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The canonical presentation of a group uses one symbol for each group element and all words that evaluate to the identity element as relators. Evaluation is a surjective group homomorphism from the free group on the underlying set; the first isomorphism theorem proves that the resulting group presentation recovers the original group. Equivalently the multiplication-table relators $s_g{s}_h{s}_{gh}^{-1}$ normally generate its kernel: they reduce every word to the symbol for its value. ^[ref]

The center of a group consists of the elements that commute with every element of that group:where $C_G(g)$ is the centralizer of $g$. These are operations whose position in any product can be changed freely. The center is an abelian group and a characteristic subgroup, hence a normal subgroup. A group is an abelian group precisely when $Z(G)=G$.

For the symmetric group $S_3$, the conjugacy classes areAn element belongs to the center precisely when its conjugacy class has one element, so $Z(S_3)=\{e\}$. Yet commutator subgroup of the symmetric group on three letters gives $S_3^{\prime }=A_3$, with three elements. Thus belonging to the commutator subgroup does not mean commuting with everything. The alternating group $A_3$ is an abelian group as a subgroup, but its nonidentity elements do not commute with the transpositions in the surrounding $S_3$.

For the dihedral group $D_8$ of group order eight in abelianization of the square symmetry group, the center is $\{e,r^2\}$. A rotation $r^i$ commutes with $s$ precisely when $r^i=r^{-i}$, which requires $i$ even. No reflection $r^{i}s$ commutes with $r$, since $(r^{i}s)r=r^{i-1}s$ whereas $r(r^{i}s)=r^{i+1}s$. ConsequentlyThe half-turn can be performed before or after any square symmetry with the same result. Its appearance both here and in the commutator subgroup is a feature of this example, rather than a general equality. See Conrad, Dihedral groups I, theorem 3.4 and example 3.5.

The center and the commutator subgroup answer different useful questions: which operations already commute with everything, and which operations must disappear for the entire group to commute. In an abelian group these answers are $Z(G)=G$ and $G^{\prime }=\{e\}$. Another way to use the center is through conjugation: the group homomorphism $g\mapsto (h\mapsto gh{g}^{-1})$ has group kernel $Z(G)$, so the first isomorphism theorem gives $G/Z(G)\cong \mathrm{Inn}(G)$, the inner automorphism group. The quotient group by the center thus records the distinct conjugation operations; it need not be an abelian quotient.

A nontrivial finite p-group always has a nontrivial center. In the class equation, each noncentral conjugacy class has size divisible by $p$, so $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$. The identity element ensures $Z(G)$ is nonempty, hence it has at least $p$ elements. See McKernan, MIT 18.703, lemmas 23.6–23.7 and definition-lemma 23.10.

In coordinates $h(x,y,z)=\left(\begin{array}{ccc}1& x& y\\ 0& 1& z\\ 0& 0& 1\end{array}\right)$, the Heisenberg group law is $h(x,y,z)h(a,b,c)=h(x+a,y+b+xc,z+c)$. Commutation for every $(a,b,c)$ forces $xc=az$, hence $x=z=0$. The center of a group is therefore the $y$-axis. The surjective group homomorphism $h(x,y,z)\mapsto (x,z)$ has exactly this kernel of a group homomorphism, giving the stated quotient group by the first isomorphism theorem.

For the symmetric group $S_3$, take the transpositions $x=(12)$ and $y=(23)$. Products of permutations apply the rightmost factor first. ThenThe last permutation sends $1\mapsto 3$, $3\mapsto 2$, and $2\mapsto 1$: two swaps fail to commute by an entire three-cycle.

Every group commutator has sign of a permutation $+1$, so $S_3^{\prime }\subseteq A_3$, the alternating group of even permutations. Conversely, the displayed group commutator generates $A_3=\{e,(123),(132)\}$. Therefore $S_3^{\prime }=A_3$. The two cosets are the even permutations and the odd permutations, and their multiplication is just multiplication of signs:This is the first isomorphism theorem applied to the sign of a permutation. Once commutativity is forced, all three transpositions become indistinguishable, and both three-cycles become the identity element. The surviving information is exactly whether the permutation is even or odd. Compare the very different center of a group $Z(S_3)=\{e\}$.

