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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.

If a group $K$ acts on a finite set $X$, it acts on the direct product of groups $H^X$ by permuting its factors. The resulting semidirect product is the permutation wreath product. For $K=S_m$ acting on $m$ two-point blocks and $H=C_2$, it describes independent within-block swaps combined with a permutation of blocks, and has order $2^{m}m!$. The action on $X$ is part of the definition; it need not be the regular action used in a regular wreath product.

The restricted direct sum of a family of groups $G_i$ is the subgroup of their direct product of groups consisting of tuples with only finitely many nonidentity coordinates. Multiplication is coordinatewise. If every factor is a finite group, each subgroup that is a finitely generated group lies in a finite subproduct, so the restricted direct sum is a locally finite group. This gives concrete infinite torsion groups without a uniform bound on element orders.

If a group homomorphism section $i:G/N\to G$ satisfies $\pi i=\mathrm{id}$, then each $g\in G$ has a unique form $i(q)n$ with $q\in G/N$ and $n\in N$: take $q=\pi (g)$ and $n=i(q{)}^{-1}g$. Thus $(q,n)\mapsto i(q)n$ is a bijection. It is a homomorphism for the ordinary direct product of groups exactly when the section image centralizes $N$. In general it describes a semidirect product with a nontrivial conjugation action.