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Lagrange's theorem states that, for a finite group $G$ and a subgroup $H$,To prove it, define an equivalence relation by $x\sim y$ when $x^{-1}y\in H$. Its classes are the left cosets $xH$, which therefore partition $G$. Left multiplication by $x$ gives a bijection $H\to xH$, so every coset has $|H|$ elements. Counting this partition gives the formula; in particular $|H|$ divides $|G|$. Applying the theorem to the cyclic subgroup generated by an element shows that its order of a group element divides $|G|$.

Write the given order-two normal subgroup as $N=\{1,z\}$. Conjugation must preserve $N$ and cannot send its nonidentity element to the identity, so $gz{g}^{-1}=z$ for all $g\in G$. Thus $z$ is in the center of a group: a normal subgroup of order two is central.

Choose $g\notin N$. By Lagrange's theorem, its order of a group element is $2$, $p$ or $2p$. It cannot have order two: since $z$ is central, the four distinct elements $1,z,g,zg$ would form a subgroup of order four, whereas $4$ does not divide $2p$ for odd $p$. If $g$ has order $2p$, it already generates $G$. Otherwise $g$ has order $p$, commutes with $z$, and $(gz{)}^k=1$ implies $g^k=z^{-k}\in ⟨g⟩\cap N=\{1\}$. Hence both $p$ and $2$ divide $k$, so $gz$ has order $2p$. In either case

For a group action of a finite group $G$ on a set and a point $x$, the orbit-stabilizer theorem iswhere $G_x=\{g:gx=x\}$ is the stabilizer subgroup. The map $g{G}_x\mapsto gx$ is well defined because elements of the same coset act identically on $x$. If $gx=hx$, then $h^{-1}g\in G_x$, so their cosets agree; every orbit point has the form $gx$, giving surjectivity. This is a bijection. Lagrange's theorem, $|G|=[G:H]|H|$ for any subgroup $H$, yields the counting formula.

Apply the conjugation action of the finite p-group $G$ to its nontrivial normal subgroup $N$. Normality ensures the action stays inside $N$. The singleton orbits of a group action are exactly the elements of $N\cap Z(G)$. Every larger orbit of a group action has size a positive power of $p$, by the orbit-stabilizer theorem and Lagrange's theorem. ThereforeSince $N$ is nontrivial, its order is divisible by $p$. The intersection contains the identity and has cardinality a positive multiple of $p$, so it contains a nonidentity element. This proves the central intersection property of normal subgroups of finite p-groups.

For the proper subgroup $H$, let $H$ act by left multiplication on the left cosets $G/H$. A coset $gH$ is fixed by every $h\in H$ exactly when $g^{-1}Hg\subseteq H$. Both groups have the same finite size, so this inclusion is equality; the fixed cosets are precisely those with $g\in N_G(H)$. Hence their number is $[N_G(H):H]$. All non-singleton orbits of a group action have size divisible by $p$, soSince $H$ is proper, $[G:H]$ is a positive power of $p$ greater than one. The fixed-coset count is nonzero, because $H$ itself is fixed, and is divisible by $p$. Therefore the normalizer strictly contains $H$:This is the normalizer condition for finite p-groups. The proof also covers $H=\{1\}$, whose action fixes all cosets.

Since $351=3^3\cdot 13$, the number of Sylow subgroups for $p=13$ divides $27$ and is $1$ modulo $13$, hence is $1$ or $27$. In the first case the unique subgroup is a nontrivial proper normal subgroup.

In the second case the $27$ subgroups of order $13$ have pairwise trivial intersection: two such subgroups sharing a nonidentity element would both be generated by it. They account for $27\cdot 12=324$ nonidentity elements of order $13$. Exactly $27$ elements, including the identity, remain. Every Sylow subgroup for $p=3$ has order $27$, and all its elements lie among these remaining elements, by Lagrange's theorem. It must therefore be precisely that set. The Sylow subgroup for $p=3$ is unique and hence normal. In either case $G$ is not a simple group.

A finite nonabelian simple group cannot be a p-group: a nontrivial finite p-group has nontrivial centre of a group, since its class equation makes every noncentral conjugacy class size divisible by $p$, and hence makes the centre size a positive multiple of $p$. That centre is a normal subgroup, so must be the whole group if $G$ is a simple group, making it abelian; an abelian simple group has prime order. Therefore every Sylow subgroup here is nontrivial and proper. It cannot be normal, so $n_p>1$.

To obtain a simple group embedding from Sylow conjugation, the conjugation action on Sylow subgroups gives a group homomorphism $\rho :G\to S_{n_p}$. This group action is transitive by the Sylow theorems and is nontrivial since $n_p>1$. Its kernel of a group homomorphism is a normal subgroup, so the defining property of a simple group makes the kernel of a group homomorphism trivial: $\rho$ is an embedding. Now compose with the sign of a permutation $S_{n_p}\to \{\pm 1\}$. A nontrivial composite would again be injective because $G$ is a simple group, embedding $G$ in a group of order two, impossible for a nonabelian group. Thus $\rho (G)$ lies in the alternating group $A_{n_p}$. By Lagrange's theorem,The argument only invokes the factorial formula once $n_p\ge 2$; the impossible case $n_p=2$ itself is also excluded by the embedding.

