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Darko Dekan
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Introduction to Group Theory

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Group theory is the mathematical study of symmetry and structure. Instead of focusing on what particular objects are made of, group theory studies the operations that can be performed on them and the rules those operations obey.

The same abstract structure appears in many different places:

Once we recognize that these examples satisfy the same axioms, we can prove results about all of them at once.

Table of contents

Open Table of contents

What is a group?

A group is a set GG together with an operation

 ⁣:G×GG\ast \colon G \times G \longrightarrow G

that combines two elements a,bGa,b\in G to produce another element abGa\ast b\in G.

The pair (G,)(G,\ast) is a group when the following four properties hold.

1. Closure

For every a,bGa,b\in G,

abG.a\ast b\in G.

The operation never takes us outside the set.

2. Associativity

For every a,b,cGa,b,c\in G,

(ab)c=a(bc).(a\ast b)\ast c=a\ast(b\ast c).

Associativity allows us to write abca\ast b\ast c without ambiguity. Notice that associativity is not the same as commutativity: it changes the placement of parentheses, not the order of the elements.

3. Identity element

There exists an element eGe\in G such that, for every aGa\in G,

ea=ae=a.e\ast a=a\ast e=a.

The element ee is called the identity of the group.

4. Inverses

For every aGa\in G, there exists an element a1Ga^{-1}\in G such that

aa1=a1a=e.a\ast a^{-1}=a^{-1}\ast a=e.

The element a1a^{-1} is called the inverse of aa.

We often suppress the operation symbol and write

abab

instead of aba\ast b. In this notation, the group axioms become

(ab)c=a(bc),ea=ae=a,aa1=a1a=e.(ab)c=a(bc),\qquad ea=ae=a,\qquad aa^{-1}=a^{-1}a=e.

The integers under addition

The most familiar example of a group is

(Z,+).(\mathbb Z,+).

Let a,b,cZa,b,c\in\mathbb Z. Then:

a+bZ,a+b\in\mathbb Z, (a+b)+c=a+(b+c),(a+b)+c=a+(b+c), 0+a=a+0=a,0+a=a+0=a,

and

a+(a)=(a)+a=0.a+(-a)=(-a)+a=0.

Therefore (Z,+)(\mathbb Z,+) is a group. Its identity is 00, and the inverse of aa is a-a.

Because the operation is addition, we usually write the inverse as a-a rather than a1a^{-1}. Similarly, repeated addition is written

na=a+a++an timesna=\underbrace{a+a+\cdots+a}_{n\text{ times}}

for n>0n>0.

A non-example: the integers under multiplication

Consider (Z,)(\mathbb Z,\cdot). Multiplication is closed and associative, and 11 is an identity:

1a=a1=a.1\cdot a=a\cdot1=a.

However, most integers do not have multiplicative inverses in Z\mathbb Z. For example, an inverse of 22 would have to satisfy

2x=1,2x=1,

so x=12x=\frac12, but 12Z\frac12\notin\mathbb Z. Therefore

(Z,)(\mathbb Z,\cdot)

is not a group.

If we instead use the nonzero rational numbers,

Q×=Q{0},\mathbb Q^\times=\mathbb Q\setminus\{0\},

then (Q×,)(\mathbb Q^\times,\cdot) is a group because every q0q\neq0 has the inverse

q1=1q.q^{-1}=\frac1q.

We exclude 00 because no number xx can satisfy 0x=10x=1.

Abelian and non-abelian groups

A group (G,)(G,\ast) is abelian if

ab=baa\ast b=b\ast a

for all a,bGa,b\in G.

The groups (Z,+)(\mathbb Z,+) and (Q×,)(\mathbb Q^\times,\cdot) are abelian. Not every group is.

For example, rotations in three-dimensional space generally do not commute. Rotating an object 9090^\circ around the xx-axis and then 9090^\circ around the yy-axis usually produces a different result from performing the rotations in the opposite order:

RxRyRyRx.R_xR_y\neq R_yR_x.

The order of operations matters in a non-abelian group.

Basic consequences of the axioms

The group axioms are short, but they imply many useful facts.

The identity is unique

Suppose ee and ff are both identity elements. Since ee is an identity,

ef=f.ef=f.

Since ff is an identity,

ef=e.ef=e.

Therefore

e=f.e=f.

A group cannot have two different identity elements.

Every inverse is unique

Suppose bb and cc are both inverses of aa. Then

ab=ba=eab=ba=e

and

ac=ca=e.ac=ca=e.

