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For a finite simple group with a subgroup of index $n\ge 3$, the coset action is nontrivial and its kernel is a normal subgroup, hence trivial. Its sign homomorphism must also be trivial: otherwise simplicity would give an isomorphism to $C_2$, which cannot act transitively on at least three points. Thus the image lies in the alternating group. In particular Lagrange's theorem gives $|G|\mid n!/2$. This can exclude simplicity with very little counting: a group of order $112$ has a subgroup of index seven by Sylow theorems, but $112$ does not divide $7!/2$.

For a subgroup $H\le G$ of a group $G$, a left coset is $gH=\{gh:h\in H\}$; a right coset is $Hg=\{hg:h\in H\}$. The name refers to the side on which the representative $g$ multiplies $H$. Think of a coset as a translated copy of a subgroup: it remembers the relative differences allowed by $H$, while moving the whole collection to another location in the group.

Two left cosets of the same subgroup are either equal or disjoint. Thus distinct left cosets form a partition of a set covering $G$; the same statement holds separately for right cosets. The useful membership test isThis also defines an equivalence relation: changing a representative by right multiplication with a member of $H$ leaves its left coset unchanged. These facts hold even for infinite groups. See MIT's coset notes, definitions 3.6 and 3.9.

For an explicit additive example, take the cyclic group $G=\mathbb{Z}/6\mathbb{Z}$, with modular arithmetic, and the subgroup $H=\{0,3\}$. Adding each representative to both members gives exactlyRepresentatives $3,4,5$ repeat these three cosets, respectively. For example, $4+H=\{4,1\}=1+H$, since $4-1=3\in H$. Every residue appears once in the displayed partition of a set. Each coset has two members, so the group order of $H$ is $2$ and its index of a subgroup is $3$. Left and right cosets coincide here because $G$ is an abelian group.

A coset need not be a subgroup. In this example, $\{1,4\}$ lacks the identity element $0$, and $1+1=2$ leaves it. In fact, the only left or right coset of $H$ that is a subgroup is $H$ itself: a subgroup must contain the identity element, and the coset containing that element is $H$. Equal sizes come instead from the bijection $h\mapsto gh$, whose inverse is $x\mapsto g^{-1}x$; for right cosets, use $h\mapsto hg$ and $x\mapsto x{g}^{-1}$.

The nonabelian group $S_3$, the symmetric group on three letters, separates the two sides. Use function composition with the rightmost permutation applied first, and let $H=\{e,(12)\}$. Then $(13)(12)=(123)$ but $(12)(13)=(132)$, giving the differing pairThe complete partition of a set into left cosets consists of $H$, $\{(13),(123)\}$ and $\{(23),(132)\}$; the complete partition of a set into right cosets consists of $H$, $\{(13),(132)\}$ and $\{(23),(123)\}$. Each lists all six permutations exactly once. Both have three parts of size two, although their parts differ.

Left cosets support a group action by $a\cdot (gH)=(ag)H$ for every subgroup, as in the coset action. Multiplication of representatives defines a quotient group operation $(aH)(bH)=(ab)H$ precisely when $H$ is a normal subgroup. Otherwise the answer can depend on the representatives: in this $S_3$ example, $e$ and $(12)$ represent the same coset $H$, yet $e(13)H=(13)H$ and $(12)(13)H=(23)H$ differ. In the additive example, $H$ is a normal subgroup, making $(1+H)+(2+H)=3+H=H$ valid. See MIT's quotient-group notes, theorem 9.1.

A group acts on the left cosets of a subgroup by left multiplication; its kernel is the core of that subgroup.

Let $H\le G$ be a subgroup. The notation $G/H$ here means the set of left cosets $\{aH:a\in G\}$, even when $H$ is not a normal subgroup. DefineThis gives a group action without needing a quotient group structure on $G/H$.^[ref]

The representative check is essential: if $aH=bH$, then $a=bh$ for some $h\in H$, and $(ga)H=(gbh)H=(gb)H$. Thus equal input cosets give equal output cosets. The identity element fixes every coset, and associativity gives $g_1\cdot (g_2\cdot aH)=(g_1{g}_2)\cdot aH$. Each induced map is a permutation, with inverse induced by $g^{-1}$. This is a transitive group action because $b{a}^{-1}$ sends $aH$ to $bH$.

