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A subgroup $N$ is a normal subgroup of $G$ if $gN{g}^{-1}=N$ for every $g\in G$. Define multiplication on its left cosets byTo check that this is well-defined, replace $g,h$ by $g{n}_1,h{n}_2$ with $n_1,n_2\in N$. Thenbecause normality gives $h^{-1}{n}_{1}h\in N$. Thus the product depends only on the two cosets. Associativity follows from that of $G$, the identity is $N$, and the inverse of $gN$ is $g^{-1}N$. These are all the group axioms, so $G/N$ is a group under coset multiplication. The natural projection $g\mapsto gN$ is a surjective group homomorphism with kernel $N$.

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 $N$.

Use additive notation and take $G=\mathbb{Z}/4\mathbb{Z}$, $N=\{0,2\}$. Its quotient group is $C_2$. A group homomorphism from $C_2$ sends its generator to an element annihilated by $2$, so the image must be either $0$ or $2$. Both project to the identity coset $N$, and neither can project to the nonidentity generator of $G/N$. Therefore this quotient projection has no homomorphic section. It is a nonsplit group extension, despite both $G$ and $N$ being nontrivial finite groups.

Unless a coefficient group is displayed, use integral singular cohomology. The standard CW complex structure on infinite-dimensional real projective space has one cell in each nonnegative dimension. Its cellular chain complex has boundary $d_j=2$ for positive even $j$ and $d_j=0$ for odd $j$. The cellular cohomology differential ${\delta }^j$ is therefore zero for even $j$ and multiplication by two for odd $j$. ConsequentlyMore generally, for an abelian group $A$, the positive odd groups are $A[2]=\{a:2a=0\}$ and the positive even groups are $A/2A$. In particular $H^j({\mathbb{RP}}^{\infty };{\mathbb{Z}}_2)={\mathbb{Z}}_2$ in every nonnegative degree.

For the required Bockstein homomorphism, use the short exact sequenceHere $m\ge 2$; for $m=1$ all three coefficient groups are zero. Since singular chains are free abelian groups, applying cochains gives a short exact sequence of cochain complexes. Its connecting homomorphism defines $\beta$ and its long exact sequence is precisely the required one, with the other maps induced by $\iota$ and $q$.

Explicitly, represent a class by a cocycle $u\in C^n(X;{\mathbb{Z}}_m)$ and choose a lift $\tilde{u}\in C^n(X;{\mathbb{Z}}_{m^2})$. Since $q(\delta \tilde{u})=0$, there is a unique cochain $v$ with $\iota (v)=\delta \tilde{u}$. Injectivity of $\iota$ and ${\delta }^2=0$ show $\delta v=0$. DefineChanging the lift by $\iota (w)$ changes $v$ by the coboundary $\delta w$. Changing the representative $u$ by a coboundary can be lifted by a coboundary as well and leaves the resulting class unchanged. Thus this is a well-defined group homomorphism, and the standard cochain lifting argument gives exactness.

Compute the Bockstein homomorphism on infinite-dimensional real projective space using its cellular cohomology complex. A generator with coefficients ${\mathbb{Z}}_2$ is represented by $1$ in degree $n$, lifted to $1\in {\mathbb{Z}}_4$. Its coboundary is $0$ for even $n$ and $2$ for odd $n$. Dividing via $\iota (1)=2$ givesThus the odd-degree maps are isomorphisms. The comparison between cellular cohomology and singular cohomology is natural with respect to coefficient maps, so this computes the same connecting homomorphism constructed above.

We use two precise facts about the formal group of an elliptic curve. At a prime $p$ of good reduction, its kernel of reduction of an elliptic curve is identified by the uniformizer $t=-x/y$ with the group $\hat{E}(p{\mathbb{Z}}_p)$. For odd $p$ this group is torsion-free. To see the second fact, the integral invariant differential of a formal group law has the form $(1+{\sum }_{j\ge 1}{c}_j{T}^j)dT$, with $c_j\in {\mathbb{Z}}_p$. Integrating constructs the formal logarithmIt is a group homomorphism to the additive group. For $0\ne t\in p{\mathbb{Z}}_p$ and $j\ge 2$,when $p$ is odd. Hence the series converges and $v_p({\log }_F(t))=v_p(t)$, so it is injective. The target has no nonzero torsion. Therefore reduction is injective on the entire rational torsion subgroup at an odd prime of good reduction, including its $p$-primary part.

For the present elliptic curve, $\Delta =64D^6$, so every odd $p\nmid D$ is a prime of good reduction. If $p\equiv 3(\mathrm{mod}4)$, the Legendre symbol of $-1$ is $-1$. The values $x$ and $-x$ cancel in the sum of the Legendre symbols of $x^3-D^{2}x$, yieldingThe rational torsion subgroup injects into each of these groups, so it is finite and its order $M$ divides every such $p+1$.