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 any group action with group kernel $K$, the quotient group $G/K$ has a faithful group action on the same set, given byTo check independence of the coset representative, replace $g$ by $gk$ with $k\in K$: then $(gk)\cdot x=g\cdot (k\cdot x)=g\cdot x$. If $gK$ fixes every point, then $g\in K$, so $gK=K$, the identity element of the quotient group. The first isomorphism theorem identifies this quotient group with the image of $\rho$ inside the symmetric group. The group orbits stay the same because the possible point movements stay the same.

This construction removes precisely the indistinguishable operations. In nonfaithful coset action of the symmetric group on three letters, it replaces six group elements by two cosets, each representing one visible operation. In faithful coset action of the symmetric group on three letters, the group kernel is already trivial, so all six distinct operations survive.

For a group homomorphism $\phi :G\to H$, two inputs give the same output exactly when $\phi (h^{-1}g)=e_H$. Consequently the fibre over an output $\phi (h)$ is the left coset $h\ker \phi$. These cosets identify precisely the inputs indistinguishable to $\phi$, which is the mechanism behind the first isomorphism theorem.

For a group homomorphism $\phi :G\to H$, the group isomorphism $g\ker \phi \mapsto \phi (g)$ givesThe group kernel records which inputs become indistinguishable, and the image records which outputs actually occur. The first isomorphism theorem says that identifying precisely those indistinguishable inputs gives a group isomorphic to the image. It is one of the general isomorphism theorems. ^[ref].

For example, reduction modulo three gives a surjective group homomorphism from the additive cyclic group $\mathbb{Z}/6\mathbb{Z}$ to $\mathbb{Z}/3\mathbb{Z}$. Its group kernel is $\{0,3\}$. The three cosets $\{0,3\}$, $\{1,4\}$, and $\{2,5\}$ become the three outputs zero, one, and two, respectively. ThusThe quotient group diagram shows both the identification and the resulting operation. More generally, fibres of a group homomorphism are cosets of its kernel explains why this map is well-defined and injective; it is surjective onto the image by definition.

The isomorphism theorems relate homomorphisms, their kernels and images, and quotient constructions in algebra. For groups, the first isomorphism theorem identifies a homomorphism's image with the quotient group by its group kernel. Analogous statements include the first isomorphism theorem for rings and the First isomorphism theorem for vector spaces: identifying precisely the inputs with the same output preserves the remaining algebraic structure.

For a subgroup $H$ of a group $G$, the normalizer isIt is the stabilizer subgroup of $H$ under the conjugation action on subgroups. Its elements can permute members of $H$; they need not commute with them. For $r=(123)$ in the symmetric group $S_3$, the cyclic subgroup $H=⟨r⟩$ has normalizer $S_3$, although the centralizer of $r$ has only three elements, as illustrated in centralizer and normalizer.

Every element of $H$ preserves $H$ under conjugation, so $H\subseteq N_G(H)$. Moreover, $H$ is a normal subgroup of $N_G(H)$, and this is the largest subgroup of $G$ with that property. Indeed, any subgroup in which $H$ is a normal subgroup consists entirely of elements preserving $H$ under conjugation. In particular, $N_G(H)=G$ exactly when $H$ is a normal subgroup of $G$.^[ref]

The orbit-stabilizer theorem counts distinct conjugate subsets of a subgroup byFor $H=⟨(123)⟩$ in $S_3$, there is just one, since $[S_3:N_{S_3}(H)]=6/6=1$. For the subgroup $K=⟨(12)⟩=\{e,(12)\}$, preserving $K$ forces its unique nonidentity element to stay fixed. Hence $N_{S_3}(K)=C_{S_3}((12))=K$, and its three distinct conjugate subsets are $\{e,(12)\}$, $\{e,(13)\}$ and $\{e,(23)\}$. This is the subgroup counterpart of counting a conjugacy class with a centralizer: the object being stabilized determines which denominator belongs in the count.^[ref]

The equality in the definition is essential. For a subgroup $H$ that is a finite group, an inclusion $gH{g}^{-1}\subseteq H$ already gives equality by equal cardinalities. For an infinite subgroup, that inference can fail: conjugation might carry it onto a proper subgroup. The normal subgroup criterion is therefore, with the explicit hypothesis $H\le G$,No finiteness or commutativity assumption is needed. A normal subgroup is preserved as a set; its elements need not lie in the center of a group.

