<<

. 13
( 14 .)



>>

where • is the Euler •-function.

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

<<

. 13
( 14 .)



>>