Algebra I: Section 6. The structure of groups. 6.1 Direct products of ...

and the subgroup N in (Aff(V ), ◦ ).
Purely Linear Operators: GL(V ) = {A : det(A) �= 0} is the set of all invertible
linear operators A : V → V . It is obviously a subgroup in Aff(V ).
We have just remarked that G = N · GL(V ). It is also clear that N ∩ GL(V ) = {I}, where I is
the identity operator on V . [In fact, every linear operator A leaves the origin in V fixed, but a
translation ta does this only when a = 0 and ta = I.] We claim that N is a normal subgroup
in G, and hence we have a semidirect product Aff(V ) = N ×φ GL(V ).
Let T be any element in G. To see that TtaT −1 ∈ N for T ∈ Aff(V ) we first apply (21) to
compute the inverse of any operator Tv = Av + b in G
(23) T −1 v = A −1 (v − b) = A −1 (v) − A −1 (b)
(Simply solve w = Tv = Av +b for v in terms of w.) Next we compute the effect of conjugation
by T on a translation ta in the special case when T = A is linear and b = 0. We get
(24) AtaA −1 = t A(a) for all a ∈ V, A ∈ GL(V )
because for every v we have
AtaA −1 (v) = A(A −1 (v) + a) = AA −1 (v) + A(a) = v + A(a) = t A(a)(v)
Thus, conjugating a translation ta by a purely linear operator produces another translation as
expected, but it is also useful to observe that
When we identify (V, +) ∼ = N via the bijection j : a ↦→ ta, the action GL(V ) × N →
N by conjugation (which sends ta ↦→ AtaA −1 ) gets identified with the usual action
of the linear operator A on vectors in V (which sends a ↦→ A(a)).
Now let T be any element in G. To see that TNT −1 ⊆ N for all T ∈ G, we write T = tb ◦ A
and compute the effect of conjugation; since t −1
b = t−b, equation (24) implies that
TtaT −1 = tb(AtaA −1 )t−b = tbt A(a)t−b = t b+A(a)−b = t A(a)
Compare with (24): throwing in the translation component tb has no effect when T acts by
conjugation on the translation subgroup N. That makes sense: N is abelian and acts trivially
on itself via conjugation. �
We have stated that Dn is the full set of symmetries of the n-gon, without going into the proof.
First you must understand what we mean by “symmetries.” The set M(2) of all bijections
f : R 2 → R 2 that are distance-preserving
dist(f(x), f(y)) = dist(x,y) where dist(x,y) = � (x1 − y1) 2 + (x2 − y2) 2
is easily seen to be a group under composition of mappings. It is called the group of rigid
motions in the plane. With some clever effort (details in Section 1.9) one can show that every
operator in M(2) is actually an invertible affine mapping, as defined in 6.2.14, whose linear
part has a special form.
Tv = Av+b where
� b ∈ R 2 and A ∈ O(2) = {A ∈ M(2, R) : AA t = I = A t A},
the group of 2 × 2 real orthogonal matrices.
Orthogonal matrices arise here because a linear operator A : R 2 → R 2 preserves distances in
the plane if and only if it corresponds to an orthogonal matrix. Since translation operators
ta : v ↦→ v + a are automatically distance preserving, the subgroup M(2) ⊆ Aff(R 2 ) of distance
preserving mappings of the plane consists precisely of those maps whose linear part lies in
O(2). Thus M(2) is the semidirect product M(2) = R 2 ×φ O(2) when we identify the normal
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