For any odd prime $\ell$, the Dirichlet theorem on primes in arithmetic progressions supplies infinitely many $p$ with $p\equiv 3(\mathrm{mod}4)$ and $p\equiv 1(\mathrm{mod}\ell )$. Discard the finitely many dividing $D$. Since $\ell \nmid p+1$, it cannot divide $M$. Similarly choose $p\equiv 3(\mathrm{mod}8)$, again avoiding $D$; then $v_2(p+1)=2$, so $M\mid 4$.

There are already four rational 2-torsion points,which are distinct because a squarefree integer $D$ is nonzero. ThusIts order is four. In fact the argument works for every nonzero integer $D$; squarefreeness is not needed for this torsion conclusion.

Work in characteristic different from two, as in the number-field application below. Nonsingularity is equivalent to $b(a^2-4b)\ne 0$. The chord through $P=(x,y)$ and $T=(0,0)$ has slope $y/x$. Using $y^2/x^2=x+a+b/x$ in the elliptic-curve addition formula givesThese formulas hold for $P\ne O,T$; addition interchanges $O$ and $T$.

It follows that $\xi =x+a+b/x=y^2/x^2$ and $\eta =y(1-b/x^2)$. The relation isFor a direct verification, observe thatTherefore the two-isogeny formula isThe target elliptic curve is nonsingular because its corresponding coefficient product is $16b(a^2-4b)\ne 0$. The rational map of projective varieties extends over the exceptional points to a morphism of smooth projective algebraic curves; at both $O$ and $T$ its affine coordinates tend to infinity, giving the displayed values. A nonconstant morphism between elliptic curves sending $O$ to $O$ is a group homomorphism, so this is an isogeny of elliptic curves.

One can also see the quotient directly: translation by $T$ leaves $\xi ,\eta$ invariant. The equation $x^2+(a-\xi )x+b=0$ makes the source function field a degree-two extension of the target function field; its nontrivial automorphism is translation by $T$. Equivalently, the degree of an isogeny from its x-coordinate map is two. The kernel of an isogeny is precisely $\{O,T\}$. Thus $\varphi$ is a separable isogeny of degree two.

Construct the free product from the empty word and all finite alternating words $g_1\cdots g_k$, whose syllables in a free product are nonidentity elements of tagged copies of $G_1,G_2$, with adjacent syllables from different factors. Multiply by concatenating, multiplying adjacent elements in the same factor, and deleting identities until the word is reduced. The normal form theorem for a free product gives a unique result; reducing three concatenated words gives the same result under either parenthesization, so multiplication is associative. The empty word is the identity, and the inverse reverses the word and inverts each syllable. Each factor embeds as words of length one.

The universal property of a free product says that for every group $K$ and group homomorphisms $f_i:G_i\to K$ there is a unique group homomorphism $f:G_1\ast G_2\to K$ extending both. Explicitly, send a reduced word to the product of its syllable images; reduction preserves this product. This gives existence, while generation by the two factors gives uniqueness.

A standard form of Klein's combination theorem, or the ping-pong lemma, is the following. Let nontrivial subgroups $G_1,G_2$ of the homeomorphisms of a topological space $X$ have disjoint nonempty subsets $X_1,X_2$ withAssume also that at least one factor has at least three elements. Then the generated subgroup is $\boxed{⟨G_1,G_2⟩\cong G_1\ast G_2}$. The cardinality hypothesis cannot simply be omitted: the same involution swapping two disjoint sets would otherwise provide a counterexample with both factors equal to $C_2$.

Here is the ping-pong lemma proof. A reduced word of odd length begins and ends in the same factor, so repeated application of the displayed inclusions sends the other factor's domain into that factor's domain. Disjointness shows that the word is not the identity. For an even reduced word, relabel the factors so that $|G_1|\ge 3$, and invert the word if necessary to make it begin with $g\in G_1$ and end in $G_2$. Choose $s\in G_1\setminus \{1,g^{-1}\}$. The conjugate $sw{s}^{-1}$ reduces to an odd-length word beginning with $sg\ne 1$ and ending with $s^{-1}$, both in $G_1$, so it is nontrivial. Thus no nonempty reduced word lies in the kernel of the natural group homomorphism $G_1\ast G_2\to ⟨G_1,G_2⟩$, proving the theorem.