Using associativity,

b=be=b(ac)=(ba)c=ec=c.b=be=b(ac)=(ba)c=ec=c.

Thus the inverse of aa is unique.

The inverse of a product

For a,bGa,b\in G,

(ab)1=b1a1.(ab)^{-1}=b^{-1}a^{-1}.

The order reverses because

(ab)(b1a1)=a(bb1)a1=aea1=aa1=e,(ab)(b^{-1}a^{-1}) =a(bb^{-1})a^{-1} =aea^{-1} =aa^{-1} =e,

and similarly,

(b1a1)(ab)=e.(b^{-1}a^{-1})(ab)=e.

This is analogous to putting on socks and then shoes: to undo the process, we remove the shoes first and the socks second.

Cancellation

If

ab=ac,ab=ac,

then multiplying both sides on the left by a1a^{-1} gives

a1(ab)=a1(ac).a^{-1}(ab)=a^{-1}(ac).

Associativity now gives

(a1a)b=(a1a)c,(a^{-1}a)b=(a^{-1}a)c,

so

eb=eceb=ec

and therefore

b=c.b=c.

This is the left cancellation law. Similarly,

ba=ca    b=c.ba=ca\implies b=c.

Finite groups and Cayley tables

A group is finite if it contains finitely many elements. The number of elements in GG is called the order of GG and is denoted by

G.|G|.

For a finite group, its operation can be displayed in a Cayley table.

Consider the integers modulo 33 under addition:

Z3={0,1,2}.\mathbb Z_3=\{0,1,2\}.

Addition is performed modulo 33, so

1+20(mod3)1+2\equiv0\pmod3

and

2+21(mod3).2+2\equiv1\pmod3.

Its Cayley table is

+3012001211202201\begin{array}{c|ccc} +_3 & 0 & 1 & 2\\ \hline 0 & 0 & 1 & 2\\ 1 & 1 & 2 & 0\\ 2 & 2 & 0 & 1 \end{array}

Every row and every column contains each group element exactly once. This follows from cancellation. For example, if two entries in the row corresponding to aa were equal, then

ab=acab=ac

would imply b=cb=c.

Subgroups

A subset HGH\subseteq G is a subgroup of GG if HH is itself a group under the operation inherited from GG. We write

HG.H\leq G.

Every group has at least two obvious subgroups:

{e}G\{e\}\leq G

and

GG.G\leq G.

The first is called the trivial subgroup.

For example, the even integers form a subgroup of (Z,+)(\mathbb Z,+):

2Z={2k:kZ}.2\mathbb Z=\{2k:k\in\mathbb Z\}.

Indeed, if 2m,2n2Z2m,2n\in2\mathbb Z, then

2m+2n=2(m+n)2Z,2m+2n=2(m+n)\in2\mathbb Z,

and

(2m)=2(m)2Z.-(2m)=2(-m)\in2\mathbb Z.

The identity 0=200=2\cdot0 also belongs to 2Z2\mathbb Z.

The subgroup test

A nonempty subset HGH\subseteq G is a subgroup if, for every a,bHa,b\in H,

ab1H.ab^{-1}\in H.

This single condition contains everything we need.

Taking a=ba=b gives

aa1=eH.aa^{-1}=e\in H.

Now that eHe\in H, taking a=ea=e gives

eb1=b1H.eb^{-1}=b^{-1}\in H.

Finally, since b1Hb^{-1}\in H, applying the condition to aa and b1b^{-1} gives

a(b1)1=abH.a(b^{-1})^{-1}=ab\in H.

Thus HH contains the identity, inverses, and products.

Cyclic groups

Let aGa\in G. The subgroup generated by aa is

a={an:nZ}.\langle a\rangle=\{a^n:n\in\mathbb Z\}.

Here,

a0=e,an=(a1)n.a^0=e,\qquad a^{-n}=(a^{-1})^n.

If

G=a,G=\langle a\rangle,

then GG is called a cyclic group, and aa is called a generator of GG.

The additive group of integers is cyclic:

Z=1=1.\mathbb Z=\langle1\rangle=\langle-1\rangle.

In additive notation,

a={na:nZ}.\langle a\rangle=\{na:n\in\mathbb Z\}.

The group Z6\mathbb Z_6 is also cyclic:

Z6=1={0,1,2,3,4,5}.\mathbb Z_6=\langle1\rangle =\{0,1,2,3,4,5\}.

