1
Abstract
In this paper, we calculate the automorphism of the Taxicab group which is a
transformation group preserving the Taxicab metric. In the process, we showed
that the automorphism of R with the standard addition is the set of all linear
transformations with respect to R over Q. Furthermore, the Aut(R) is isomorphic to the ring of R × R column-finite matrices with entries in Q . We go
further to characterize the center of Aut(R) and show GL(n, Q) is a subgroup
of Aut(R) for all finite n.
2
The Taxicab Group
The Taxicab group consists of the isometries of the plane with respect to the
Taxicab metric defined by d(u, v) =| ux − vx | + | uy − vy | where u = (ux , uy ),
v = (vx , vy ) are vectors in R2 [2]
It has been shown in [2] that the group of isometries of the plane with respect
to the taxicab metric is the semi-direct product of the dihedral group D4 and
T (2) where T (2) is the translation group. It has also been shown in [3] that
Aut (D4 ) ' Af f (Zn ) = ax + b | a ∈ U, b ∈ Zn
6 ∃φ ∈ Aut (R, +) such that ∀d ∈ D4 , ∀t ∈ T2 we have φ(d) = t or φ(t) = d
T (2) is isomorphic to R2 which is isomorphic to R
3
Proven Already
The cardinality of H is the cardinality of any hamel basis is the cardinality of
R
Definition 1. The Cauchy functional equation is a function f : X → Y such
that f (x + y) = f (x) + f (y)∀x, y ∈ X
Theorem 1. If an additive function f satisfies the Cauchy functional equation
and satisfies one of the following conditions, then there exists a real constant c
such that f (x) = cx for all x ∈ R [1]:
1. f is continuous at a point;
2. f is monotonic on an interval of positive length;
3. f is bounded from above or below on an interval of positive length;
4. f is integrable;
5. f is Lebesgue measurable;
6. f is a Borel function.
1
4
What We Think:
Aut (R, +) ' Aut (H, +) where H is the collection of all Hamel bases.
5
Automorphism of R
Definition 2. Definition: A set S finitely spans a vector space X if ∀ x ∈ X ∃
N ∈ N such that
x=
N
X
αi si where αi are scalars and si ∈ S ∀ i
(5.1)
i=1
Definition 3. A Hamel basis for an infinite vector space X is a linearly independent set that finitely spans X.
Definition 4. An indexed Hamel basis is a list whose elements come from a
Hamel basis such that the union of all elements in the indexed basis is the Hamel
basis.
Theorem: Let H be a Hamel basis. If h ∈ H, then h ∈ R / Q
Definition 5. Let H and H 0 be two indexed Hamel bases. Now let TH 0 : R → R
be given by
!
Nx
Nx
X
X
αi hi =
αi TH 0 (hi ) where TH 0 (H) = H 0
(5.2)
TH 0 (x) = TH 0
i=1
i=1
We will call T a Hamel Basis Transformation. We will call H the Central basis
for TH 0 . Now, denote the set of all Hamel Basis Transformations having a given
Central Basis H by TH .
Lemma 1. If T is a Hamel Basis Transformation, then T is a Linear Operator.
proof: Let x, y ∈ R and px , py ∈ Q

T (px x + py y) = T px
Nx
X
qx,i hi + py
i=0
Ny
X

qy,i hi 
(5.3)
i=0
Without loss of generality, we can assume Nx = Ny = N . If not, for M =
min{Nx , Ny } set qm,i = 0∀i ∈ (M, max{Nx , Ny }). Now,
!
N
X
T (px x + py y) = T
(px qx,i + py qy,i ) hi
i=0
=
N
X
(px qx,i + py qy,i ) T (hi )
i=0
= px T (x) + py T (y)
2
This shows that T is a linear operator.
Lemma 2. If T is a Hamel Basis Transformation, then T ∈ Aut (R, +)
Proof. Let H be the Central Basis for a Hamel Basis Transformation TH 0 .
Take TH 0 ∈ TH and for simplicity call it T .
Assume r 6= 0, then r =
Nx
X
qi hi where qi 6= 0∀i
i=0
T (r) =
Nr
X
qi T (hi ) =
i=0
Nr
X
qi h0i where h0i ∈ H 0
(5.4)
i=0
and qi h0i 6= 0 since 0 ∈
/ H 0 . But if
Nr
X
qi h0i = 0, then h0i is not linearly indepen-
i=0
0
r
dent since qi 6= 0∀i, therefore {h0i }N
i=0 is not subset of any basis. Hence, H is
not a basis which is a contradiction. Thus, T (r) 6= 0. By 2.2-10, [4], and since
T is a linear operator, the inverse map exists. Therefore T is bijective.