There is a second useful count, distinct from counting conjugate subgroups. The normalizer acts on $H$ itself by group automorphisms, giving the group homomorphismIts group kernel is $C_G(H)$, so the centralizer of a subset is a normal subgroup of $N_G(H)$ and the first isomorphism theorem givesThis quotient group records the different rearrangements of $H$ induced by the surrounding group. It gives a faithful group action on $H$, but need not be the entire automorphism group of $H$. For example, take the ambient group $G=H=⟨(123)⟩$. Since $G$ is an abelian group, $N_G(H)=C_G(H)=G$, so the quotient group is trivial; nevertheless $r\mapsto r^{-1}$ is a nonidentity group automorphism of $H$. Returning to the ambient symmetric group $S_3$ makes that group automorphism available through a reflection, and $N_{S_3}(H)/C_{S_3}(H)\cong C_2$. Its two cosets are the three rotations and the three reflections.

The subgroup $H$ itself need not be contained in $C_G(H)$: their intersection is the center of a group $Z(H)$. It is contained precisely when $H$ is an abelian group. Thus $N_G(H)/H$, which exists because $H$ is a normal subgroup of its normalizer, and $N_G(H)/C_G(H)$ answer different questions and should not be substituted for one another.

The conjugacy classes of the symmetric group on three letters have sizes $1,3,2$: the identity, the three transpositions and the two 3-cycles. A normal subgroup is a union of whole conjugacy classes and contains the identity. Its candidate orders are therefore $1,4,3,6$. Lagrange's theorem excludes four, leaving precisely $\{e\}$, $A_3$ and $S_3$. The order-three subgroup is the cyclic group of the two 3-cycles and the identity; it is also the kernel of a group homomorphism of the sign map.

The first isomorphism theorem gives the corresponding quotients $S_3$, $C_2$ and the trivial group. Thus these are exactly the possible images of surjective group homomorphisms from $S_3$. Each occurs, by the identity map, the sign map and the constant identity map, respectively.

Use the matrices of solution. Their projective classes satisfy $[-M]=[M]$, since $-I_2\in N$. The group homomorphism is therefore explicitlyThese four classes are distinct: the two diagonal representatives are not scalar multiples of one another, the two off-diagonal representatives are not scalar multiples of one another, and a diagonal representative cannot be a nonzero scalar multiple of an off-diagonal one.

Although $A$ and $B$ anticommute as matrices, their projective classes commute:Furthermore $[A{]}^2=[B{]}^2=[I_2]$ and $[C{]}^2=[-I_2]=[I_2]$. Thus every nonidentity element of the image has order two. An explicit group isomorphism to the Klein four-group isIndeed reordering products introduces only a sign, which disappears in the projective linear group, so this map turns multiplication into addition modulo two in each coordinate. Equivalently, $\ker \tilde{R}=G\cap N=\{\pm I_2\}$ and the first isomorphism theorem givesThis projective image of the square symmetry group illustrates how a noncommutative group can have an Abelian projective image: its scalar commutator becomes the identity class.

First the group homomorphism is surjective. For any $W\in U(n)$, choose $z\in U(1)$ with $z^n=\det W$ and set $U=z^{-1}W$. Then $U$ is unitary and $\det U=z^{-n}\det W=1$, so $W=f(z,U)$.

Next $f(z,U)=I_n$ is equivalent to $U=z^{-1}{I}_n$, and the special unitary group condition becomes $z^n=1$. With $\zeta =e^{2\pi i/n}$, the kernel of a group homomorphism is thereforeIt is the cyclic central subgroup generated by $(\zeta ,{\zeta }^{-1}{I}_n)$; the scalar and special unitary group factors are paired with inverse phases. The first isomorphism theorem now givesExplicitly the isomorphism sends the coset $[(z,U)]$ to $zU$. The $n$ possible choices of the root $z$ differ precisely by multiplication by elements of $K$, which explains why the product before quotienting is not a unique factorization.

First prove surjectivity. For any $A\in U(n)$, its determinant has modulus one. Choose $z\in U(1)$ with $z^n=\det A$ and define $U=z^{-1}A$. Then $U^{†}U=I$ and $\det U=z^{-n}\det A=1$, so $A=f(z,U)$.

Next, $f(z,U)=I$ is equivalent to $U=z^{-1}I$. This lies in $SU(n)$ precisely when $z^n=1$. Therefore the kernel of a group homomorphism is the central cyclic groupIts generator is $(\zeta ,{\zeta }^{-1}{I}_n)$, which has order $n$. The inverse phases in the two factors are essential: this is an explicitly embedded subgroup, not a quotient by scalar matrices in just one factor. The first isomorphism theorem now givesThis also holds as a Lie group isomorphism: the induced map is smooth and a continuous bijection from a compact quotient space to the Hausdorff unitary group, and its differential is the Lie algebra isomorphism $(it,X)\mapsto itI+X$. For $n=1$ the kernel of a group homomorphism is trivial, consistently with $SU(1)=\{1\}$.

