Theorem 3.12 Parts (a), (b), (c), & (d) in Gilbert –
Let G be a group.
Part (a)
Show that for any x in G, xn x-n = e for any integer n.
Proof:
We will need the following related claim: x xn = xn x for any integer n.
Since x x1 = x x = x1 x, the claim holds if n = 1. Now assume it holds for
some k . Thenx xk+1 = x (xk x) = (x xk) x = xk+1 x, so it holds for k + 1
and therefore, the principle induction, it holds for all n > 0.
Now it’s clear the claim holds for n = 0, since x x0 = x e = x = e x = x0 x.
Recall: if n < 0, then xn has been defined to be (x*)-nfor x* the inverse of x.
This means x-1 is defined as (x*)-(-1) = x*which is consistent with our usage
of that symbol to date, i.e. x-1 for the inverse of x. Thus, the claim holds for
n = -1, since x x-1 = e = x-1 x.
Suppose n < 0. Then –n > 0 andper the above, we have:
x xn = x (x*)-n = x (x*)-n (x* x) = x ((x*)-n x*) x =
x (x* (x*)-n) x = ((x x*)(x*)-n) x = (x*)-n x = xn x
Therefore, x xn = xn x holds for all integers n any x in G.
Finally, as regardsPart (a) –
For x in G, we have x0 x-0, so the claim holds if n = 0.
We also have that x1 x-1 = x x* = e, so it holds if n = 1.
Suppose now that the claim holds for some k > 0. Then
xk+1 x-(k+1) = (xk x) (x*)k+1 = (xk x) ((x*)k x*) =
(xk x) (x* (x*)k) = (xk(x x*))(x*)k = (xke) (x*)k = xk x-k = e.
Then, by the principle of induction, it holds for all n ≥ 0.
If k < 0, then –k > 0 and,by what we’ve shown,we have
x-k x-(-k) = x-k xk = e, implying(xk)* is x-k so that xk x-k = e.
Part (b)
For any x in G and n & m integers, xn xm = xn+m.
Proof:
Cf. Gilbert (8th Edition) Pg. 157.
Part (c)
For any x in G and integers n & m, (xn)m = xnm.
Proof:
Fix n in Z.
Then since (xn)0 = e = x0 = xn0, the claim holds for m = 0.
Assumethe statement holds for some k . Then from (b),
(xn)k+1 = (xn)k xn = xnk xn = xnk+n = xn(k+1), so it holds for k + 1.
Therefore, by mathematical induction, it must hold for all m > 0.
Finally, let m < 0. Then (xn)m = ((xn)-1)-m = (x-n)-m = x(-n)(-m) = xnm.
Part (d)
Let G be abelian. Then for any x & y in G, (xy)n = xn yn.
Proof:
Since (xy)0 = e =e e = x0 y0, the claim holds for n = 0.
Now assume that the statement holds for some k . Then,because G is abelian,
(xy)k+1 =(xy)k xy = (xkyk) xy =xk(yk x)y =(xk(x yk)y = (xkx) (yky)= xk+1 yk+1,
andso it holds for k + 1. Then by the induction principle, itmust hold for all n≥ 0.
Finally, let n < 0. Then –n > 0 (x y)n= ((x y)-1)-n = (x-1 y-1)-n = (x-1)-n (y-1)-n =xnyn.
For n = 18
Divisors 1, 2, 3, 6, 9, 18
Quotients 18, 9, 6, 3, 2, 1
Form the 18 fractions:
1/18, 2/18, 3/18, 4/18, 5/18, …, 17/18, 18/18
Now reduce these to lowest terms:
1/18, 1/9, 1/6, 2/9, 5/18, 1/3, …, 8/9, 17/18, 1
Since e/d is in the reduced list if & only if gcd(e,d) =1,we have
(18) k/18’s, (9) k/9’s, (6) k/6’s,(3) k/3’s,(2) k/2’s, & (1) 1’s
Therefore, 18 = (18) + (9) + (6) + (3) +(2) + (1)
Suppose N is normal in G and take G/N to
denote the set of alldistinct left cosets of N.
Define the following operation on G/N: (g1N)(g2N) = g1g2N.
This is well-defined since if g1N = g1’N and g2N = g2’N, then
(g1g2)-1(g1’g2’) = g2-1(g1-1g1’)g2’∈ N (g1-1g1’∈N & N is normal.
Therefore, g1g2N = g1’g2’N.
The operation is associative since
the group operation in G is associative.
N is an identity and (gN)-1 is just g-1N.
Therefore, G/N is group under this operation.
The order of G/N is [G : N]. It can be shown that
this operation is well-defined iff N is normal in G.
Let G be a finite symmetry group (G = Sym(X) for some
Subset X of R2). Take G to be {f1, f2, …, fn} (so |G| is n).