The stabilizer subgroup of $H$ is $H$ itself. At a general coset, the stabilizer subgroup and the group kernel areConsequently the coset action is a faithful group action exactly when the subgroup core is trivial.^[ref]

Indeed, $g\cdot (aH)=aH$ exactly when $a^{-1}ga\in H$, or $g\in aH{a}^{-1}$. Intersecting these membership conditions over all cosets gives the group kernel. Conjugation permutes the subgroups in this intersection, making it a normal subgroup of $G$ contained in $H$. Conversely, any normal subgroup of $G$ contained in $H$ lies in every $aH{a}^{-1}$. This explains why the subgroup core is the largest such normal subgroup: it is exactly the portion of $H$ invisible at every label, rather than merely at the label $H$.

For a normal subgroup $H$, every $aH{a}^{-1}$ equals $H$, so $\ker {\rho }_H=H$. For $H=\{e\}$, the construction is the regular group action behind Cayley's theorem. Between these extremes, a nontrivial subgroup with trivial subgroup core can give a faithful group action on fewer labels. If the index of a subgroup is finite, $[G:H]=n$, the first isomorphism theorem givesThus a subgroup of small index of a subgroup exposes a small permutation group associated with $G$. The two examples below and in nonfaithful coset action of the symmetric group on three letters show exactly which information can survive.

For a finite group $G$, transitive finite group actions are classified up to equivalence by conjugacy classes of subgroups: the class of $H$ gives the coset action on $G/H$. Every finite action decomposes uniquely as a disjoint union of its orbits, so its equivalence type is a multiset of these transitive types. Consequently the coefficient of $t^n$ in the displayed product, with one factor for every subgroup conjugacy class, counts actions on $n$ points up to relabeling. Each factor expands as a geometric series recording any multiplicity of that orbit type. For both $C_6$ and $S_3$ there is exactly one transitive type of each size $1,2,3,6$; their coefficient at degree six is $4+2+1+1=8$.

Take the symmetric group $G=S_3$ and the subgroup $H=⟨(12)⟩=\{e,(12)\}$ of group order two. With the rightmost permutation applied first, its three left cosets areThe subscripts record the image of $3$: every member of $C_i$ sends $3$ to $i$. Because $H$ is the stabilizer subgroup of $3$ in the usual group action, $aH\mapsto a(3)$ is a bijection onto $\{1,2,3\}$. It is an equivariant map, since $(ga)(3)=g(a(3))$. The coset action therefore becomes the familiar group action on three letters. In particular,

The three stabilizer subgroups areTheir intersection is $\{e\}$, so the subgroup core and the group kernel are trivial. This proves that the coset action is a faithful group action even though every point has a nontrivial stabilizer subgroup. The orbit-stabilizer theorem reads $6=3\cdot 2$, while the induced group homomorphism gives six distinct permutations of the three cosets. For example, $(12)$ and $e$ agree at $C_3$, but disagree at $C_1$; inspecting one label would miss their difference.

The subgroup need not be a normal subgroup for the coset action to exist: $(123)H(123{)}^{-1}=\{e,(23)\}\ne H$. Here the group action is informative because the three conjugate stabilizer subgroups have no common nonidentity element. For the normal subgroup $A_3$ instead, nonfaithful coset action of the symmetric group on three letters shows that only the sign of a permutation survives.

A group action is faithful when only the identity acts as the identity on every point.

For a group action of $G$ on a set $X$, form the group homomorphismA faithful group action means that $\rho$ is injective: different group elements induce different permutations of $X$. Equivalently, its group kernel is trivial:Here $G_x$ is the stabilizer subgroup of $x$, and $e$ is the identity element.^[ref]

The test is global. A faithful group action requires each nonidentity group element to move at least one point; a free group action requires it to move every point. In the faithful coset action of the symmetric group on three letters, $(12)$ fixes one coset and exchanges the other two. Thus a nontrivial stabilizer subgroup does not rule out faithfulness. The distinction follows directly from the order of the quantifiers:On a nonempty set, a free group action is therefore a faithful group action. A transitive group action instead asks whether every point can reach every other; it allows several group elements to perform exactly the same rearrangement. The nonfaithful coset action of the symmetric group on three letters is a counterexample to the claim that every transitive group action is a faithful group action.

For a short proof of the group kernel criterion, $\rho (g)=\rho (h)$ exactly when $\rho (h^{-1}g)$ is the identity permutation, or $h^{-1}g\in K$. If $K=\{e\}$, this forces $g=h$; if $K$ contains a nonidentity element, that element and $e$ act identically. Thus a faithful group action turns $G$ into an actual subgroup of a symmetric group, retaining its group operation as function composition. Choosing a suitable coset action can give a much smaller set of labels than the regular group action used in Cayley's theorem.