Starting from 11 and repeatedly adding 11 modulo 66 gives

0, 1, 2, 3, 4, 5, 0,0,\ 1,\ 2,\ 3,\ 4,\ 5,\ 0,\ldots

The element 55 is another generator because

5={0,5,4,3,2,1}=Z6.\langle5\rangle =\{0,5,4,3,2,1\} =\mathbb Z_6.

However,

2={0,2,4}\langle2\rangle=\{0,2,4\}

is only a proper subgroup of Z6\mathbb Z_6.

Every cyclic group is abelian. If x=amx=a^m and y=any=a^n, then

xy=aman=am+n=an+m=anam=yx.xy=a^ma^n=a^{m+n}=a^{n+m}=a^na^m=yx.

The order of an element

The order of an element aGa\in G, written ord(a)\operatorname{ord}(a), is the smallest positive integer nn such that

an=e.a^n=e.

If no such positive integer exists, then aa has infinite order.

In Z6\mathbb Z_6, using additive notation, the order of 22 is 33 because

2+2+20(mod6),2+2+2\equiv0\pmod6,

while neither 22 nor 2+22+2 is congruent to 00 modulo 66.

Thus

ord(2)=3.\operatorname{ord}(2)=3.

More generally, in the additive group Zn\mathbb Z_n,

ord(k)=ngcd(n,k).\operatorname{ord}(k)=\frac{n}{\gcd(n,k)}.

For example,

ord(8 in Z12)=12gcd(12,8)=124=3.\operatorname{ord}(8\text{ in }\mathbb Z_{12}) =\frac{12}{\gcd(12,8)} =\frac{12}{4} =3.

Indeed,

8,8+84,8+8+80(mod12).8,\quad 8+8\equiv4,\quad 8+8+8\equiv0\pmod{12}.

Permutation groups

A permutation of a set is a bijection from the set to itself. The set of all permutations of

{1,2,,n}\{1,2,\ldots,n\}

forms the symmetric group SnS_n under composition.

The group S3S_3 has

S3=3!=6|S_3|=3!=6

elements.

One permutation is

σ=(123231).\sigma= \begin{pmatrix} 1&2&3\\ 2&3&1 \end{pmatrix}.

This means

σ(1)=2,σ(2)=3,σ(3)=1.\sigma(1)=2,\qquad \sigma(2)=3,\qquad \sigma(3)=1.

In cycle notation,

σ=(1 2 3).\sigma=(1\ 2\ 3).

Applying σ\sigma three times returns every element to its starting position:

σ3=e.\sigma^3=e.

Therefore

ord(σ)=3.\operatorname{ord}(\sigma)=3.

Permutation composition is generally not commutative. Let

α=(1 2),β=(2 3).\alpha=(1\ 2),\qquad \beta=(2\ 3).

Using the convention that the rightmost permutation acts first,

αβ=(1 2)(2 3)=(1 2 3),\alpha\beta=(1\ 2)(2\ 3)=(1\ 2\ 3),

whereas

βα=(2 3)(1 2)=(1 3 2).\beta\alpha=(2\ 3)(1\ 2)=(1\ 3\ 2).

Hence

αββα,\alpha\beta\neq\beta\alpha,

so S3S_3 is non-abelian.

Symmetry groups

Group theory gives a precise language for symmetry.

Consider an equilateral triangle. The transformations that leave it looking unchanged are:

These six symmetries form the dihedral group

D3.D_3.

Let rr denote rotation by 120120^\circ and let ss denote a reflection. Then

r3=e,s2=e,r^3=e,\qquad s^2=e,

and the interaction between rotation and reflection is described by

srs=r1.srs=r^{-1}.

Equivalently,

sr=r1s.sr=r^{-1}s.

The group can be written as

D3={e,r,r2,s,sr,sr2}.D_3=\{e,r,r^2,s,sr,sr^2\}.

Since

srrs,sr\neq rs,

the group D3D_3 is non-abelian. In fact,

D3S3.D_3\cong S_3.

The symbol \cong means that the two groups have the same abstract group structure even though their elements are described differently.

Homomorphisms

A group homomorphism is a function

φ ⁣:GH\varphi\colon G\longrightarrow H

that preserves the group operation:

φ(ab)=φ(a)φ(b)\varphi(ab)=\varphi(a)\varphi(b)

for all a,bGa,b\in G.