Lemma 3. Let TH have H as a Central Basis. Now, if φ ∈ Aut (R, +), then
φ ∈ TH
Proof. Now, we show that if φ ∈ Aut (R, +) then φ ∈ TH
By work of Cauchy and others[1], it is well known that for any q1 ∈ Q and
any r1 ∈ R we have
φ (qr1 ) = qφ (r1 )
(5.5)
furthermore, if r1 is not a rational multiple of r2 ∈ R then ∀q2 ∈ Q
φ (q1 r1 + q2 r2 ) = q1 φ (r1 ) + q2 φ (r2 )
(5.6)
We must show that φ (r1 ) 6= qφ (r2 ) for some q ∈ Q. Assume φ (r1 ) = φ (qr2 ),
then φ (r1 ) − φ (qr2 ) = φ (r1 − qr2 ) = 0 But since φ is an automorphism, then
r1 − qr2 = 0 which implies r1 = qr2 which is a contradiction to the assumption
that r1 is not a rational multiple of r2 .
By induction, if S is a linearly independent subset of R and ri ∈ S∀i and
qi ∈ Q ∀i
!
N
N
X
X
φ
qi ri =
qi φ (ri )
(5.7)
i=1
i=1
We must show that φ (S) is a linearly independent set. Assume φ (S) is not
M
X
linearly independent, then for some r ∈ S, φ (r) =
qi φ (ri ) where ∀i qi 6= 0
i=1
PM
and r 6= ri . Then by above equation, φ r − i=1 qi ri = 0. This implies that
3
PM
r =
i=1 qi ri since φ is an automorphism. Which is a contradiction to the
assumption that S is a linearly independent subset of R.
By Zorn’s lemma, there exists a maximal such S such that S finitely spans
R. Therefore, S is a Hamel basis for R
By the properties of φ, it is a linear operator and therefore is uniquely
determined by the images of the basis elements. Let H be a indexed hamel
basis. Assume H 0 = φ (H) does not span R, then ∃r0 ∈ R such that r0 is not a
linear combination of H 0 but φ is completely determined by φ (H) which implies
there is no r in R such that r0 = φ (r) and therefore φ is not onto. This is a
contradiction and implies that φ (H) spans R
Now, since φ (H) is linearly independent and spans R, it forms a hamel basis
for R.
Theorem 2. From a indexed hamel basis H to H’, there exists a unique indexed
hamel basis Transformation.
Theorem: The group of all permutations of any indexed Hamel basis is a
subgroup of Aut (R, +) is isomorphic to S (H)
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Matrix Representation
Definition 6. The matrix index set I for a matrix M = [mij ] is the set from
which i and j come.
Definition 7. Let G be a matrix group and S be any set subset of the matrix
index set for G. Now, the filter of G over S is the set.
GS = {M = [mij ] ∈ G | mij = 0 ∀ i or j ∈
/ S except when i = j}
(6.1)
Theorem 3. The filter H of a matrix group G over S is a subgroup of G.
Proof. Let A = [aij ], B = [bij ], and C = [cij ] = AB. Let I be the index set for
Now, the identity is in H and H is associative. The proof is left to the reader.
We must now show that H is closed. IfPi = j or i, j ∈ S, then cij can be anything.
Assume i 6= j and i ∈
/ S, then cij = k∈I aik bkj = aii bij since aik = 0 ∀ k 6= i.
Since i 6= j, then cij = 0. Similarly, if i 6= j and j ∈
/ S, then cij = 0.
Now we show A−1 = [αij ] inverse of A ∈ H. It is known that for any matrix
1
FijT where Fij = [fij ] is the (i, j)th minor of A. Now, as before
M , M −1 = det(A)
if i = j or i, j ∈ S, then αij can be anything. Now, assume i 6= j and i ∈
/ S,
1
then αij = det(A)
fji but fji is the determinant of a matrix D whose row coming
from the removal of the ith entry from ith row of A is all zeros. Therefore
fji = det(D) = 0 → αij = 0. Thus, A−1 ∈ H. Similarly, if i 6= j and j ∈
/ S,
then αij = 0 → A−1 ∈ H.
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Theorem 4. If there exists a maximal H subgroup of Aut(R), then H is uncountable.
Theorem: The Inn(R) is the identitity map.
Theorem: The Out(R) = Aut(R)
Theorem: The center of R is the set of all linear maps.
TO DO: 1.) identify subgroups of Aut(R) - are they maximal? - are they
normal? if so Aut(R)/N is a group 2.) identify elements of order 2 in Aut(R) centralizer of these elements : Ca (G) = g ∈ G|ga = ag which is the Fixed group
of a in G
5
References
[1] S.-M. Jung, Hyers Ulam Rassias.
Stability of Functional Equations in Nonlinear Analysis.
Springer Science+Business Media, LLC 2011.
Published in USA by publication service.
[2] Doris J. Schattschneider
The Taxicab Group The American Mathematical Monthly; Vol. 91, No.
7 (Aug. - Sep., 1984) , pp. 423-428 Mathematical Association of America
[3] Abigail Bishop, Christopher Cyr, John Hutchens, Clover May, Nathaniel
Schwartz, Bethany Turner On Involutions and Generalized Symmetric
Spaces of Dicyclic Groups arXiv:1310.0121v1
[4] Erwin Kreyszing Introductory Functional Analyssi with Applications; 1st
ed, 1989 John Wiley & Sons, Inc
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