For the preliminary index assertion, the coset action gives a homomorphism $G\to S_n$. Its kernel is a normal subgroup and cannot be all of $G$, because the action is transitive on $n>1$ points. Simplicity makes the kernel trivial. The sign homomorphism of this faithful action must be trivial: otherwise simplicity would identify $G$ with $C_2$, impossible for a transitive action of degree at least three. Thus $G$ embeds in $A_n$, and Lagrange givesA group of order $112=16\cdot 7$ has a Sylow two-subgroup of index seven. If the group were simple, the preceding result would give $112\mid 7!/2=2520$. This fails because $112$ is divisible by $16$ while $2520$ has only three factors of two. Thus there is no simple group of order $112$.

For order $56$, Sylow gives $n_7\in \{1,8\}$. If it is one, the order-seven Sylow is a normal subgroup. Otherwise its eight Sylows of order seven intersect only at the identity, giving $8\cdot 6=48$ nonidentity elements of order seven. Only eight elements remain including the identity. Every Sylow two-subgroup has order eight and consists entirely of these remaining elements. Hence all Sylow two-subgroups coincide and the order-eight subgroup is a normal subgroup. Either case gives a nontrivial proper normal subgroup, so no group of order $56$ is simple.

Use the following standard facts: $A_5$ has order $60$ and is a nonabelian simple group; the coset action on the cosets of an index-$j$ subgroup is a transitive homomorphism into $S_j$. If $H<A_5$ has index $j>1$, that action is nontrivial. Its kernel is a normal subgroup, hence is trivial by simplicity. It would therefore embed $A_5$ in $S_j$. For $j=2,3,4$ this is impossible since $j!<60$. There are no subgroups of indices 2, 3 or 4.

An index-5 subgroup has order $12$ by Lagrange's theorem. Consider its natural action on the five letters. It cannot be transitive, since an orbit of size 5 would divide its order by the orbit-stabilizer theorem. If it had no fixed letter, its orbits would have sizes 2 and 3. It would then be a subgroup of the permutations preserving that partition, namely $S_2\times S_3$. Exactly half of those 12 permutations are even, so $(S_2\times S_3)\cap A_5$ has order 6, too small to contain $H$.

Thus $H$ fixes a letter $i$. The stabilizer of $i$ in $A_5$ is the copy of $A_4$ consisting of even permutations of the other four letters; it has order 12. Containment and equality of orders force $H$ to equal that stabilizer. Conversely each such stabilizer has index 5. ConsequentlyThese five subgroups are distinct: for $i\ne j$, a three-cycle on letters other than $i$ can move $j$, so it belongs to the stabilizer of $i$ but not that of $j$.

The requested representation is the left coset actionIt is well defined because equal cosets remain equal after multiplication by $g$, and $\rho (g{g}^{\prime })=\rho (g)\rho (g^{\prime })$. It is transitive. For $H=1$ it is the left regular action; for general $H$ it is the associated transitive coset representation.

Its kernel is the normal core of a subgroupIndeed, $g$ fixes every $xH$ precisely when $x^{-1}gx\in H$ for every $x$. Thus $N⊲G$ and $N\le H$. If $[G:H]=i<\infty$, the image lies in the finite symmetric group $S_i$, soIn particular $N$ has finite index in $H$.

For Higman group as an amalgam of iterated HNN extensions, use the cyclic squaring presentationFirst form $L=⟨b,c\mid b^{-1}cb=c^2⟩$, an HNN extension of the infinite cyclic group $⟨c⟩$. Its base embeds, so $c$ has infinite order; $b$ has infinite order as well, since the homomorphism $L\to \mathbb{Z}$ sending $b\mapsto 1,c\mapsto 0$ is onto. Consequently $⟨b⟩$ and $⟨b^2⟩$ are isomorphic infinite cyclic subgroups. Adjoining the stable letter $a$ with $a^{-1}ba=b^2$ givesWe claim $⟨a,c⟩$ is a rank-two free group. No nonzero power of $c$ belongs to $⟨b⟩$: applying the map $b\mapsto 1,c\mapsto 0$ would make that power equal to $1$, contradicting the infinite order of $c$. Thus no nonzero power of $c$ belongs to $⟨b^2⟩$ either. Any nonempty freely reduced word in $a,c$ that contains $a$ is therefore a reduced HNN sequence: between inverse $a$ letters, a nonzero $c$ power cannot create a pinch. Britton's lemma makes it nonidentity. A word containing only $c$ is nonidentity by the base embedding. This proves the claim.