Choose any x0 in X and let xi = fi(x0) for i = 1 to n. Set z
Tobe the ‘centroid’ of these points, i.e. z = ∑xi/n. We
claimthat each f in G fixes z:
Let fj be in G with fj(x) = Ax + b. Then
fj(z) = Az + b = ∑Axi/n + b = ∑(Axi+b)/n = ∑fj(fi(x0)/n = z
since the mapping fi fjfi is a bijection of G to G so the terms
fj(fi(x0)) for i from 1 to n must constitute a reordering of the xi’s.
Note: The mapping f g for g(x) = f(x+z) – z is an isomorphism of
Sym(X) to Sym(X - z) with the g’s all fixing 0. Therefore, Sym(X – z)
is isomorphic to a finite subgroup of O(2) (henceforth, denoted by G’).
From Example 12.10, any A in G’ is either a rotation or a reflection.
There aretwo cases to be considered:
(1)All the A’s in G’ are rotations (I is a rotation);
(2)There exists at least one reflection, T, in G’.
In case (1), either G’ = {I} and is trivial (therefore trivial cyclic)
or G’ includes a non-trivial rotation, say R. This means that G’
includes a rotation matrix with a positive rotation angle, say .
Assume this angle to be the least positive rotation angle of all
therotations in G’(finiteness is needed here). We claim that
the rotation matrix Rgenerates G’: If not, there exists an R
in G’ such that n < (n+1)for some integer n. But this
means that Rnis a rotation in G’ through a positive angle
less than , a contradiction. Therefore, G’ is cyclic.
In case (2), G’ includes at least one reflection, say T. Consider the
mapping from G’ to the multiplicative group {-1, 1} defined by
(A)= det(A). Then is an epimorphism (det(T) = -1 & det(I) = 1).
By the first isomorphism theorem, G’/Ker is isomorphic to {-1,1}.
Since Ker includes only the rotations in G’, it is cyclic by case (1).
Therefore, either G’ = {I, T} in which case G’ is cyclic of order 2, or
G’ contains a rotation matrix, say R, with positive rotation angle
such that <R> = Ker . In this case, G’ = Ker U T Ker and
G’ = {I, R, R2, …, Rn-1, T, TR, TR2, …, TRn-1}
and, since TR is a reflection, (TR)2 = I showing that TRT = R-1.
This all shows that G’ is isomorphic to the dihedral group Dn/2.
In D4, the only proper nontrivial subgroups are –
H1 = {e, r, r2, r3}, H2 = {e, s, sr2, r2}, H3 = {e, sr3, sr, r2},
K1 = {e, r2}, K2 = {e, s2}, K3 = {e, sr},and K4 = {e, sr3},
where e is the identity, r 90o CCW rotation, and s is reflection
across the y-axis (with the square being in ‘standard position’).
Extras –
For f(x) = Ax + b,
fn(x) = Anx + An-1b + An-2b + … Ab + b
equals Ax + n/2 b + n/2 Ab, if n is even,
orAx + (n+1)/2 b + (n-1)/2 Ab, if n is odd.
<x, y> <-x + 1, y> is a reflection across x = ½
<x, y> <-x + c, y> is a reflection across x = c/2
<x, y> <-x, y + c> is aglide-reflection across x = 0 for c ≠ 0
Suppose X is a bounded set in R2 and f an isometry mapping X into X.
Then does it necessarily follow that f(X) = X? Answer: No, it does not…
Let be apositive real so that ≠ r for any rational r (eg. = 2 or…).
Now define x0 = (1, 0) and xi+1 = Rxi= Ri+1x0for i ≥ 0 with R the CCW
rotation matrix determined by . Finally, take X to be the set {x0, x1, x2, …}.
We claim that themapping f : x Rx (an isometry) maps X into X properly:
From construction, it’s clear that f maps X into X. But x0 is not the image
of any x in X. If it were, then x0 = Rkx0 for some k > 0 which implies that
sin(k) = 0 or that k = n for some integer n, violating the assumption
that ≠ r for any rational r. This shows the set X is countably infinite,
i.e. that the xi’s are distinct. This construction can be taken further…
OTOH, if the isometry has finite order, then the answer is YES:
Suppose f(x) = Ax + b has order n. Then for any y in X,y = fn(y)
implying y = f(fn-1(y)) and, since fn-1(y) is in X, y belongs to f(X).
Define D1 = Sym(“T”) Z2 and D2 = Sym(‘I”) Klein 4-Group
U(9) = {e = 1, 2, 4, 5, 7, 8}. Take a = 7 and b = 8.
Then o(a) is 3 & o(b) is 2 and U(9) = {e, a, a2, b, ba, ba2}
U(16) = {e = 1, 3, 5, 7, 9, 11, 13, 15}. Take a = 5 and b = 7.
Then o(a) is 4 & o(b) is 2 and U(16) = {e, a, a2, a3, b, ba, ba2, ba3}
U(22) = {e = 1, 3, 5, 7, 9, 13, 15, 17, 19, 21}. Take a = 3 and b = 21.
Then o(a) is 5 & o(b) is 2 and U(22) = {e, a, a2, a3, a4, b, ba, ba2, ba3,ba4}