A subnormal series is a finite chain $1=G_r⊴G_{r-1}⊴\cdots ⊴G_0=G$, in which each term is normal in the preceding term. Its factors are the quotient groups $G_i/G_{i+1}$. A normal series of a group imposes the stronger condition that every term be normal in $G$. These conditions should not be conflated: inserting a subgroup normal in an intermediate term need not preserve normality in the whole group. A refinement inserts additional terms, and two series are equivalent when their nontrivial factors can be paired by group isomorphisms.

The basic refinement mechanism is the Zassenhaus butterfly lemma. For $A^{\prime }⊴A$ and $B^{\prime }⊴B$, let $C=A\cap B$ and $D=(A^{\prime }\cap B)(A\cap B^{\prime })$. Both factors of $D$ are normal in $C$, so $D⊴C$. The map from $C$ onto $A^{\prime }C/A^{\prime }(A\cap B^{\prime })$ has kernel $D$: if an element of $C$ equals $a^{\prime }{b}^{\prime }$ with $a^{\prime }\in A^{\prime }$ and $b^{\prime }\in A\cap B^{\prime }$, then $a^{\prime }=c{b}^{\prime -1}\in A^{\prime }\cap B$. Conversely each product in $D$ is in the kernel. The first isomorphism theorem therefore givesThe denominators are normal in their numerators: conjugation by $C$ preserves them, while conjugation by $A^{\prime }$ or $B^{\prime }$ changes their elements only by a factor already in that normal subgroup.

Given two subnormal series $(G_i)$ and $(H_j)$, refine the first by $G_{i+1}(G_i\cap H_j)$, and the second by $H_{j+1}(H_j\cap G_i)$. The endpoints are $G_i,G_{i+1}$ and $H_j,H_{j+1}$ respectively, and the Zassenhaus butterfly lemma pairs the successive factors indexed by $(i,j)$ and $(j,i)$. This proves the Schreier refinement theorem.

A composition series is a strict subnormal series all of whose factors are simple groups. Every finite group has one: choose a maximal proper normal subgroup, repeat inside it, and use decreasing orders to terminate. A simple factor cannot admit a nontrivial refinement. Applying the preceding refinement theorem to two composition series proves the Jordan–Hölder theorem: their lengths and multisets of simple factors agree, up to group isomorphism. The factors are the Jordan–Hölder factors, and their number is the composition length. If $N⊴G$, lifting a composition series of $G/N$ and adjoining one for $N$ gives

These factors do not determine the whole group, because they do not determine how its group extensions are assembled. For example, $C_4$ and the Klein four-group both have two factors $C_2$, but one contains an element of order four and the other does not. The symmetric group $S_3$ has a composition series $1<C_3<S_3$, with factors $C_3,C_2$, whereas the simple group $A_5$ has the single factor $A_5$.

A chief series is a maximal strict normal series of a group; each chief factor is a minimal nontrivial normal subgroup in the appropriate quotient of $G$. Finite chief factors are characteristically simple. To justify their structure, let $N$ be any nontrivial finite characteristically simple group and choose a minimal nontrivial normal subgroup $M$ of $N$. Its distinct images under automorphisms of $N$ are minimal normal subgroups, so distinct ones intersect trivially and commute. Choose a maximal family giving an internal direct product of groups. If any other image were not contained in that product, its intersection with the product would be trivial by minimality, and it could be added, a contradiction. Hence this product is characteristic and equals $N$. Each factor $M$ is simple: a proper nontrivial normal subgroup of $M$ would be normal in the whole direct product $N$, contrary to the choice of $M$. Thus $N$ is a direct product of groups of isomorphic finite simple groups. This applies to chief factors by conjugation from $G$.

Chief and composition factors need not coincide. In $A_4$, the chain $1<V_4<A_4$ is a chief series, since conjugation cycles the three order-two subgroups of $V_4$. A composition series instead inserts any one of those subgroups, giving factors $C_2,C_2,C_3$. Its inserted order-two subgroup is normal in $V_4$ but not in $A_4$. Chief series retain the ambient group's conjugation information that composition series discard.