Likewiseis obtained from $⟨d,a\mid d^{-1}ad=a^2⟩$ by adjoining $c$, and its subgroup $⟨c,a⟩$ is free of rank two by the same argument. Form the amalgamated free product $P{\ast }_{⟨a,c⟩}Q$, identifying these two free subgroups generator by generator. Its presentation is exactly that of $\mathcal{H}$. The component embeddings allowed in the question show that it contains $P$, hence is infinite. It has four group generators and four relators, so it is finitely presented.

To show that this Higman group has no nontrivial finite quotient, consider any finite image and the orders of the four generator images. A relation $x^{-1}yx=y^2$ makes $y$ and $y^2$ have the same order, so every generator order is odd. Suppose one is nontrivial, and let $p$ be the smallest prime dividing any of the four orders. Choose a generator $y$ whose order is divisible by $p$, with predecessor $x$ satisfying the displayed relation. Iterating conjugation $|x|$ times givesThe multiplicative order $d$ of $2$ modulo $p$ thus divides $|x|$. Also $d\mid p-1$ and $d>1$, since $2\not\equiv 1(\mathrm{mod}p)$. Any prime divisor of $d$ is smaller than $p$ and divides the order of $x$, contradicting minimality of $p$. Hence every generator image is trivial, so every finite image is trivial. A proper finite-index subgroup would give a nontrivial transitive finite coset image, which is impossible. Therefore $\mathcal{H}$ is infinite and finitely presented, with no proper finite-index subgroups.

A maximal normal subgroup is a proper normal subgroup $M⊲G$ with no normal subgroup strictly between $M$ and $G$. To prove that finitely generated groups have maximal proper normal subgroups, fix a proper normal subgroup $K$ and order the proper normal subgroups containing $K$ by inclusion. For a chain, its union is normal and contains $K$. It is still proper: if the union were all of $G$, each element of a finite generating set would lie in some chain member, and the chain order would put all these finitely many group generators in one member. That member would equal $G$, a contradiction. The Zorn lemma therefore supplies a maximal proper normal subgroup $M$ containing $K$.

Apply this to $\mathcal{H}$ and $K=1$. By the normal-subgroup correspondence, $\mathcal{H}/M$ is nontrivial and simple. It is finitely generated as a quotient of $\mathcal{H}$. It cannot be finite, since $\mathcal{H}$ has no nontrivial finite quotients. Thus

The final assertion for arbitrary proper subgroups is false. Take $G=S_3$ and $H=⟨(12)⟩$. If a normal subgroup contained $H$, it would contain all conjugate transpositions, which generate $S_3$, so it would equal $G$. Hence this proper subgroup lies in no proper normal subgroup, and in particular in no maximal normal subgroup. The maximal-normal existence result requires the starting subgroup itself to be normal.

For a normal subgroup $H⊴G$, define multiplication of left cosets byTo check that this is well-defined, replace $g,k$ by $g{h}_1,k{h}_2$ with $h_1,h_2\in H$. Their product is $g{h}_{1}k{h}_2=gk(k^{-1}{h}_{1}k)h_2$, and normality puts $(k^{-1}{h}_{1}k)h_2$ in $H$. Thus the product coset is independent of the representatives. Associativity follows from multiplication in $G$, the identity is $H$, and the inverse of $gH$ is $g^{-1}H$. These operations form the quotient group $G/H$.

For the unheaded finite-index conclusion, let $G$ act on the $n$ left cosets of its proper subgroup $H$ by $g\cdot (xH)=gxH$. This is well-defined even if $H$ is not normal: replacing $x$ by $xh$ leaves $gxH$ unchanged. The maps are bijections with inverses supplied by $g^{-1}$, and their compositions give a group homomorphism $\phi :G\to S_n$. Its kernel of a group homomorphism $N$ is normal, by the calculation in part (ii).

The action is not trivial, because some $g\notin H$ sends the coset $H$ to $gH\ne H$; hence $N\ne G$. If $N$ were trivial, $\phi$ would be injective, implying $|G|\le |S_n|=n!$. Under the given strict inequality this is impossible. Thus $N$ is a nontrivial proper normal subgroup, andMore precisely, $N={\bigcap }_{x\in G}xH{x}^{-1}$ is the subgroup core: $g$ fixes every coset if and only if $x^{-1}gx\in H$ for every $x$. This identifies the normal subgroup produced by the coset action.

Suppose $G$ is topologically generated by $d$ elements. An open subgroup $H$ of index $n$ gives a continuous transitive coset actionA continuous homomorphism $G\to S_n$ is determined by the images of the $d$ topological generators, so there are at most $|S_n{|}^d$ such homomorphisms. Each has only finitely many point stabilizers. Therefore $G$ has only finitely many open subgroups of index $n$.

