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A nilpotent group has a finite central seriesEquivalently the upper central series, defined by $Z_{j+1}/Z_j=Z(G/Z_j)$, reaches $G$.

We first prove the normalizer condition for nilpotent groups. If $H<G$, choose the least $j$ with $Z_j⊈H$. Then $Z_{j-1}\le H$, and an element $z\in Z_j\setminus H$ satisfies $[z,H]\le Z_{j-1}\le H$. Thus $z$ normalizes $H$, giving $H<N_G(H)$.

Let $P$ be a Sylow subgroup of a finite nilpotent $G$, and write $N=N_G(P)$. If $N<G$, the normalizer condition gives some $g\in N_G(N)\setminus N$. The subgroups $P$ and $P^g$ are both Sylow subgroups of $N$, so Sylow theorem supplies $n\in N$ with $P^{gn}=P$. But then $gn\in N_G(P)=N$, forcing $g\in N$, a contradiction. Therefore every Sylow subgroup is a normal subgroup of $G$.

For normal Sylow subgroups $P,Q$ at different primes, $[P,Q]\le P\cap Q=1$. They commute, have coprime orders, and their product has order $|G|$. Multiplication consequently gives an isomorphism from their external direct product of groups onto $G$.

Conversely, a nontrivial finite p-group has nontrivial center of a group: its class equation has every noncentral conjugacy-class size divisible by $p$, so $|Z(G)|$ is a positive multiple of $p$. Induction on order, applied to $G/Z(G)$, gives a central series for $G$. Finite direct products of these central series give a central series for the product, padding shorter series by the full factor. Thus the finite nilpotent group decomposition is

Lagrange's theorem states that for a subgroup $L$ of a finite group $G$, $|G|=[G:L]|L|$. In particular $|L|$ divides $|G|$; the equal-sized cosets partition $G$.

Apply the theorem to $H\cap K$ inside both subgroups. Its order divides both coprime orders, so $H\cap K=\{1\}$. For $h\in H$ and $k\in K$, the group commutator $hk{h}^{-1}{k}^{-1}$ lies in $K$ by $K$ being a normal subgroup, and in $H$ because $k{h}^{-1}{k}^{-1}\in H$. It is therefore the identity. Thus every element of $H$ commutes with every element of $K$.

Define $\theta :H\times K\to G$ by $\theta (h,k)=hk$. The cross-commutation just proved gives $\theta (h,k)\theta (h^{\prime },k^{\prime })=h{h}^{\prime }k{k}^{\prime }=\theta (h{h}^{\prime },k{k}^{\prime })$, so this is a group homomorphism from the direct product of groups. The product assumption makes it surjective. If $hk=1$, then $h=k^{-1}\in H\cap K$, so both entries are the identity and the kernel of a group homomorphism is trivial. HenceThis proves the internal direct product theorem in the present coprime-order setting, rather than merely matching the numbers of elements.

Place the cube at the origin with face normals $\pm e_1,\pm e_2,\pm e_3$, and label opposite faces by $(1,2),(3,4),(5,6)$. Every cube symmetry is a signed permutation matrix: there are $2^{3}3!=48$ choices. Its group action on faces is faithful because fixing all faces fixes all their normal directions. Central inversion $-I$ exchanges every opposite pair and therefore acts as $h$. Since $-I$ commutes with every linear cube symmetry, the face-action image lies in $C_{S_6}(h)$. The preceding count gives $|C_{S_6}(h)|=48$, so the faithful face action is onto this centraliser:For the remaining isomorphism, let $R$ be the rotational symmetry group of a cube. Half the signed permutation matrices have determinant $1$, so $|R|=24$. Central inversion has determinant $-1$, and every orientation-reversing symmetry is $(-I)r$ for a unique $r\in R$. Since $-I$ is central and $R\cap \{I,-I\}=\{I\}$, multiplication gives a direct product of groups:The group action of $R$ on the four body diagonals is faithful. To prove this, choose direction vectors $(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)$. They span ${\mathbb{R}}^3$, and their only linear relation is that their sum is zero. A rotation fixing all four diagonal lines sends each listed vector to itself or its negative. Applying the relation forces all four signs to agree. The all-negative choice is $-I$, which is not in $R$, so the rotation is the identity. Thus $R$ embeds in $S_4$; both have order $24$, giving $R\cong S_4$. Finally $C_{S_6}(g)\cong C_2\times S_4$ from part (i). ConsequentlyThe direct-product decomposition of the cube symmetry group supplies the second isomorphism, while the first comes from the specified face action.

