,2$. For example, the powers of two return through $1,2,4,8,1$. Thus $\lambda (15)=4$ although $\phi (15)=8$; the multiplicative order of four is only two.

The three quantities are related byA common return exponent must be a multiple of every multiplicative order, hence of their least common multiple. Conversely, that least common multiple makes every unit return to one. Lagrange's theorem makes each multiplicative order divide the group order $\phi (n)$, proving the final divisibility. The Carmichael function applies to all units at once, while the multiplicative order concerns a chosen unit.

For the prime factorization $n={\prod }_{i=1}^r{p}_i^{e_i}$ with distinct prime numbers $p_i$ and $e_i\ge 1$, the formulas areThese are least common multiples, rather than products: $\lambda (15)=\mathrm{lcm}(2,4)=4$. The convention $\lambda (1)=1$ handles the trivial case separately.^[ref]

The Chinese remainder theorem for unit groups explains the least common multiple: a power returns to one in every prime power component exactly when it is a multiple of each component's exponent of a finite group. For an odd prime power, the unit group is a cyclic group of group order $p^{e-1}(p-1)$. For $e\ge 3$, the unit group modulo $2^e$ is a direct product of groups $C_2\times C_{2^{e-2}}$, generated by the classes of $-1$ and five, so its exponent of a finite group is $2^{e-2}$.^[ref]

This gives a uniform reduction for modular exponentiation: for a unit $a$ and a nonnegative integer $k$, reduce $k$ modulo $\lambda (n)$. For example, $2^{1001}\equiv 2(\mathrm{mod}15)$ because $1001\equiv 1(\mathrm{mod}4)$. The unit hypothesis matters: $3^2\equiv 1(\mathrm{mod}8)$, but $2^2\not\equiv 1(\mathrm{mod}8)$. Computing the formula assumes the prime factorization is available. The same common-exponent viewpoint explains Carmichael numbers: a composite number $n$ passes the Fermat primality test for every unit base exactly when $\lambda (n)\mid n-1$. For example, $\lambda (561)=\mathrm{lcm}(2,10,16)=80\mid 560$.

The category of groups has groups as objects and group homomorphisms as morphisms. Its product is the direct product of groups, and its coproduct is the free product, as expressed by the universal property of a free product.

Category theory studies mathematical structures through the morphisms between them. A category specifies which objects of a category and arrows are allowed; a functor translates these objects and arrows into another category; a natural transformation compares two such translations coherently. For example, the category of groups records both groups and group homomorphisms, so it remembers which maps respect multiplication.

The usefulness appears in constructions we already know. A Cartesian product combines two sets, while a direct product of groups combines two groups. Both satisfy the same universal property of a categorical product: any pair of maps into the factors comes from exactly one map into the product. This gives a reusable way to construct maps and establish isomorphisms, without depending on a particular choice of coordinates or element labels. In a partially ordered set regarded as a category, the corresponding categorical product is a greatest lower bound. The shared definition exposes a connection between pairing data and finding the best common bound.

Start with the computed examples under category, functor, natural transformation, and product in a category. Their arrow equations are practical tests: do both routes in a commutative diagram produce the same result? Later, the same language organizes tensor products, adjunctions, and the fundamental group of a topological space. Emily Riehl's textbook develops these basic notions before using them to compare constructions across different subjects. ^[ref]

In their 1945 paper, Samuel Eilenberg and Saunders Mac Lane set out the notions of category, functor, and natural transformation. The introduction uses the relation between a vector space and its dual space to ask which comparisons are natural, setting up a precise language for compatibility with maps. ^[ref]

The nontrivial center of a finite p-group theorem implies that the center of a group of order $p^2$ has order $p$ or $p^2$. If it has order $p$, the quotient by the center has prime order and is cyclic; a cyclic quotient by the center forces the whole group to be abelian, contradicting the assumed size of its center. Thus the group is abelian. An element of order $p^2$ makes it cyclic. Otherwise choose $a\ne 1$ and $b\notin ⟨a⟩$; their order-$p$ cyclic subgroups intersect trivially and generate the direct product of groups $C_p\times C_p$. The two possibilities are nonisomorphic because only the cyclic one has an element of order $p^2$.