A Sylow subgroup of a finite group $G$ is a subgroup whose order is the largest power of a prime $p$ dividing $|G|$. The Sylow theorems say that such subgroups exist, that every $p$-subgroup lies in one, that all Sylow $p$-subgroups are conjugate, and that their number $n_p$ divides $|G|/p^a$ and obeys $n_p\equiv 1(\mathrm{mod}p)$.

For the last congruence, let one Sylow $p$-subgroup $P$ act by conjugation on the set of all Sylow $p$-subgroups. Every orbit other than a fixed point has size divisible by $p$. If $Q$ is fixed, then $P,Q\le N_G(Q)$. They are Sylow subgroups of this normalizer, so they are conjugate within $N_G(Q)$; because every element of $N_G(Q)$ fixes $Q$ under conjugation, this forces $P=Q$. Thus there is exactly one fixed point and $n_p\equiv 1(\mathrm{mod}p)$.

Now let $H<A_n$ have index $m$ with $1<m<n$. The coset action gives a homomorphismIts kernel is a normal subgroup of the simple group $A_n$. It cannot be all of $A_n$, since the action is transitive and nontrivial, so it is trivial. This would embed $A_n$ into $S_m$, contrary to $|A_n|=n!/2>m!=|S_m|$. Hence no such subgroup exists.

Suppose finally that a group $G$ of order $90$ were simple. The Sylow count satisfiesand simplicity excludes $n_5=1$, so $n_5=6$. Conjugation on these six subgroups gives a nontrivial homomorphism $G\to S_6$, which simplicity makes injective. Its image lies in $A_6$, because the composite with the sign homomorphism $S_6\to \{\pm 1\}$ must be trivial. It would therefore be a subgroup of $A_6$ of index $360/90=4$, contradicting the result just proved. Thus

Because $n$ is not a prime power, distinct primes $p<q$ divide $n$. By Cauchy's theorem, $G$ has subgroups $H,K$ of orders $p,q$. Combine the two coset actions to obtainThe kernel of the first action is contained in $H$, and that of the second is contained in $K$. Their intersection is trivial because $H\cap K=\{1\}$, so the combined action is faithful.

Since $p\ge 2$, $q\ge 3$, and $n\ge 6$,Adding fixed points embeds this symmetric group into $S_{n-1}$. Thus every group of non-prime-power order $n$ is a subgroup of $S_{n-1}$, as summarized by symmetric-group embedding at one less than the group order.

Let $k=|G/H|$. The coset action gives a nontrivial homomorphism $G\to S_k$ whose kernel is normal. Simplicity makes it injective. The sign map is trivial on the nonabelian simple image, so $G$ embeds in $A_k$. For $k\le 4$, $A_k$ is solvable, as are its subgroups, whereas a nonabelian simple group is not. Hence $k\ge 5$.

On the set of left cosets $G/H$, define $g\cdot (xH)=(gx)H$. Given $xH$ and $yH$, the element $y{x}^{-1}$ sends the first to the second. Thus the coset action is transitive. In particular the orbit of $H$ is the entire coset space, and its stabilizer is $H$.

The symmetric group $S_3$ has subgroups of orders $1,2,3,6$. There is one trivial subgroup, three conjugate subgroups generated by transpositions, one order-three subgroup $A_3$, and the whole group. These are all subgroups by Lagrange's theorem. Thus again there is one conjugacy class for each index $6,3,2,1$.

The coset action classification in part (c) supplies exactly one transitive orbit type of each of these sizes. The same orbit-multiplicity equation as in part (i) applies, givingThe equality of the two counts comes from their identical lists of transitive orbit types, rather than from the groups being isomorphic.

On the set of left cosets $G/H$, define $g\cdot (xH)=(gx)H$. Given $xH$ and $yH$, the element $y{x}^{-1}$ sends the first to the second. Thus the coset action is transitive. In particular the orbit of $H$ is the entire coset space, and its stabilizer is $H$.

The symmetric group $S_3$ has subgroups of orders $1,2,3,6$. There is one trivial subgroup, three conjugate subgroups generated by transpositions, one order-three subgroup $A_3$, and the whole group. These are all subgroups by Lagrange's theorem. Thus again there is one conjugacy class for each index $6,3,2,1$.

The coset action classification in part (c) supplies exactly one transitive orbit type of each of these sizes. The same orbit-multiplicity equation as in part (i) applies, givingThe equality of the two counts comes from their identical lists of transitive orbit types, rather than from the groups being isomorphic.

For a subgroup $H\le G$, the permutation character of the coset action on $G/H$ is the induced trivial characterand