The derived series gives a canonical criterion for a group to be soluble: put $G^{(0)}=G$ and $G^{(i+1)}=[G^{(i)},G^{(i)}]$. Its terms are characteristic, and each factor is Abelian. A finite group is soluble exactly when this series reaches the identity, equivalently when all its composition factors are cyclic of prime order. For one direction, the derived series supplies Abelian factors, whose simple refinements are cyclic of prime order. Conversely an extension of two soluble groups is soluble: if $(G/N{)}^{(r)}=1$, then $G^{(r)}\subseteq N$; if $N^{(s)}=1$, then $G^{(r+s)}=1$. Induction up a composition series with cyclic prime factors proves the other direction. In a finite soluble group, each minimal normal subgroup is elementary Abelian: its characteristic commutator subgroup must be trivial, its primary components force a single prime, and its characteristic subgroup of $p$th powers must be trivial. Thus every chief factor is a vector space over some ${\mathbb{F}}_p$. In a general finite group the Jordan–Hölder factors can instead be nonabelian simple groups.

The central series demand more than Abelian factors: each factor must be central in the corresponding quotient of the whole group. Define the upper central series by $Z_0=1$ and $Z_{i+1}/Z_i=Z(G/Z_i)$, and the lower central series by ${\gamma }_1=G$ and ${\gamma }_{i+1}=[{\gamma }_i,G]$. A group is nilpotent exactly when the upper series reaches $G$, equivalently when the lower series reaches the identity. Indeed, for any ascending central series $1=C_0\le \cdots \le C_c=G$, induction gives $C_i\subseteq Z_i$ and ${\gamma }_{c+1-i}\subseteq C_i$; reversing the lower series gives an ascending central series when it terminates. The least such $c$ is the nilpotency class. A central series has Abelian factors, so every nilpotent group is soluble; the converse fails, as $S_3$ is soluble but has trivial group center.

For a finite p-group, the class equation shows that the group center is nontrivial, since every noncentral conjugacy class has size divisible by $p$. Applying the same argument to successive central quotients makes the upper central series reach the group. More generally a finite group is nilpotent exactly when it is the direct product of groups of its Sylow subgroups. Here is the nontrivial direction. A central series gives the normalizer condition for nilpotent groups: for a proper subgroup $H$, choose the first $Z_i$ not contained in $H$. Then $[Z_i,H]\subseteq Z_{i-1}\subseteq H$, so $Z_i\subseteq N_G(H)$ and $N_G(H)>H$. If a Sylow subgroup $P$ had proper normalizer $N$, then $N_G(N)>N$; but $P$ is the unique Sylow subgroup of $N$, hence characteristic there, and $N_G(N)\subseteq N_G(P)=N$, a contradiction. Thus all Sylow subgroups are normal, different ones commute because their intersection is trivial, and they form an internal direct product. Conversely each finite Sylow subgroup is nilpotent, and products of their central series give a central series for their direct product.

Composition series record simple building blocks, chief series record ambient normal structure, derived series detect solubility, and central series detect nilpotence. Their associated factors and extension data explain why the same group can have very different useful series.

For the transfer homomorphism, choose representatives $T$ of the right cosets $H\backslash G$. For $t\in T$ and $g\in G$, write $tg=h(t,g)t_g$, where $h(t,g)\in H$ and $t_g\in T$. The permutation $t\mapsto t_g$ represents right multiplication on the cosets. DefineThe product has no order ambiguity in the abelianization $H/H^{\prime }$. Replace each representative $t$ by $a_{t}t$, with $a_t\in H$. The new factor is $a_{t}h(t,g)a_{t_g}^{-1}$. In the Abelian quotient, the extra products cancel because $t\mapsto t_g$ permutes $T$. Thus the transfer is independent of the transversal. Moreover,Multiplying and using the same permutation of representatives gives $V(g_1{g}_2)=V(g_1)V(g_2)$. This proves that group transfer is a well-defined group homomorphism.

For the Burnside transfer theorem, let $P\subseteq Z(N_G(P))$ be a Sylow $p$-subgroup. In particular $P$ is Abelian. We first prove the fusion fact that two elements of $P$ which are conjugate in $G$ must be equal. Suppose $u\in P$ and $v=g^{-1}ug\in P$. Both $P$ and $g^{-1}Pg$ centralize $v$, and they are Sylow subgroups of $C_G(v)$: they have the maximal $p$-order possible even in $G$. By Sylow theorems inside this centralizer, choose $h\in C_G(v)$ with $h^{-1}{g}^{-1}Pgh=P$. Thus $gh\in N_G(P)$ andBut $gh$ centralizes $P$ by hypothesis, so $v=u$.

