Selected problems on groups
Problem 32 class handout: Let G be the set of all matrices in M 3 ( R) that have the form
a 0 0
 0 b 0 a, b, c  R \{0}


 0 0 c 
Prove or disprove that G is a group with respect to multiplication.
We will prove that G is a group with respect to multiplication.
 ai 0 0 
Let Ai   0 bi 0   G, i  1, 2,3.


 0 0 ci 
0
0 
 a1a2

Notice that G is closed under the multiplication: indeed, A1 A2  0 b1b2
0   G .

 0
0 c1c2 
Now, we already know that the multiplication of matrices is associative, therefore
A1 ( A2 A3 )  ( A1 A2 ) A3 .
1 0 0 
The identity element is the identity matrix, I  0 1 0 . Indeed, AI  IA  A for all


0 0 1 
AG .
 a 1 0
0


1
1
0  , and AA1  A1 A  I .
Every matrix in G has an inverse: indeed, A   0 b
0
0 c 1 

Therefore, G is a group. Moreover, it is an abelian group since for these matrices
A1 A2  A2 A1 .
Problem 41 class handout: An element x in a multiplicative group G is called idempotent
if x 2  x . Prove that the identity element, e, is the only idempotent element in a group.
Suppose x  G : x 2  x . Then, by multiplying by the inverse of x we get: x 1 x 2  x 1 x  e ,
but x 1 x 2  x so x  e .
Problem 48 class handout: Prove that if (ab)1  a 1b1 for all a,b in a group G, then G is
abelian.
Using the definition of the inverse, we must have e  (ab)(ab)1  (ab)a 1b1 . “Multiply”
by ba to the right on both sides of the above equation, to get:
(ba)e  (ab)a 1b1 (ba)  (ab)a 1 (b1b)a  (ab)a 1a  ab . Therefore, ab  ba for all
a, b  G so the group is abelian.
Problem 14 (Nicodemi): Let G be a group and let g  G . Let Tg : G  G be the function
defined by Tg ( x)  gx .
i) We give all the values of Tg ( x) when G  U 8 , g  3 .
x
1 3 5 7
T3 ( x) 3 1 7 5
ii) We prove that Tg ( x ) is a bijection for any group. In order to do so we need to check it
is one-to-one and onto.
We first check it is one-to-one. Let x, y  G : Tg ( x)  Tg ( y ) . Then gx  gy . Since g has an
inverse in G, by multiplying to the left we obtain: g 1 gx  ex  g 1 gy  ey, so x  y , and
thus Tg ( x ) is one-to-one.
Now to check it is onto, let y  G . We need to find x  G such that Tg ( x)  gx  y .
Multiply again by the inverse of g to obtain: g 1 gx  g 1 y, or x  g 1 y proving that
Tg ( x ) is onto.
iii) Suppose that an element appears twice in a row of the operation table for G. To be
more exact, suppose in the row of an element g we have ga  gb  c for two distinct
elements a, b  G. Define Tg ( x)  gx as above. Then
ga  gb  c  Tg (a )  Tg (b)  c , which contradicts the fact that Tg ( x ) is one-to-one.
Therefore, each element must appear exactly once in the table for G.
iv) You try this one!