Let $r$ have order $2m$ with $m$ odd, and let $s^2=1$, $srs=r^{-1}$. The subgroup $H=⟨r^2,s⟩$ is a dihedral group of order $2m$, while $K=⟨r^m⟩$ is a central cyclic group of order two. Since $m$ is odd, $H\cap K=\{1\}$ and $r$ lies in $HK$ by Bézout's identity. Thus the original group of order $4m$ is the direct product of groups $H\times K$. This statement specifies group orders and avoids the two competing conventions for dihedral notation.

The symmetry group of a cube splits into its central inversion and its rotational symmetry group of a cube. Inversion has determinant $-1$, the rotation subgroup has determinant $1$, and together they generate the full group with trivial intersection. The rotation subgroup acts faithfully on the four body diagonals and is isomorphic to $S_4$. Consequently the full group is this direct product of groups.

This abelian group has eight elements $c^r{a}^s$, with $r$ taken modulo four and $s$ modulo two. Its group operation adds these exponent pairs separately. The elements of order two are $c^2,a,c^{2}a$, and the other four nonidentity elements $c,c^3,ca,c^{3}a$ have order four. It is a direct product of groups rather than a cyclic group, since no element has order eight.

Two subgroups give an exact factorization of a group when every element has a unique expression $h_1{h}_2$ with $h_i\in H_i$. Existence and trivial intersection are equivalent to this uniqueness: if $h_1{h}_2=k_1{k}_2$, then $k_1^{-1}{h}_1=k_2{h}_2^{-1}\in H_1\cap H_2$, so both sides are the identity. Conversely, an element $a$ in the intersection gives the two factorizations $a\cdot e=e\cdot a$, forcing $a=e$. For finite groups the multiplication map $H_1\times H_2\to G$ is therefore a bijection and $|G|=|H_1||H_2|$. It need not be a group homomorphism: an exact factorization is not automatically a direct product of groups. If $H_1$ is normal, it gives an internal semidirect product.

The only nontrivial proper normal subgroups of this direct product of groups are its two factors. Indeed a normal subgroup meeting $S\times 1$ nontrivially contains it by the simple group property. Otherwise each of its commutators with $S\times 1$ lies in that trivial intersection. Its first coordinates therefore lie in $Z(S)=1$, so it is contained in $1\times C_p$. Both factors are simple and each quotient by one factor is isomorphic to the other. Hence precisely two composition series occur, one through each factor. This is compatible with the Jordan–Hölder theorem, since the two series have the same simple factors in opposite orders. For $S=A_5,p=2$, the group has order $120$.

A finite abelian group combines a finite set of reversible operations with commutativity. All finite cyclic groups qualify, but direct products of groups supply examples that need more than one generator of a group. The Fundamental theorem of finitely generated abelian groups says that every finite abelian group is a direct product of groups of finite cyclic groups; the finite case has no infinite cyclic factors.^[ref]

These factors distinguish different ways to have the same total number of states. A single four-position dial gives $C_4$, while two independent binary switches give $C_2\times C_2$. Both have four states and commutative composition, yet only the dial permits a single operation whose repetition visits all four states. The following Cayley tables make that distinction visible without relying on the labels alone.

A finite group is nilpotent exactly when it is the direct product of groups of its Sylow subgroups. The normalizer condition for nilpotent groups and Sylow conjugacy force each Sylow normalizer to be the whole group. Distinct normal Sylow subgroups commute because their commutators lie in their trivial intersection. Conversely finite p-groups are nilpotent by the nontrivial center and induction; direct products preserve nilpotence.

An internal direct product finds independent coordinates already present inside a group. Let $H,K$ be normal subgroups of $G$ withThen the multiplication mapis a group isomorphism. Thus every element of $G$ has a unique factorization into one element of each subgroup. Conversely, the embedded factors $G\times \{e_H\}$ and $\{e_G\}\times H$ inside an external direct product of groups satisfy these conditions.

For a concrete example, take the additive cyclic group $C_6=\mathbb{Z}/6\mathbb{Z}$ and its subgroups $H=\{0,3\}$ and $K=\{0,2,4\}$. They are normal subgroups because $C_6$ is an abelian group. Their intersection is $\{0\}$, and the sums $h+k$ give every residue class exactly once. For instance, $1=3+4\mathrm{mod}6$. The table makes the unique decomposition visible.