Evaluate transfer to $P$ on $u\in P$. Partition the right cosets of $P$ into cycles under multiplication by $u$. For a cycle of length $r$ beginning at $Pt$, the product of its transfer factors telescopes to $t{u}^r{t}^{-1}\in P$. It is conjugate to $u^r\in P$, so the fusion fact makes it equal to $u^r$. The cycle lengths sum to $m=[G:P]$. Since $P$ is Abelian, all factors commute andThe integer $m$ is prime to $p$, and this power map is an automorphism of the finite Abelian $p$-group $P$: choose an inverse exponent modulo the exponent of $P$. Therefore $V_{G,P}:G\to P$ is surjective. Its kernel $K$ is normal and the first isomorphism theorem givesThus $K$ is the required normal p-complement; it also has trivial intersection with $P$.

Now suppose $P$ is cyclic of order $p^a$, where $p$ is the smallest prime divisor of $|G|$. Conjugation embeds $N_G(P)/C_G(P)$ into $\mathrm{Aut}(P)$, whose order is $p^{a-1}(p-1)$. Since $P\subseteq C_G(P)$ and $P$ is Sylow in $N_G(P)$, the quotient has order prime to $p$. Any remaining prime factor would divide $p-1$ and hence be smaller than $p$, but would also divide $|G|$, an impossibility. Thus $N_G(P)=C_G(P)$ and $P\subseteq Z(N_G(P))$. The Burnside transfer theorem proves that a cyclic Sylow subgroup at the least prime divisor gives a normal complement.

Apply this to a finite nonabelian simple group. If $p^3\nmid |G|$, its Sylow $p$-subgroup has order $p$ or $p^2$. A cyclic Sylow subgroup would give a normal complement; simplicity would then force either a trivial complement and a $p$-group, or an improper complement, neither possible for a nonabelian simple group. Groups of order $p^2$ are Abelian: their nontrivial group center, supplied by the class equation, is either the whole group or has cyclic quotient, which also forces the group Abelian. Thus the only remaining Sylow possibility is $P\cong C_p^2$.

The conjugation quotient $N_G(P)/C_G(P)$ now embeds in $G{L}_2(p)$ and is still prime to $p$. Its possible prime divisors come fromIf $p$ is odd, every prime divisor of $p-1$ is smaller than $p$, and every prime divisor of the even number $p+1$ is at most $(p+1)/2<p$. Since $p$ was the least prime divisor of $|G|$, the quotient must be trivial. Then the Burnside transfer theorem again contradicts simplicity. If $p=2$, the only nontrivial odd divisor of $|G{L}_2(2)|=6$ is three. Trivial conjugation would again give a normal complement, so three must divide $|G|$; the Sylow subgroup has order four. We have proved the least-prime divisibility constraint for a finite simple group:

For infinitely many instances of the latter alternative only, takeThese are simple, as permitted in the question (and also covered by the next solution). For $q=5^{2m+1}$, the group order is $q(q^2-1)/2$. Since $q\equiv 5(\mathrm{mod}8)$, $q-1$ is divisible by four but not eight and $q+1$ by two but not four. Hence the order is divisible by four but not eight. Also $q\not\equiv 0(\mathrm{mod}3)$, so $3\mid q^2-1$. Their smallest prime divisor is two andTheir orders strictly increase with $m$, so they give infinitely many distinct examples.

The identity belongs to the center of a group. If $a,b$ are central, then for every $g\in G$,so the center is a subgroup. For a central $a$, $ga{g}^{-1}=a$ for every $g$, giving $gZ(G)g^{-1}=Z(G)$. Thus the center is a normal subgroup.

Write a matrix of the Heisenberg group as $h(x,y,z)$. Direct matrix multiplication and inversion giveThe identity is $h(0,0,0)$ and each matrix has determinant one, so these formulas show closure under products and inverses inside $G{L}_3(\mathbb{R})$. Hence $H$ is a subgroup.