If the groups use additive notation, this becomes

φ(a+b)=φ(a)+φ(b).\varphi(a+b)=\varphi(a)+\varphi(b).

For example, define

φ ⁣:ZZn\varphi\colon\mathbb Z\longrightarrow\mathbb Z_n

by

φ(k)=[k]n.\varphi(k)=[k]_n.

Then

φ(a+b)=[a+b]n=[a]n+[b]n=φ(a)+φ(b),\varphi(a+b) =[a+b]_n =[a]_n+[b]_n =\varphi(a)+\varphi(b),

so φ\varphi is a homomorphism.

Every homomorphism sends the identity to the identity:

φ(eG)=eH.\varphi(e_G)=e_H.

To see this, observe that

φ(eG)=φ(eGeG)=φ(eG)φ(eG).\varphi(e_G) =\varphi(e_Ge_G) =\varphi(e_G)\varphi(e_G).

Cancellation in HH gives

φ(eG)=eH.\varphi(e_G)=e_H.

Homomorphisms also preserve inverses:

φ(a1)=φ(a)1.\varphi(a^{-1})=\varphi(a)^{-1}.

Indeed,

φ(a)φ(a1)=φ(aa1)=φ(eG)=eH.\varphi(a)\varphi(a^{-1}) =\varphi(aa^{-1}) =\varphi(e_G) =e_H.

Kernel and image

For a homomorphism φ ⁣:GH\varphi\colon G\to H, its kernel is

kerφ={gG:φ(g)=eH},\ker\varphi =\{g\in G:\varphi(g)=e_H\},

and its image is

imφ={φ(g):gG}.\operatorname{im}\varphi =\{\varphi(g):g\in G\}.

For the reduction map

φ ⁣:ZZn,φ(k)=[k]n,\varphi\colon\mathbb Z\longrightarrow\mathbb Z_n, \qquad \varphi(k)=[k]_n,

we have

kerφ=nZ\ker\varphi=n\mathbb Z

because

[k]n=[0]n    nk.[k]_n=[0]_n \iff n\mid k.

Also,

imφ=Zn\operatorname{im}\varphi=\mathbb Z_n

because every congruence class [r]n[r]_n is the image of the integer rr.

The kernel measures which elements become indistinguishable under the homomorphism. A homomorphism is injective exactly when

kerφ={eG}.\ker\varphi=\{e_G\}.

Cosets

Let HGH\leq G and gGg\in G. The left coset of HH determined by gg is

gH={gh:hH}.gH=\{gh:h\in H\}.

The corresponding right coset is

Hg={hg:hH}.Hg=\{hg:h\in H\}.

In an additive group, we write

g+H={g+h:hH}.g+H=\{g+h:h\in H\}.

Consider the subgroup

H=3ZH=3\mathbb Z

of Z\mathbb Z. Its cosets are

0+H=3Z,0+H=3\mathbb Z, 1+H={3k+1:kZ},1+H=\{3k+1:k\in\mathbb Z\},

and

2+H={3k+2:kZ}.2+H=\{3k+2:k\in\mathbb Z\}.

Every integer belongs to exactly one of these three cosets. They are precisely the three congruence classes modulo 33.

Every coset gHgH has the same number of elements as HH because the function

HgH,hghH\longrightarrow gH,\qquad h\longmapsto gh

is a bijection.

Lagrange’s theorem

One of the first major results in finite group theory is Lagrange’s theorem:

If GG is a finite group and HGH\leq G, then

HG.|H|\mid|G|.

In other words, the order of every subgroup divides the order of the group.

The reason is that the left cosets of HH partition GG, and every coset contains exactly H|H| elements. If there are [G:H][G:H] distinct cosets, then

G=[G:H]H.|G|=[G:H]|H|.

The number [G:H][G:H] is called the index of HH in GG.

As an immediate consequence, the order of every element divides the order of the group:

ord(a)=aG.\operatorname{ord}(a)=|\langle a\rangle|\mid|G|.

For example, a group of order 1212 cannot contain an element of order 55, because

512.5\nmid12.

If G=p|G|=p for a prime number pp, then every non-identity element has order pp. Therefore every group of prime order is cyclic:

G=p    GZp.|G|=p\implies G\cong\mathbb Z_p.

Normal subgroups and quotient groups

A subgroup NGN\leq G is normal if

gN=NggN=Ng

for every gGg\in G. We write

NG.N\trianglelefteq G.

An equivalent condition is

gNg1=NgNg^{-1}=N

for every gGg\in G.