Use the class equation for the action of $G$ on itself by conjugation:where the sum runs over noncentral conjugacy classes. By Lagrange's theorem, each noncentral class has size a positive power of $p$ greater than one. Thus $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$. Since the identity element belongs to the center of a group,For $|G|=p^3$, the possible orders of the center of a group are $p,p^2,p^3$. Order $p^3$ would make $G$ an abelian group. Order $p^2$ would make the quotient group $G/Z(G)$ have prime order and hence be a cyclic group. But a cyclic quotient by the center forces $G$ to be Abelian: if $gZ(G)$ generates the quotient, any two elements are $g^{a}z$ and $g^{b}w$ with $z,w\in Z(G)$, and their products commute. Both larger orders contradict the hypothesis. Consequently

Let $G$ act by left multiplication on the set $G/H$ of $n$ left cosets. This gives a group action homomorphismIts kernelis a normal subgroup of $G$, and the first isomorphism theorem gives $G/K\cong \mathrm{im}\phi$. By Lagrange's theorem, $|G/K|$ divides $n!$. The image acts transitively on the $n$ cosets, so the orbit-stabilizer theorem shows that $n$ divides $|G/K|$; in particular,

Now suppose $G$ is nonabelian and simple. Since $H$ is proper, the coset action is nontrivial, so $K\ne G$. Simplicity gives $K=\{e\}$, and $\phi$ embeds $G$ into $S_n$. Composing with the sign homomorphism givesIts kernel is normal. A nontrivial map would embed the simple group $G$ into the abelian group of order two, which is impossible because $G$ is nonabelian. The sign is therefore always $+1$, so

Lagrange's theorem states that for a subgroup $H$ of a finite group $G$,where $[G:H]$ is the index of a subgroup, the number of its left cosets. In particular, $|H|$ divides $|G|$.

The cyclic subgroup $⟨g⟩$ consists of the distinct elements $e,g,\dots ,g^{n-1}$, where $n=o(g)$. Every power reduces to one of these by the Euclidean division, and a repetition among them would contradict minimality of $n$. Thus $|⟨g⟩|=o(g)$. Applying Lagrange's theorem gives

By Lagrange's theorem, the index of a subgroup is $[G:H]=75/25=3$. Thus the set of left cosets has three elements, and the preceding group action yields a group homomorphism $\Phi :G\to S_3$ after labelling them.

Its image is a subgroup of $S_3$, so its order divides $6$. By the first isomorphism theorem, $|\mathrm{im}\Phi |=|G|/|\ker \Phi |$ also divides $75$. Hence its order divides $\gcd (6,75)=3$, giving possibilities $1$ and $3$. The action on cosets is transitive: $g_2{g}_1^{-1}$ sends $g_{1}H$ to $g_{2}H$. It cannot be the trivial action on a three-element set, so the image does not have order $1$. ThereforeIt follows that $|\ker \Phi |=75/3=25$. Part (b)(i) gives $\ker \Phi \le H$, and $|H|=25$; equal finite orders force equality. A kernel of a group homomorphism is a normal subgroup, soThis is an instance of the fact that a subgroup of index the smallest prime divisor is normal.

The order of a group element $g$ is the smallest positive integer $d$ with $g^d=e$. Such an integer exists in a finite group: two powers in $e,g,g^2,\dots$ must agree, and cancellation gives a positive power equal to $e$.

By minimality, $e,g,\dots ,g^{d-1}$ are distinct. Every integer power of $g$ is one of them, by division of its exponent by $d$. They form the cyclic subgroupApplying Lagrange's theorem to this subgroup gives

Lagrange's theorem states that if $H$ is a subgroup of a finite group $G$, thenwhere the index of a subgroup $[G:H]$ is the number of distinct left cosets.

Every $g\in G$ lies in $gH$, since $g=ge$. By i, distinct left cosets are disjoint, so these sets form a partition of a set $G$. Each has cardinality $|H|$: the function $h\mapsto gh$ from $H$ to $gH$ is a bijection, with inverse $gh\mapsto h$, by cancellation in a group. Counting the elements in this finite partition of a set gives $|G|=[G:H]|H|$, proving the theorem.

Let $|G|=p$, with $p$ a prime number, and choose $g\ne e$. By Lagrange's theorem, the order of a group element $g$ divides $p$ and is greater than one. It is therefore $p$, and $G=⟨g⟩$. Hence a group of prime cardinality is a cyclic group.

The converse is false: the cyclic group $C_4$ has four elements, and $4$ is not a prime number.

The complete classification up to group isomorphism isHere $C_1$ is the trivial group. The prime number cases follow from i. For four elements, an element of order of a group element four generates $C_4$. If there is none, Lagrange's theorem makes every nonidentity element have order of a group element two. Then iv produces a four-element subgroup, which must be all of $G$, giving the Klein four-group.