If a nonidentity power $x^j$ belongs to $H$, reduce $j$ modulo $p$ so that $1\le j<p$. There is an integer $a$ with $aj\equiv 1(\mathrm{mod}p)$, and therefore $(x^j{)}^a=x\in H$, a contradiction. ThusNow suppose that $G$ is abelian and finite. Start with $H_0=\{1\}$, and whenever $H_k\ne G$, choose $x_{k+1}\notin H_k$. The multiplication mapis a group homomorphism because all elements commute. It is surjective by construction and injective because the intersection of its two factors is trivial: $hx=h^{\prime }{x}^{\prime }$ implies $h^{\prime -1}h=x^{\prime }{x}^{-1}$ lies in that intersection. Hence $H_{k+1}\cong H_k\times C_p$, and its order is $p$ times the previous order. Strict growth and finiteness ensure that the construction ends at $G$. Thereforewhere $n$ is the number of factors, possibly zero. This is an elementary abelian group, obtained by an explicit direct product of groups construction.

Take $G=S_3$ and its normal subgroup $N=A_3=⟨r⟩$, where $r=(123)$. Normality follows either from the sign of a permutation or directly from conjugation sending $r$ to $r$ or $r^{-1}$. The quotient group has order two, and $t=(12)$ provides a section $i:C_2\to S_3$ because $t^2=1$ and $tN$ is its nonidentity coset.

However, the map $\Phi (q,n)=i(q)n$ from the direct product of groups is not a group homomorphism. In its domain $(1,r)$ commutes with $(tN,1)$, whereas their images $r$ and $t$ do not commute: $tr{t}^{-1}=r^{-1}\ne r$. This is a split group extension giving a genuine semidirect product instead of a direct product. The pair $S_3,A_3$ supplies the required split but non-direct example.

For the general bijectivity assertion, let $\pi :G\to G/N$ be the quotient projection and assume $\pi \circ i$ is the identity. Given $g\in G$, put $q=\pi (g)$ and $n=i(q{)}^{-1}g$. Then $\pi (n)=q^{-1}q=1$, so $n\in N$ and $g=i(q)n$. This proves surjectivity of $\Phi$. If $i(q)n=i(q^{\prime })n^{\prime }$, applying $\pi$ gives $q=q^{\prime }$, after which cancellation gives $n=n^{\prime }$. ThusThis section normal form for a split group extension does not require finiteness. The obstruction to its being a homomorphism is the nontrivial conjugation action of the section on $N$.

Take $G=C_2\times C_2$ and $N=C_2\times \{1\}$. Both are nontrivial, and $N$ is normal because $G$ is abelian. The second-coordinate projection identifies the quotient group $G/N$ with $C_2$. Swapping the two coordinates identifies $(G/N)\times N$ with $G$, soThis is a direct product of groups with its first factor as the specified normal subgroup.

A soluble group, also called a solvable group, has a terminating derived series:For a subgroup $S\le G$, induction gives $S^{(j)}\le G^{(j)}$, so subgroups of soluble groups are soluble. For a surjective group homomorphism $q:G\to Q$, $q(G^{(j)})=Q^{(j)}$, so quotients of soluble groups are soluble. Finally, in a group extension $1\to K\to E\to Q\to 1$, suppose $K^{(r)}=1$ and $Q^{(s)}=1$. Then $E^{(s)}\le K$ and $E^{(s+r)}=1$. Soluble groups are closed under subgroups, quotients and group extensions.

A virtually soluble group contains a soluble group as a finite-index subgroup. The finite-index facts proved in parts (i)–(iii) imply closure under subgroups and quotients: intersect a finite-index soluble subgroup with the chosen subgroup, or take its image under the quotient map.

For group extensions, no finite-generation hypothesis may be inserted. We first establish the finite-index characteristic soluble subgroup lemma. If $K$ is a virtually soluble group, the kernel of its action on the cosets of a finite-index soluble subgroup is a soluble normal subgroup $K_0$ of finite index. Among soluble normal subgroups containing $K_0$, choose $R$ with maximal $|R/K_0|$, possible because $K/K_0$ is finite. If $S$ is any soluble normal subgroup of $K$, then $RS$ is soluble: it is an extension of $R$ by $S/(R\cap S)$. Maximality forces $S\le R$. Thus $R$ is the unique largest soluble normal subgroup of $K$, making it a characteristic subgroup, and it has finite index.