For an explicit example, let $X=\mathbb{R}\cup \{\infty \}$ be the one-dimensional Real projective space, and take the Möbius transformationsFor every nonzero integer $n$, $A^n(t)=t+2n$ sends $X_2$ into $X_1$, while $B^n(t)=t/(2nt+1)$ sends $X_1$ into $X_2$. Indeed $|2nt+1|>|t|$ when $|t|>1$, and $B^n(\infty )=1/(2n)$. Both transformations have infinite order. Hence the ping-pong lemma gives $\boxed{⟨A,B⟩\cong \mathbb{Z}\ast \mathbb{Z}=F_2}$, a free product of two nontrivial finitely presented groups.

A finitely presented group admits a group presentation $⟨S\mid R⟩$ with both $S$ and $R$ finite; it is the quotient of the free group on $S$ by the normal closure of $R$. Ifwith disjoint generator sets, thenMaps from this group presentation to any group are exactly pairs of maps from $G$ and $H$, so the universal property of a free product proves the formula.

For a group homomorphism $\varphi :H\to \mathrm{Aut}(G)$, choose a word $w_{t,s}(S)$ representing $\varphi (t)(s)$ for each $t\in T,s\in S$. The presentation of a semidirect product isThe presentation maps onto the specified semidirect product. Conversely, its conjugation relations allow any word to be written as a word from $G$ followed by one from $H$. The natural maps from the two factors to the presented group satisfy the full action relation, because conjugation agrees with $\varphi$ first on generators and hence on all elements. They define the reverse group homomorphism $(g,h)\mapsto gh$. The two maps are inverse on every generator, proving the group isomorphism. There are finitely many cross-relations, so the result is again a finitely presented group.

Apply this to the specified permutation action. The presentation isEliminate $y=cx{c}^{-1}$ and $z=c^{2}x{c}^{-2}$. The last cross-relation becomes $c^{3}x{c}^{-3}=x$, already implied by $c^3=1$. ThusBoth factors are nontrivial finitely presented groups, as required.

The map of coset setsis well defined and surjective, because the group homomorphism $\theta$ is surjective. ThereforeNo normality assumption on $H$ is required.

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.

Let $n=[G:H]<\infty$. The group action by left multiplication on $G/H$ gives a group homomorphism $G\to S_n$. Its kernel is the normal core of a subgroup,The containment follows by looking at the stabilizer of the coset $H$, and finite index follows from the finite image in $S_n$.

The Higman group isIt is visibly a finitely presented group. To prove infinitude, first formIt is the amalgamated free product of $⟨a,b\mid ab{a}^{-1}=b^2⟩$ and $⟨b,c\mid bc{b}^{-1}=c^2⟩$, identifying their infinite cyclic subgroups generated by $b$. Each factor is an HNN extension of an infinite cyclic group, so its base and stable letter both have infinite order. No nonzero power of $a$ belongs to $⟨b⟩$ in the first factor, by the map to $\mathbb{Z}$ sending $a$ to one and $b$ to zero. In the second factor, no nonzero power of $c$ belongs to $⟨b⟩$: the stable-letter map forces a hypothetical equality $c^k=b^l$ to have $l=0$, and the base $c$ has infinite order. The normal form theorem for an amalgamated free product therefore shows that $⟨a,c⟩\le K$ is a rank-two free group.

Similarly,contains $⟨c,a⟩$ as a rank-two free group. Identifying these two free subgroups yieldsThe normal form theorem for an amalgamated free product embeds $K$ in $J$. In particular, $J$ contains a free group of rank two and is infinite.

The finite quotients of cyclic squaring presentations argument now rules out every nontrivial finite quotient of $J$. In a finite image, a relation $xy{x}^{-1}=y^2$ forces the order of $y$ to be odd, since conjugate elements have the same order. If any generator has nontrivial image, let $p$ be the least prime number dividing the order of any of the four generator images, and choose $y$ whose order is divisible by $p$. Its predecessor $x$ conjugates it to its square. If $r$ is the order of $x$, iterating conjugation gives $2^r\equiv 1(\mathrm{mod}p)$. Hence the multiplicative order of $2$ modulo $p$ divides $r$. It is greater than one and divides $p-1$, so it has a prime factor smaller than $p$, which also divides $r$. This contradicts the minimal choice of $p$. Thus all four generator images are trivial. $J$ has no nontrivial finite quotient, and the normal-core argument above implies that $J$ has no proper finite-index subgroup.

For the final argument, Conjugation preserves the order of an element. Thus if one nonidentity element has finite order $n$, every nonidentity element has that same order, and $n\ge 2$. Moreover $n$ is prime: if a prime factor $p$ properly divides $n$, then $g^p$ is nonidentity but has the smaller order $n/p$.

