Example 6.4. The automorphism group of G = Z/2 — Z/2 has order 6 and is isomorphic

to S3 .

Proof. Let

0 = (0, 0), u = (1, 0), v = (0, 1), w = (1, 1).

Then for an endomorphism θ of G, the e¬ect of θ on any element is given by its e¬ect on u and

v since every element is of the form ru + sv and hence

θ(ru + sv) = rθ(u) + sθ(v).

Moroever, such an endomorphism is an automorphism if and only if it is injective, which is true

if and only if

θ(u) = 0 and θ(v) ∈ {θ(u), 0}.

/

Equivalently, an automorphism must permute the three elements u, v, w. In fact every permu-

tation of these extends uniquely to an automorphism (it must send (0, 0) to itself). Thus, we

see that there is a one-one correspondence between Aut(Z/2 — Z/2) and Perm({u, v, w}), which

is in fact a group isomorphism. But the latter has 6 elements.

The group Aut(Z/2 — Z/2) is called the 2 — 2 general linear group over Z/2 and is usually

written GL2 (Z/2) or just GL2 (2).

Definition 6.5. For a prime p, the automorphism group of (Z/p)n = Z/p — · · · — Z/p (p

factors) is called the n — n general linear group over Z/p and is denoted GLn (Z/p) or GLn (p).

The following result will not be used in it general form, but is nevertheless important.

1. AUTOMORPHISMS 61

Theorem 6.6. For a prime p, GLn (p) is isomorphic to the group of all n — n matrices over

Z/p. The order of GLn (p) is

n

|GLn (p)| = (pn ’ 1)(pn ’ p)(pn ’ p2 ) · · · (pn ’ pn’1 ) = p( 2 ) (pn ’ 1)(pn’1 ’ 1) · · · (p ’ 1).

To see where the ¬rst part of this comes from, notice that if we write elements of (Z/p)n as

column vectors with entries in Z/p. Let ek be the column vector with all entries 0 except for 1p

in the kth place. Then for any • ∈ GLn (p), we can form the n — n matrix [•] with

[•]ij = aij

where

n

•(ej ) = arj er .

r=1

It turns out that

[• —¦ • ] = [•][• ]

and so this gives a homomorphism from GLn (p) to the group of all invertible n — n matrices;

this is in fact an isomorphism. The order of this group is obtained by a counting argument.

One reason for being interested in such groups is the following result. There is a determinant

homomorphism det : GLn (p) ’’ Z/p— , where the latter is the group of units in the ring Z/p

and is of order (p ’ 1). The kernel of det, ker det is called special linear group of n — n matrices

over Z/p and denoted SLn (p). Notice that the order of SLn (p) is |GLn (p)|/(p ’ 1) since det is

surjective.

Theorem 6.7. For each prime p and n 1, the centre of SLn (p) is given by

Z(SLn (p)) = {»In ∈ SLn (p) : » ∈ (Z/p)— , »n = 1}.

which is cyclic of order equal to the highest common factor of n and (p ’ 1), (n, p ’ 1).

For n 2 and p = 2, 3, the quotient group SLn (p)/ Z(SLn (p)) is simple.

Notice that the order of SLn (p)/ Z(SLn (p)) is |GLn (p)|/(p ’ 1)(n, p ’ 1). For example, if

p = 7 and n = 2, the order is

2

7(2) (72 ’ 1)(7 ’ 1)/((7 ’ 1) — 2) = 7 — 48/2 = 168.

Definition 6.8. A subgroup H G is called a characteristic subgroup of G if •(H) = H

for each automorphism • ∈ Aut(G).

Proposition 6.9. A characteristic subgroup H G is a normal subgroup.

Proof. For each g ∈ G, conjugation by g gives an automorphism conjg : G ’’ G. But

then conjg (H) = H. Expanding the de¬nition of conjg (H), we get

conjg (H) = {ghg ’1 : h ∈ H} = gHg ’1 ,

hence gHg ’1 = H for every g ∈ G. Thus H G.

Proposition 6.10. The centre Z(G) and commutator subgroup [G, G] are characteristic

subgroups. More generally, the subgroups G(r) , G(r) , Zr (G) in the derived, lower central and

upper central series are all characteristic subgroups.

Proof. Let c ∈ Z(G). Then for any • ∈ Aut(G) and g ∈ G we have

•(c)g = •(c)• •’1 (g) = • c•’1 (g) = • •’1 (g)c = • •’1 (g) •(c) = g•(c).

Hence •(Z(G)) Z(G). A similar argument with •’1 gives •’1 (Z(G)) Z(G), hence Z(G)

•(Z(G)). Combining these results gives •(Z(G)) = Z(G).