Every subgroup of an abelian group is normal because

gng1=ngg1=ngng^{-1} =ngg^{-1} =n

for all gGg\in G and nNn\in N.

When NGN\trianglelefteq G, the set of cosets

G/N={gN:gG}G/N=\{gN:g\in G\}

becomes a group under

(gN)(hN)=(gh)N.(gN)(hN)=(gh)N.

For example,

Z/nZ\mathbb Z/n\mathbb Z

is the quotient of Z\mathbb Z by the subgroup nZn\mathbb Z. Its elements are

[0]n,[1]n,,[n1]n.[0]_n,[1]_n,\ldots,[n-1]_n.

This connects modular arithmetic directly with group theory.

A compact example

Consider the additive group

G=Z12.G=\mathbb Z_{12}.

Let

H=4.H=\langle4\rangle.

Repeatedly adding 44 gives

0, 4, 8, 0,0,\ 4,\ 8,\ 0,

so

H={0,4,8}H=\{0,4,8\}

and

H=3.|H|=3.

The distinct cosets of HH are

0+H={0,4,8},1+H={1,5,9},2+H={2,6,10},3+H={3,7,11}.\begin{aligned} 0+H&=\{0,4,8\},\\ 1+H&=\{1,5,9\},\\ 2+H&=\{2,6,10\},\\ 3+H&=\{3,7,11\}. \end{aligned}

Therefore

[G:H]=4[G:H]=4

and Lagrange’s theorem appears concretely as

G=[G:H]H=43=12.|G|=[G:H]|H|=4\cdot3=12.

Because GG is abelian, HH is normal, and the quotient group has order

G/H=4.|G/H|=4.

In fact,

Z12/4Z4.\mathbb Z_{12}/\langle4\rangle\cong\mathbb Z_4.

Exercises

Exercise 1

Show that

3Z={3k:kZ}3\mathbb Z=\{3k:k\in\mathbb Z\}

is a subgroup of (Z,+)(\mathbb Z,+).

Exercise 2

Find every generator of Z10\mathbb Z_{10}.

Exercise 3

Calculate the order of every element of Z8\mathbb Z_8.

Exercise 4

Let

φ ⁣:ZZ6,φ(k)=[k]6.\varphi\colon\mathbb Z\longrightarrow\mathbb Z_6, \qquad \varphi(k)=[k]_6.

Find kerφ\ker\varphi and imφ\operatorname{im}\varphi.

Exercise 5

Let GG be a finite group with G=15|G|=15. What element orders are permitted by Lagrange’s theorem?

Answers

For Exercise 1, the set is nonempty, and for 3m,3n3Z3m,3n\in3\mathbb Z,

3m3n=3(mn)3Z.3m-3n=3(m-n)\in3\mathbb Z.

Therefore the subgroup test gives

3ZZ.3\mathbb Z\leq\mathbb Z.

For Exercise 2, an element kZ10k\in\mathbb Z_{10} generates the group exactly when

gcd(k,10)=1.\gcd(k,10)=1.

Hence the generators are

1, 3, 7, 9.1,\ 3,\ 7,\ 9.

For Exercise 3, use

ord(k)=8gcd(8,k).\operatorname{ord}(k)=\frac8{\gcd(8,k)}.

Thus

k01234567ord(k)18482848\begin{array}{c|cccccccc} k&0&1&2&3&4&5&6&7\\ \hline \operatorname{ord}(k)&1&8&4&8&2&8&4&8 \end{array}

For Exercise 4,

kerφ=6Z\ker\varphi=6\mathbb Z

and

imφ=Z6.\operatorname{im}\varphi=\mathbb Z_6.

For Exercise 5, the order of an element must divide 1515. Therefore the only possibilities are

1, 3, 5, 15.1,\ 3,\ 5,\ 15.

Lagrange’s theorem permits these orders, although it does not by itself guarantee that elements of every permitted order exist.

Where to go next

The basic objects of group theory now fit into one picture:

elementssubgroupscosetsquotient groups.\text{elements} \longrightarrow \text{subgroups} \longrightarrow \text{cosets} \longrightarrow \text{quotient groups}.

At the same time, homomorphisms compare different groups:

φ ⁣:GH.\varphi\colon G\longrightarrow H.

The next natural topics are:

The main habit to develop is to move constantly between abstract axioms and concrete examples. Groups become intuitive when we see the same structure in integers, permutations, matrices, and geometric symmetries.


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