The two products $h(x,y,z)h(a,b,c)$ and $h(a,b,c)h(x,y,z)$ differ only in their upper-right entries, by $xc-az$. Commutation with every $h(a,b,c)$ therefore requires $xc=az$ for all real $a,c$. Setting $a=0,c=1$ forces $x=0$; setting $a=1,c=0$ forces $z=0$. These conditions are also sufficient. ThusDefine the surjective group homomorphism $F:H\to ({\mathbb{R}}^2,+)$ by $F(h(x,y,z))=(x,z)$. The multiplication formula gives $F(h{h}^{\prime })=F(h)+F(h^{\prime })$, and its kernel of a group homomorphism is precisely $Z(H)$. The first isomorphism theorem yieldsThis center quotient of the real Heisenberg group can also be checked directly: two elements have the same $(x,z)$ coordinates exactly when they differ by multiplication by a central element, and quotient multiplication adds these coordinates.

Let $G$ act on its $n$ left cosets by $g\cdot (xH)=gxH$. This is well-defined and gives a group homomorphism $\rho :G\to S_n$. Its kernel $K$ is a normal subgroup. An element lies in this kernel exactly when it fixes every coset, soThis is the normal core of a subgroup. By the first isomorphism theorem, $G/K\cong \rho (G)$, a subgroup of $S_n$. The Lagrange theorem therefore givesIn particular $K$ has finite index even when $G$ is infinite.

Now let $H$ be generated by an element of order $q$ in a group of order $pq$. Then $|H|=q$ and $[G:H]=p$. The preceding result gives a normal $K\subset H$ with $[G:K]\mid p!$. Since $H$ has prime order, its subgroup $K$ is either $\{e\}$ or all of $H$. The first choice would imply $pq\mid p!$, which is impossible because the prime $q>p$ divides none of the factors of $p!$. Therefore $K=H$, proving that every subgroup generated by an element of order $q$ is normal. No classification of groups of order $pq$ is needed.

Here $A_1(K)$ denotes the adjoint Chevalley group of type A1; the isogeny form matters, since the simply connected form is $S{L}_2(K)$ itself. Over $\mathbb{C}$, take the sl2 Lie algebra with Chevalley basisTheir Lie brackets satisfy $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. Conjugation defines a group homomorphism $\rho :S{L}_2(\mathbb{C})\to \mathrm{Aut}({\mathfrak{sl}}_2(\mathbb{C}))$. Since $e^2=f^2=0$, $u(t)=\exp (te)$ and $l(s)=\exp (sf)$. Differentiating conjugation, or expanding its finite polynomial on this basis, givesThe adjoint Chevalley group is generated by these two families. Part (i) therefore proves that it is exactly the image of $\rho$. A matrix commuting with $e$ has the form $\left(\begin{array}{cc}a& b\\ 0& a\end{array}\right)$; also commuting with $f$ forces $b=0$. Thus the kernel of a group homomorphism $\ker \rho$ consists of scalar matrices of determinant one, namely $\{I,-I\}$. The first isomorphism theorem now gives $A_1(\mathbb{C})\cong S{L}_2(\mathbb{C})/\{\pm I\}=PS{L}_2(\mathbb{C})$.

For a general field, first compute the Chevalley basis root operators as integral polynomials, and only then reduce their coefficients to $K$. Explicitly,They are still precisely conjugation by $u(t),l(s)$ on the three-dimensional trace-zero matrix space. These formulas remain valid in characteristic two, where $h=I$ but $e,h,f$ remain linearly independent. Part (i) still identifies the generated image with the conjugation image of $S{L}_2(K)$. The same computation with $e,f$ gives kernel of a group homomorphism $\{aI:a^2=1\}$, exactly its scalar center. ConsequentlyIn characteristic two the scalar center is trivial. The integral formulas avoid the invalid operation of dividing by factorials inside a field of small positive characteristic.

There are two necessary qualifications to the literal formulation. The whole group $G$ is itself a closed normal subgroup, so the claim about containment in the centre must concern proper closed normal subgroups. Also, the homomorphism conclusion needs a nontrivial target $H$: for example, $SU(2)\to \{1\}$ has surjective differential onto the zero Lie algebra, but its group kernel is all of $SU(2)$ and no quotient by a finite subgroup is a one-point group. We prove the precise assertions, including these exceptional cases.