Now suppose $1\to K\to G\stackrel{\pi }{\to }Q\to 1$ has both $K$ and $Q$ virtually soluble. Replace $G$ by the preimage $G_0$ of a finite-index soluble subgroup of $Q$. The subgroup $R$ just constructed is characteristic in $K$ and therefore normal in $G_0$. In $E=G_0/R$, the subgroup $A=K/R$ is a finite normal subgroup, and $E/A$ is a soluble group. The centralizer $C_E(A)$ has finite index in $E$, since conjugation gives a map $E\to \mathrm{Aut}(A)$ with finite image. Its intersection with $A$ is the center of a group $Z(A)$, an abelian group, while its quotient by $Z(A)$ embeds in the soluble group $E/A$. Thus $C_E(A)$ is a soluble group. Its preimage in $G_0$ is an extension by $R$, so it too is soluble and has finite index in $G$. Virtually soluble groups are closed under group extensions.

For the final example take the restricted direct sum of groupswhere $A_5$ is the nonabelian simple group of even permutations on five letters. Every finite collection of elements lies in a product of finitely many finite factors, so $D$ is a locally finite group and hence a torsion group. It cannot contain a nonabelian free group, which is a torsion-free group.

To show that $D$ is not a virtually soluble group, let $H$ be any finite-index subgroup and let $N\le H$ be the kernel of the finite coset action. Each coordinate $A_5$ maps either injectively or trivially into the finite quotient $D/N$, by Simplicity of the alternating group A5. The nontrivial images of distinct factors commute, and each has trivial centre, so any $k$ of them generate a direct product of groups of order $60^k$. Only finitely many such images can occur in a finite quotient. Therefore $N$, and hence $H$, contains a whole coordinate copy of $A_5$, which is not soluble: its nontrivial commutator subgroup is normal and therefore equals $A_5$. No finite-index subgroup of $D$ is soluble.

In a direct product of groups, $(a,b{)}^r=(a^r,b^r)$. Thus a finite order for $(a,b)$ makes both coordinate orders finite. Conversely, if their orders are $p$ and $q$, the least positive exponent annihilating both is their least common multiple. Hencewhen both orders are finite; if either coordinate has infinite order, so does $(a,b)$.

Every finitely presented group $P$ embeds in $P\times C_2$ by $p\mapsto (p,1)$. This direct product of groups is finitely presented: combine finite presentations of the factors and add the finitely many cross-commutation relations between their generators. Projection onto $C_2$ gives a nontrivial finite quotient of a group.

Thus no finitely presented $P$ can serve as the forbidden witness $A_{-}$ for this property. Hence

A subgroup $H$ of a group $G$ is a normal subgroup when $gH{g}^{-1}=H$ for every $g\in G$, equivalently $gH=Hg$. The quotient group $G/H$ has the left cosets as elements, with $(gH)(kH)=gkH$, identity $H$ and inverse $(gH{)}^{-1}=g^{-1}H$. Normality makes this multiplication independent of representatives.

The first isomorphism theorem for groups says that for a group homomorphism $\phi :G\to K$, its kernel of a group homomorphism is normal andThe image is taken with the induced group operation; surjectivity onto all of $K$ is not required.

For the given upper triangular matrices, take diagonal entries:The target is a direct product of groups, with multiplication of nonzero real numbers in each coordinate. Indeed the product of two such matrices has diagonal $(a{a}^{\prime },d{d}^{\prime })$, proving the group homomorphism property. Any pair of nonzero diagonal entries is attained by a diagonal matrix, so $\phi$ is surjective. Its kernel of a group homomorphism consists precisely of the matrices with $a=d=1$, namely $H$. Hencewith coordinatewise multiplication, not addition on ${\mathbb{R}}^2$.

Let $a$ and $b$ generate $C_m$ and $C_n$. The order of $(a,b)$ in the direct product of groups isWhen $\gcd (m,n)=1$, this is $mn=|C_m\times C_n|$. Hence $(a,b)$ generates the whole product and

For subgroups $H,P\le G$, define $x\sim y$ when $y\in HxP$. The identity elements show reflexivity, inverses show symmetry, and multiplication shows transitivity. Thus the double cosets $HxP$ partition $G$.

Let $H$ act by left multiplication on the set of left cosets $G/P$. The orbit of $xP$ consists of the cosets contained in $HxP$, so its size is $|HxP|/|P|$. Its stabilizer isThe orbit-stabilizer theorem therefore gives

Suppose $P$ is a Sylow subgroup of $G$, with $|P|=p^a$, and write the largest power of $p$ dividing $|H|$ as $p^b$. If every $H\cap xP{x}^{-1}$ had order at most $p^{b-1}$, the displayed formula would make every double-coset size divisible by $p^{a+1}$. Their sum $|G|$ would then also be divisible by $p^{a+1}$, contradicting the choice of $P$. Hence some $H\cap xP{x}^{-1}$ has order $p^b$ and is a Sylow $p$-subgroup of $H$. This is the Sylow subgroup of a subgroup from double cosets argument.