The proof explains each hypothesis. Normality puts the group commutator $[h,k]=h^{-1}{k}^{-1}hk$ in both $H$ and $K$, so it is the identity element. Therefore the factors commute, and $\Phi$ preserves the group operation. The condition $HK=G$ makes it surjective. If $hk=e$, then $h=k^{-1}$ belongs to the intersection, forcing both to be the identity element; thus its group kernel is trivial and it is injective.

Normality cannot simply be omitted. In the symmetric group $S_3$, take $H=⟨(123)⟩$ and $K=⟨(12)⟩$. Their intersection is trivial and $HK=S_3$, but $K$ is not a normal subgroup. With permutations composed right to left, $(123)(12)=(13)$ while $(12)(123)=(23)$, so the factors do not commute. The setwise decomposition is unique, but its multiplication map is not a group homomorphism from $H\times K$. An independent-coordinate interpretation requires the group operations to agree, not merely a bijection of the underlying sets.

Group commutators are taken coordinatewise in a direct product of groups, giving ${\Gamma }_j(A\times B)={\Gamma }_j(A)\times {\Gamma }_j(B)$. A finite direct product of nilpotent groups is nilpotent, with class the maximum of the factor classes.

// Target: algebra.bigb

If finite normal subgroups $H,K$ of a group have coprime orders, their intersection is trivial by Lagrange's theorem. For $h\in H$ and $k\in K$, the commutator $hk{h}^{-1}{k}^{-1}$ belongs to both subgroups: write it as $h(k{h}^{-1}{k}^{-1})$ to use normality of $H$, and as $(hk{h}^{-1})k^{-1}$ to use normality of $K$. Hence it is the identity and $hk=kh$. The product subgroup is therefore isomorphic to the direct product of groups $H\times K$.

In odd dimension, the map $A\mapsto (\det A)A$ is a surjective group homomorphism from the orthogonal group to the special orthogonal group, with kernel of a group homomorphism $\{I,-I\}$. Scalar matrices commute, and $\det [(\det A)A]=(\det A{)}^{2m+2}=1$. Together with the determinant sign, it identifies $O(2m+1)$ with the direct product of groups $SO(2m+1)\times C_2$. In even dimension, scalar multiplication by $\det A$ does not repair the determinant, so this construction does not give the same splitting.

If $|G|=p^{a}m$ with $p\nmid m$, a Sylow p-subgroup has order $p^a$. The Sylow theorems say that such subgroups exist; every p-subgroup is contained in one; they are all conjugate; and their number $n_p$ divides $m$ and satisfies $n_p\equiv 1(\mathrm{mod}p)$.

Suppose $|G|=p^{2}q$ with distinct primes and $G$ were simple. A Sylow subgroup of order $q$ has order $q$, and $n_q>1$ divides $p^2$, so $n_q=p$ or $p^2$. If $n_q=p$, conjugation gives a nontrivial homomorphism $G\to S_p$. Its kernel is normal and, by simplicity, trivial. Hence $p^{2}q$ would divide $p!$ by Lagrange's theorem, impossible because $p!$ has only one factor of $p$.

If $n_q=p^2$, distinct Sylow subgroups of order $q$ meet only in the identity and together contain $p^2(q-1)$ nonidentity elements. Only $p^2$ elements of $G$ remain, including the identity. Every Sylow p-subgroup lies entirely in that remaining set and itself has $p^2$ elements. They must therefore all be the same subgroup, making it normal, again a contradiction. Thus no group of order $p^{2}q$ is simple. If $p=q$, the preceding p-group argument applies.

For distinct $p,q$, a finite Abelian group is the direct product of groups of its primary subgroups. A group of order $p^2$ is either cyclic or has every nonidentity element of order $p$, in which case choosing two independent elements gives $C_p\times C_p$. Consequently the possibilities areIf the primes coincide, the Fundamental theorem of finitely generated abelian groups gives instead $C_{p^3}$, $C_{p^2}\times C_p$, and $C_p^3$. This classification is used only for the requested list, not for the proof above.