When $n\ge 3$, the element $g^2$ is nonidentity, so choose $x$ with $xg{x}^{-1}=g^2$. The conjugator is not the identity, since $g^2\ne g$, and therefore $x^n=1$. Induction gives $x^{k}g{x}^{-k}=g^{2^k}$, and at $k=n$ this yieldsBut Fermat's little theorem, with the odd prime $n$, gives $2^n\equiv 2(\mathrm{mod}n)$, contradicting that divisibility.

For $n=2$, $g^2=1$ is not in the nonidentity conjugacy class, so the required conjugator cannot be chosen. Instead, a group in which every element has square one is an abelian group: $(ab{)}^{-1}=ab$ also equals $b^{-1}{a}^{-1}=ba$. In an abelian group every conjugacy class is a singleton, so one nonidentity class permits only one nonidentity element, giving a group of order two. This contradicts infinitude. Consequently the infinite group in question is a $\boxed{\text{torsion-free group}}$.

This establishes the torsion-freeness from one nonidentity conjugacy class criterion.

Both displayed matrices have determinant one and integer entries, so the generated group $M$ lies in ${\mathrm{SL}}_2(\mathbb{Z})$. For any nonidentity matrix $g\in M$, some entry of $g-I$ is a nonzero integer $k$. Choose a prime number $p$ not dividing $k$. The reduction modulo a prime in an integral matrix group gives a group homomorphismto a finite group in which $g$ is nonidentity. $\boxed{M\text{ is residually finite}.}$ No determination of the abstract subgroup generated by the two matrices is needed.

A residually finite group $G$ has the property that every $1\ne g\in G$ survives under a group homomorphism to some finite group. Equivalently, the intersection of its finite-index normal subgroups is trivial. A Hopfian group is a group for which every surjective endomorphism is an automorphism.

Suppose $G$ is generated by $d$ elements. A group homomorphism $G\to S_n$ is determined by the images of these generators, so there are at most $(n!{)}^d$ such maps. Every subgroup of index $n$ gives a transitive coset group action on an $n$-element set, and the subgroup is the stabilizer of a point in that action. There are at most $n$ point stabilizers per action. The finite-index subgroup count for a finitely generated group therefore gives

Now let $\psi :G\to G$ be a surjective endomorphism. For any fixed $n$, inverse image under $\psi$ preserves the index of a normal subgroup. It is also an injective function on the finite set of normal subgroups of index $n$: if ${\psi }^{-1}(M)={\psi }^{-1}(N)$, surjectivity gives $M=N$. It is therefore a permutation of that finite set. Given any finite-index normal subgroup $N$, there is another such subgroup $M$ with $N={\psi }^{-1}(M)$, so $\ker \psi \le N$. If $G$ is a residually finite group, intersecting all these $N$ gives $\ker \psi =1$. Hence $\psi$ is an automorphism. Every finitely generated group that is a residually finite group is a Hopfian group.

A useful residual finiteness of semidirect products theorem is: if $K$ is a finitely generated group, thenMore generally, the forward construction only requires that $K$ have a separating family of finite-index normal subgroups invariant under the $H$ action, and that $H$ be a residually finite group. Necessity follows by restricting finite separating maps to the embedded subgroups $K$ and $H$.

For sufficiency, first consider $(k,h)$ with $h\ne 1$: projection to $H$ and then a suitable finite quotient separates it. If $h=1$ and $k\ne 1$, choose a finite-index normal subgroup $U⊴K$ with $k\notin U$. Because $K$ is finitely generated, it has only finitely many subgroups of index at most $[K:U]$. Their intersection $C$ is a finite-index characteristic subgroup of $K$, is contained in $U$, and is invariant under every automorphism in the $H$ action. The quotient $K/C$ is finite. Let $\rho :H\to \mathrm{Aut}(K/C)$ be the induced action. The mapis a group homomorphism to a finite group and separates $(k,1)$. This proves the theorem and the more general invariant-subgroup criterion. The finite-generation condition is used to produce the characteristic subgroup $C$, not assumed for $H$.

Let $N$ be a normal subgroup and let $\pi :G\to G/N$ be the quotient group homomorphism. By the first part, $G=⟨a,b⟩$ for involutions $a,b$, so the quotient group is generated by $\pi (a),\pi (b)$, whose squares are the identity. If both images are nonidentity, they are involutions and the second part applies. If only one image is nonidentity, it generates $C_2$; choose it twice. Both images cannot be trivial because the quotient group is nontrivial. Hence every nontrivial quotient is dihedral in the stated extended convention. This is the nontrivial-quotient closure of two-involution groups.

True. The determinant gives a group homomorphismFor an orthogonal matrix, $A^{T}A=I$ implies $(\det A{)}^2=1$, and a reflection has determinant $-1$, so the target has order two and the map is onto. Its kernel of a group homomorphism is exactlyThis is the defining special orthogonal group inside the orthogonal group.