Write $\mathfrak{g}=\mathrm{Lie}(G)$. The centre $\mathfrak{z}(\mathfrak{g})$ is invariant under the Adjoint representation, so irreducibility makes it either zero or all of $\mathfrak{g}$. In the latter case $\mathfrak{g}$ is Abelian. For connected $G$, the adjoint action is then trivial: the group is generated by exponentials, and ${\mathrm{Ad}}_{\exp X}=e^{\mathrm{ad}X}=I$. A nonzero trivial representation is irreducible only in dimension one. A connected compact one-dimensional Lie group is a circle, so $G\cong U(1)$.

For every other $G$, the Lie algebra of its centre is zero. The centre is a closed compact Lie subgroup; a zero-dimensional compact Lie group is finite. Let $N$ be a closed normal subgroup. The closed-subgroup theorem identifies $\mathfrak{n}=\mathrm{Lie}(N)$ as an adjoint-invariant subspace of $\mathfrak{g}$, hence $\mathfrak{n}=0$ or $\mathfrak{n}=\mathfrak{g}$. If the latter holds, $N$ contains an identity neighbourhood and is open, so connectedness forces $N=G$. If $N$ is proper, it therefore has zero Lie algebra and is finite. For any $a\in N$, the continuous map $g\mapsto ga{g}^{-1}$ takes connected $G$ into the finite set $N$, so it is constant and equal to $a$. ThusProper closed subgroups of the circle are finite as well, and of course central.

Now let $d\Phi$ be surjective. The submersion theorem makes $\Phi (G)$ contain an identity neighbourhood in $H$, so it is an open subgroup. Since $H$ is connected, $\Phi (G)=H$. Its group kernel $F$ is closed and normal. If $H$ is nontrivial, $F$ is proper and the preceding argument gives a finite central group kernel; this also holds for $G\cong U(1)$, whose proper closed subgroups are finite. The Lie-group first isomorphism theorem givesIf $H$ is trivial, the correct statement is instead $H=G/G$.

For the Spin groups, use the real Clifford algebra with $v^2=-\Vert v{\Vert }^2$. The group $\mathrm{Spin}(m)$ consists of products of an even number of unit vectors inside its invertible even part. Conjugation on the embedded ${\mathbb{R}}^m$ defines the standard double covering $\mathrm{Spin}(m)\to SO(m)$ with group kernel $\{\pm 1\}$. The Clifford relations show that a unit-vector reflection, with the usual twisted conjugation, is a hyperplane reflection; the even products give orientation-preserving transformations. The Cartan–Dieudonné theorem states that every orthogonal transformation is a product of reflections, proving surjectivity. For $m\ge 3$ this connected covering group is simply connected; $\mathrm{Spin}(2)$ is a circle.

Here is an explicit exterior-square construction of the requested six-dimensional map. Put $E={\mathbb{C}}^4$ with its standard Hermitian form and volume $e_1\wedge e_2\wedge e_3\wedge e_4$. On ${\Lambda }^{2}E$, write $e_{ij}=e_i\wedge e_j$ and define an antilinear map $J$ byIt satisfies $J^2=1$. Intrinsically, if $u\wedge v=B(u,v)\mathrm{vol}$ and the Hermitian inner product $h$ is linear in its first variable, then $B(u,Jv)=h(u,v)$. Because $SU(4)$ preserves both $B$ and $h$, it commutes with $J$. This is the real structure of the SU4 exterior square.

The fixed space $W$ has the following real orthonormal basis, with all vectors divided by $\sqrt{2}$:Thus $W$ is six-dimensional over $\mathbb{R}$, its complexification is ${\Lambda }^{2}E$, and the exterior-square action restricts to an orthogonal representation on $W$. Its determinant is one because $SU(4)$ is connected. We obtain $R:SU(4)\to SO(W)\cong SO(6)$.

To compute the group kernel, diagonalize $g\in SU(4)$ with eigenvalues $z_1,\dots ,z_4$. If $R(g)=I$, its complexified exterior-square action is the identity, hence $z_i{z}_j=1$ for every $i<j$. Comparing pairs shows that all $z_i$ are equal, and their common value has square one. A unitary matrix with that single eigenvalue is scalar, so $\ker R=\{\pm I_4\}$. Conversely, those two scalar matrices plainly act trivially on the exterior square.

The differential of $R$ is injective because its group kernel is finite. Both Lie algebras have dimension $15$, so the differential is an isomorphism. Its image is consequently open in connected $SO(6)$, hence all of $SO(6)$. ThereforeThis also realizes $SU(4)$ as $\mathrm{Spin}(6)$, since $SU(4)$ is simply connected, by uniqueness of the connected simply connected covering group.