62 6. AUTOMORPHISMS AND EXTENSIONS

The commutator subgroup [G, G] is generated by all elements of the form [x, y] = xyx’1 y ’1

for x, y ∈ G. For • ∈ Aut(G), we have

•([x, y]) = •(xyx’1 y ’1 ) = •(x)•(y)•(x’1 )•(y ’1 ) = •(x)•(y)•(x)’1 •(y)’1 = [•(x), •(y)].

Hence, •([G, G]) [G, G]. Similarly, we have •’1 ([G, G]) [G, G]. Combining these we obtain

•([G, G]) = [G, G].

The remaining cases follow by inductive arguments and are left as exercises.

2. Extensions

Let G be a group and N G a normal subgroup. Then every element g ∈ G acts on N by

conjugation,

g · n = gng ’1 .

Equivalently, the conjugation map conjg : G ’’ G restricts to an automorphism conjN : N ’’

g

N of the group N . This will not be an inner automorphism of the group N in general. Thus

we have a homomorphism

conjN : G ’’ Aut(N ); g ’ conjN .

g

From now on, to ease the technical details we will assume that N = A is an abelian group.

Definition 6.11. Given groups Q and N together with a homomorphism • : Q ’’ Aut(N ),

we say that a group G is an extension of N by Q with action • and sometimes write

j π

1 ’’ N ’ G ’ Q ’’ 1,

’’

if the following conditions apply.

Extn-1 there is an injective homomorphism j : N ’’ G whose image j(N ) is a normal sub-

group of G, hence we can interpret N as a normal subgroup of G with j being the

inclusion map;

Extn-2 there is a surjective homomorphism π : G ’’ Q with kernel equal to N ;

Extn-3 the action • satis¬es the equation

conjN = •(π(g)) (g ∈ G).

g

Given such an extension, notice the following. The homomorphism π gives rise to an iso-

morphism π : G/N ∼ Q, thus we will often identify Q with G/N . If we choose for each el-

=

ement gN = N g ∈ G/N a representative s(gN ) ∈ gN , this amounts to de¬ning a function

s : G/N ’’ G such that π —¦ s = IdQ , i.e., π(s(gN )) = gN ; we can even choose s so that

s(eN ) = e. Then every element g ∈ G can be uniquely written in the form g = ns(gN ) where

n ∈ N . Let us consider what happens when we write the product of two elements in this form.

We have

(n1 s(g1 N ))(n2 s(g2 N )) = n1 s(g1 N )n2 s(g1 N )’1 s(g1 N )s(g2 N )

= n1 •(s(g1 N ))(n2 ) s(g1 N g2 N )

= n1 •(s(g1 N ))(n2 )ν(g1 N, g2 N )s(g1 g2 N ),

where ν(g1 N, g2 N ) ∈ N is an element depending on g1 N and g2 N , which we will view as

determined by a function ν : Q — Q ’’ N .

In fact, given the data consisting of two groups N, Q, a homomorphism • : Q ’’ Aut(N )

and a function ν : Q — Q ’’ N , we can build an extension E of N by Q with action • as

follows, provided that ν satis¬es certain conditions which will be discussed in the next section.

Here is an example, where we take ν(b1 , b2 ) = eN for all b1 , b2 ∈ Q. To simplify the notation,

we will often put b a = •(b)(a). We take

E• = N — Q = {(a, b) : a ∈ N, b ∈ Q}

2. EXTENSIONS 63

and de¬ne a product on E• by

(a1 , b1 )(a2 , b2 ) = (a1 b1 a2 , b1 b2 ).

(2.1)

This is associative since

((a1 , b1 )(a2 , b2 )) (a3 , b3 ) = (a1 b1 a2 , b1 b2 )(a3 , b3 )

= (a1 b1 a2 b1 b2 a3 , b1 b2 b3 )

= (a1 b1 (a2 b2 a3 ), b1 b2 b3 )

= (a1 , b1 )(a2 b2 a3 , b2 b3 )

= (a1 , b1 ) ((a2 , b2 )(a3 , b3 )) .

We also have

(eN , eQ )(a, b) = (eN eQ a, eQ b) = (a, b),

and similarly

(a, b)(eN , eQ ) = (a, b),

thus (eN , eQ ) is an identity. Finally,

’1 ’1 ’1

(b (a’1 ), b’1 )(a, b) = (b (a’1 )b a, b’1 b)

’1 ’1

= (b (a’1 )b a, eQ )

’1

= (b (a’1 a), eQ )

’1

= (b eN , eQ )

= (eN , eQ )

= (aa’1 , eQ )

= (aeQ (a’1 ), eQ )

’1

= (abb (a’1 ), eQ )

’1

= (ab (b (a’1 )), bb’1 )

’1

= (a, b)(b (a’1 ), b’1 ),

which implies that (a, b) has inverse

’1

(a, b)’1 = (b (a’1 ), b’1 ).