For six elements, an element of order of a group element six gives $C_6$. Suppose there is none. If there were no element of order of a group element three either, all nonidentity elements would have order of a group element two. Part iv would then produce a subgroup of cardinality four, contradicting Lagrange's theorem. Thus choose $a$ of order of a group element three. There is also an involution $b$: otherwise the five nonidentity elements could be paired into distinct pairs $\{g,g^{-1}\}$, impossible for an odd number of elements.

The subgroup $H=⟨a⟩$ has index of a subgroup two, so it is a normal subgroup. Indeed, for $g\notin H$, both $gH$ and $Hg$ are the complement of $H$, and for $g\in H$ both equal $H$. Consequently $ba{b}^{-1}$ is either $a$ or $a^{-1}$. The first possibility makes $a,b$ commute, and $ab$ then has order of a group element six: $(ab{)}^k=e$ forces $a^k=b^{-k}\in ⟨a⟩\cap ⟨b⟩=\{e\}$, so both $3$ and $2$ divide $k$. This contradicts our assumption. ThereforeEvery element is uniquely $a^i$ or $a^{i}b$, with $0\le i<3$, and the relation $ba=a^{-1}b$ determines all their products. Sending $a$ to $(123)$ and $b$ to $(12)$ gives a group isomorphism with the symmetric group $S_3$. This completes the classification of groups of order at most six.

Of these groups, exactly $C_1,C_2,C_3,C_4,C_2\times C_2,S_3$ embed in $S_4$. Concrete subgroups areThere is no $C_5$ subgroup, since $5$ does not divide $24$. There is no $C_6$ subgroup either: the order of a group element in a symmetric group is the least common multiple of the lengths in its disjoint cycle decomposition, and on four letters the only possibilities are $1,2,3,4$.

The orbit-stabilizer theorem identifies the orbit of a group action of $s$ with the set of left cosets of $G_s$. Explicitly, there is a bijectionHere $G/G_s$ denotes a set of cosets; it is not asserted to be a quotient group. To prove the identification, observe thatThis equivalence proves both well-definedness and injectivity. Every element of the orbit of a group action has the form $g\cdot s$, proving surjectivity.

In particular, for a finite group, the cardinalities satisfyThe last equality follows from Lagrange's theorem applied to the stabilizer subgroup.

Lagrange's theorem states that for a subgroup $K$ of a finite group $G$,in particular, $|K|$ divides $|G|$. Here $[G:K]$ is the subgroup index, the number of left cosets of $K$.

The order of a group element $g\in G$ is the least positive integer $k$ such that $g^k=e$. It exists because $G$ is finite. The cyclic subgroup $⟨g⟩$ consists of exactly $e,g,\dots ,g^{k-1}$, so has order $k$. Applying Lagrange's theorem to this subgroup gives

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

Let $P$ and $Q$ be subgroups of orders five and seven, which exist by i. Their normalizers contain $P$ and $Q$, respectively. By Lagrange's theorem, $N_G(P)$ has order $5$ or $35$, and $N_G(Q)$ has order $7$ or $35$. If either has order $35$, its subgroup is normal by iv.

Suppose neither is normal. Then iv gives at least seven distinct subgroups of order five and at least five distinct subgroups of order seven. Distinct subgroups of the same prime order intersect only in the identity, and a subgroup of order five and one of order seven also intersect only in the identity by Lagrange's theorem. Thus $G$ would have at leastnonidentity elements, contradicting $|G|-1=34$. Hence at least one of $P,Q$ is a normal subgroup.

By Lagrange's theorem, every nonidentity group element has order $5$, $7$, or $35$. If some element has order $35$, its seventh and fifth powers have orders $5$ and $7$, respectively, and we are done.

Otherwise every nonidentity element has order $5$ or $7$. Distinct subgroups of a given prime number order intersect only in the identity: a shared nonidentity element generates both. If there were no element of order $5$, the $34$ nonidentity elements would therefore partition into sets of six, the nonidentity elements of the cyclic subgroups of order seven. This would force $6\mid 34$, a contradiction. If there were no element of order $7$, the analogous partition into sets of four would force $4\mid 34$, again a contradiction. Thus elements of both orders exist.

Let the finite set $G/H$ contain $r$ distinct left cosets. By i, they form a disjoint partition of a set $G$, and by ii each has $|H|$ elements. If $H$ is also finite, counting this finite disjoint set union givesThus $G$ is a finite group. This is the counting argument behind Lagrange's theorem.

The multiplicative group of a finite field $p$ has group order $p-1$: its elements are exactly the nonzero residue classes. For any prime factor $p$ of $2^q-1$, i and Lagrange's theorem therefore implySince $q$ and $p-1$ are positive integers, $q\le p-1$, and henceThis applies to every prime factor, including repeated prime factors viewed as the same prime number.

A smallest example is the Klein four-groupwith coordinatewise addition modulo $2$. Every nonidentity element has order of a group element $2$, so none generates all four elements; hence this group is not a cyclic group.

A group of group order $1$ is a cyclic group. A group of group order $2$ or $3$ is also a cyclic group: choose a nonidentity element; by Lagrange's theorem, its order of a group element is a nontrivial divisor of that prime number, so equals the whole group order. Thus no smaller example exists, and the smallest possible group order is $4$.