To count the general linear group over a finite field $G{L}_n({\mathbb{F}}_p)$, choose its columns successively. There are $p^n-1$ choices for the first, $p^n-p$ for the second, and $p^n-p^k$ for column $k+1$. ConsequentlyNone of the factors $p^j-1$ is divisible by $p$, so the upper unitriangular group is a Sylow $p$-subgroup: it has one arbitrary field entry in each of the $n(n-1)/2$ positions above the diagonal and hence order $p^{n(n-1)/2}$. The permutation matrices form a subgroup isomorphic to the symmetric group $S_n$.

By Cayley theorem, every finite group $G$ embeds in $S_{|G|}$, and permutation matrices embed this symmetric group in $G{L}_{|G|}({\mathbb{F}}_p)$. The latter has the explicit Sylow $p$-subgroup just described, so the result proved in the first part, applied to the embedded copy of $G$, proves that every finite group has a Sylow $p$-subgroup.

The counting part of the Sylow theorems says that if $p^a$ is the largest power of $p$ dividing $|G|$, then the number $n_p$ of Sylow $p$-subgroups satisfies

Finally, let $|G|=pq$ with prime numbers $p>q$. The Sylow counts giveso $n_p=1$ and the Sylow $p$-subgroup is a normal subgroup. Also $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. If $n_q=1$, both Sylow subgroups are normal; their elements commute, so $G$ is their direct product and is abelian. In the nonabelian case one must therefore have $n_q=p$, whenceor equivalently $q\mid p-1$, as recorded by nonabelian group of order pq.

Choose distinct $a,b\ne e$, possible because $|G|>2$. By iii, $a$ and $b$ commute, and $a^2=b^2=e$. The four elementsare distinct: $ab=e$ would give $a=b$, while $ab=a$ or $ab=b$ would make $b=e$ or $a=e$. Commutativity and $a^2=b^2=e$ show that products stay in $H$, and every element is its own inverse. More explicitly,is a group isomorphism, with addition of exponents modulo two. Hence $G$ contains a subgroup isomorphic to the Klein four-group.

The converse is false. The direct product of groups $C_2\times C_2\times C_3$ contains $C_2\times C_2\times \{e\}$, but it has elements of order of a group element three. Thus the existence of the specified subgroup does not force every nonidentity element to have order of a group element two.

Acting separately on disjoint sets of four and eight symbols embeds the direct product of groups $S_4\times S_8$ into $S_{12}$. The action is injective since fixing all twelve symbols forces both factors to be identities.

For the lower bound, $S_4$ contains a cycle of order four and $S_8$ contains disjoint cycles of lengths five and three, whose product has order fifteen. The corresponding element of $S_4\times S_8$ has order of a group element $\mathrm{lcm}(4,15)=60$. Every isomorphic subgroup of $S_n$ must therefore contain a permutation of order sixty. We can bound the minimum number of symbols for a permutation of prescribed order directly here.

In the disjoint cycle decomposition of such a permutation, all cycle lengths divide $60$, and their least common multiple must contain the factors $4$, $3$, and $5$. If these factors occur in three distinct cycles, those cycles use at least $4+3+5=12$ symbols. If two occur in the same cycle, that cycle has length at least their product: combining $4,3$ costs at least $12$, combining $4,5$ costs at least $20$, and combining $3,5$ costs at least $15$. Each possibility already requires at least twelve symbols. Thus $n\ge 12$, and

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

Cosets and Lagrange. A left coset is $gH=\{gh:h\in H\}$. The left cosets of a subgroup partition a group, each with $|H|$ elements. Lagrange's theorem states $\boxed{|G|=[G:H]|H|}$ for a finite group, so every subgroup order divides the group order.

The 64-element group. Work in the commutative ring $R=\mathbb{Z}/4\mathbb{Z}$, which has zero divisors. Denote the displayed lower triangular matrix by $A(a,b,c)$, where $a,b,c$ are its three entries below the diagonal. Direct matrix multiplication givesAll arithmetic is modulo four, so the set is closed. Associativity comes from matrix multiplication, the identity element is $A(0,0,0)$, andThe formula works in either multiplication order and requires no division. Hence these matrices form a group of order $\boxed{|G|=4^3=64}$. This is the lower triangular version of the Heisenberg group over a commutative ring.