Combining these results we see that E• is a group. There is an inclusion map j : N ’’ E•

de¬ned by

j(a) = (a, eQ ),

and a quotient map π : E• ’’ Q de¬ned by

π(a, b) = b,

both of which are homomorphisms.

This extension E• is called the semi-direct product of N and Q for the action • and written

E• = N H.

•

If •(b) = IdN for all b ∈ Q we have E• = N — Q, the direct product of N and Q. So the

semi-direct product generalises the direct product.

Proposition 6.12. Let G be a group, N G, H G and suppose that G = HN = N H and

H © N = {e}. Then G is an extension of H by N and moreover, G ∼ N • H.

=

64 6. AUTOMORPHISMS AND EXTENSIONS

Proof. Here we use the notation

AB = {ab : a ∈ A, b ∈ B}

for two subgroups A, B G. Warning: in general this is not a subgroup of G! If A or B is

normal in G, then AB = BA and this is a subgroup of G.

We will de¬ne a homomorphism • : H ’’ Aut(N ). Notice that since N is normal in G, for

each h ∈ H, hN h’1 = N , hence we can take

•(h) = conjN ∈ Aut(N ).

h

We take j : N ’’ G to be the inclusion map and de¬ne q : G ’’ H by using the fact that

there is a composite homomorphism

θ : H ’’ G ’’ G/N

de¬ned by

h ’’ hN.

This is an injection since hN = eN if and only if h ∈ H © N = {e}. It is surjective since for

every g ∈ G, g = hn with h ∈ H and n ∈ N , hence gN = hnN = hN . Thus we can take

q : G ’’ H to be the composite

θ’1

G ’’ G/N ’ ’ H

’

which is surjective since the quotient map G ’’ G/N is. It is now straightforward to verify

that the above conditions for an extension hold.

Given an action homomorphism • : Q ’’ Aut(N ), two extensions

j1 j2

π π

1 ’’ N ’ G1 ’1 Q ’’ 1, 1 ’’ N ’ G2 ’2 Q ’’ 1,

’ ’ ’ ’

are said to be equivalent if there is a homomorphism θ : G1 ’’ G2 for which the following

diagram commutes

j1 π

’1

N ’ ’ ’ G1 ’ ’ ’ Q

’’ ’

¦ ¦ ¦

¦ ¦ ¦

= =

θ

j2 π

’2

N ’ ’ ’ G2 ’ ’ ’ Q

’’ ’

in the sense that θ —¦ j1 = j2 and π2 —¦ θ = π1 . Note that such a homomorphism is necessarily an

isomorphism. An extension which is equivalent to a semi-direct product is called split.

Example 6.13. The symmetric group S3 is a split extension of Z/3 by Z/2 with the non-

trivial action of Z/2 ∼ (Z/3)— .

=

Proof. We will use Proposition 6.12. For the normal subgroup N take N = (123) ∼ Z/3,

=

and for H take H = (12) ∼ Z/2. Then the action of H on N is given by

=

(12)

((123)k ) = (123)2k = (123)’k .

Clearly we have N © H = {e}. Hence S3 ∼ N H with the above action.

=

Example 6.14. The dihedral group of order 2n is generated by D2n two elements ±, β of

orders |±| = n and |β| = 2 which satisfy the relation

β±β = ±’1 .

Then D2n is split extension of Z/n ∼ ± by Z/2 ∼ β with the action given by

= =

β

(±k ) = ±’k .

3. CLASSIFYING EXTENSIONS [OPTIONAL EXTRA MATERIAL] 65

Proof. Again we will use Proposition 6.12.

For the normal subgroup N take N = ± ∼ Z/n, and for H take H = β ∼ Z/2. Clearly

= =

we have N © H = {e}. Hence D2n ∼ ± β.

=

Example 6.15. Let Q8 be the quaternion group of order 8, discussed in Subsection 7.1,

with elements

±1, ±i, ±j, ±k.

In this group, the centre is Z(Q8 ) = {±1}.

If we let Z/2 — Z/2 act trivially on Z(Q8 ) ∼ Z/2, then Q8 is a non-split extension of Z/2 by

=

Z/2 — Z/2.

Proof. If Q8 = {±1} H for some H Q8 , we would have H ∼ Z/2 — Z/2. Notice that