The abelian subgroup. If $a=c$, write $h(a,b)=A(a,b,a)$. ThenThe identity is $h(0,0)$ and the inverse is $h(-a,a^2-b)$, proving an abelian subgroup. Its two independent entries give $\boxed{|H|=16}$ and $\boxed{[G:H]=4}$, in accordance with Lagrange's theorem. This is the abelian plane subgroup of the Heisenberg group modulo four.

More explicitly, $A(a,b,c)\mapsto a-c$ is a surjective group homomorphism to the additive group of $R$, with kernel $H$. The four left cosets are exactly the four sets with fixed $a-c=t$, $t=0,1,2,3$; representatives are $A(t,0,0)$. Thus $H$ is also a normal subgroup.

Matrices squaring to the identity. The product formula givesThe first and third coordinates vanish only when $a,c\in \{0,2\}$. Then $ca=0$ modulo four, so the middle coordinate vanishes exactly when $b\in \{0,2\}$. Therefore the complete set isIt is a subgroup: multiplying two such elements keeps all three entries even, and their cross term $cd$ vanishes modulo four, so the product simply adds the three even coordinates. The identity is included and each element is its own inverse. In fact $K$ is isomorphic to the direct product of groups $C_2\times C_2\times C_2$. The seven nonidentity elements are involutions.

Finally, the last subset in the PDF is defined by $a\ne 0\ne b$, with no restriction on $c$. It has $3\cdot 3\cdot 4=36$ elements, which cannot be a subgroup because $36$ does not divide $64$. It also excludes the identity element. Thus

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,

Write the cyclic factors additively. The order of a group element $(r,s)$ in a direct product of groups is the least common multiple of its component orders, since a power is the identity exactly when it is the identity in both coordinates.

In $C_2\times C_3$, $(1,1)$ has order $\mathrm{lcm}(2,3)=6$, so it generates all six pairs. An explicit group isomorphism isIt is well defined and multiplicative in additive notation; its kernel is zero because an integer divisible by both two and three is divisible by six. Equal finite orders then give bijectivity.

In $C_2\times C_4$, every element has order dividing four. A generator of $C_8$ has order eight, and isomorphisms preserve element orders, soFor positive integers $n,m$, the maximum element order in $C_n\times C_m$ is $\mathrm{lcm}(n,m)=nm/\gcd (n,m)$, attained by a pair of generators. The product is cyclic of order $nm$ exactly when that number is $nm$. Thus cyclicity of a product of two finite cyclic groups givesFor coprime factors the map $[k{]}_{nm}\mapsto ([k{]}_n,[k{]}_m)$ gives the isomorphism, the group form of the Chinese remainder theorem.

Assume (I). Existence of a decomposition is part of the hypothesis. If $h_1{h}_2=k_1{k}_2$ are two decompositions, multiply on the left by $k_1^{-1}$ and on the right by $h_2^{-1}$ to obtainThe left side belongs to $H_1$ and the right side to $H_2$, since they are subgroups. Their common value belongs to $H_1\cap H_2=\{e\}$, so $h_1=k_1$ and $h_2=k_2$. This proves uniqueness, hence (II).

Conversely, assume (II). Existence immediately gives $G=H_1{H}_2$. If $a\in H_1\cap H_2$, thenare two allowed factorizations, because the identity is in each subgroup. Uniqueness makes their first components equal, so $a=e$. Thus $H_1\cap H_2=\{e\}$ and (I) follows.

Therefore (I) and (II) are equivalent. This is an exact factorization of a group. For finite groups it gives $|G|=|H_1||H_2|$, but it does not by itself identify $G$ with a direct product of groups; multiplication between the two factors can be noncommutative.

By Lagrange's theorem, the possible values of the order of a group element areThese are possibilities across groups of order eight; they need not all occur in one group.

If every element has order at most two, every element equals its own inverse. Thus for any $x,y$,so $G$ is an abelian group. Choose $a\ne e$, and then $b\notin ⟨a⟩$. The four elements $e,a,b,ab$ are distinct and form a subgroup $H$: commutativity and $a^2=b^2=e$ prove closure and inverses. Choose $c\notin H$. The elements of $H$ and $Hc$ are disjoint, since $hc\in H$ would imply $c\in H$. These two four-element sets therefore exhaust $G$.

Every element consequently has a unique expression $a^i{b}^j{c}^k$, with $i,j,k\in \{0,1\}$. The map from triples to these products is bijective and respects multiplication by addition of exponents modulo two. Hencea direct product of groups and an elementary abelian group.