Mathematics 1214: Introduction to Group Theory
Tutorial exercise sheet 2
1. Let ∗ be an operation on a set S. Suppose that e is an identity element for ∗, and
that there are elements x, y1 , y2 ∈ S such that
x ∗ y1 = e = x ∗ y2 ,
y1 ∗ x = e = y2 ∗ x and y1 6= y2 .
(a) Explain why ∗ must fail to be associative.
(b) Find the Cayley table of an operation with these properties on S = {e, f, g}.
Solution (a) If ∗ were associative, then the hypothesis would contradict Proposition 3. So ∗ cannot be associative.
(b) For example, let ∗ be the operation with the following Cayley table:
∗
e
f
g
e
e
f
g
f
f
e
e
g
g
e
e
Clearly, e is an identity element for ∗. Taking x = f , y1 = f and y2 = g, we have
x ∗ y1 = e = x ∗ y2 , y1 ∗ x = e = y2 ∗ x and y1 6= y2 .
2. Let ∗ be an associative operation on a set S, and suppose that e is an identity
element for ∗. Show that if x, y ∈ S with
x ∗ y = e and y ∗ x 6= e
then x is not invertible with respect to ∗. [Hint: try a proof by contradiction.]
Solution Suppose that x is invertible with respect to e. Then there is z ∈ S
such that x ∗ z = e = z ∗ x. Hence, using the associativity of ∗, we have
y ∗ x = (z ∗ x) ∗ (y ∗ x) = z ∗ (x ∗ y) ∗ x = z ∗ (e ∗ x) = z ∗ x = e,
which contradicts the hypothesis that y ∗ x 6= e. So x cannot be invertible with
respect to ∗.
3. Let ∗ be an operation on a set S with an identity element e ∈ S, and let us write
T = {x ∈ S : x 6= e}. Prove that ∗ is associative if and only if
x ∗ (y ∗ z) = (x ∗ y) ∗ z
for all x, y, z ∈ T .
[Note: the statement “x, y, z ∈ T ” is shorthand for “x ∈ T , y ∈ T and z ∈ T ”.]
Solution Suppose that S is associative. Then x ∗ (y ∗ z) = (x ∗ y) ∗ z for all
x, y, z ∈ S. Since x, y, z ∈ T =⇒ x, y, z ∈ S, this shows that x ∗ (y ∗ z) = (x ∗ y) ∗ z
for all x, y, z ∈ T .
Now suppose that x ∗ (y ∗ z) = (x ∗ y) ∗ z for all x, y, z ∈ T . If x, y, z ∈ S then
either x, y, z ∈ T (that is, they are all in T ), or at least one of x, y, z is e. So there
are four cases to consider:
(i) If x, y, z ∈ T then x ∗ (y ∗ z) = (x ∗ y) ∗ z by hypothesis.
(ii) If x = e then x ∗ (y ∗ z) = e ∗ (y ∗ z) = y ∗ z and (x ∗ y) ∗ z = (e ∗ y) ∗ z = y ∗ z,
so x ∗ (y ∗ z) = (x ∗ y) ∗ z.
(iii) If y = e then x ∗ (y ∗ z) = x ∗ (e ∗ z) = x ∗ z and (x ∗ y) ∗ z = (x ∗ e) ∗ z = x ∗ z,
so x ∗ (y ∗ z) = (x ∗ y) ∗ z.
(iv) If z = e then x ∗ (y ∗ z) = x ∗ (y ∗ z) = x ∗ y and (x ∗ y) ∗ z = (x ∗ y) ∗ e = x ∗ y,
so x ∗ (y ∗ z) = (x ∗ y) ∗ z.
In every case, x ∗ (y ∗ z) = (x ∗ y) ∗ z. Hence ∗ is associative.
4. Let ∗ be an operation on a set S, and let ~ : S × S → S be given by
x~y =y∗x
for x, y ∈ S.
(a) Explain how the Cayley tables of ∗ and ~ are related.
(b) Prove that ∗ is commutative ⇐⇒ ∗ = ~ ⇐⇒ ~ is commutative.
(c) If (S, ∗) is a group, prove that (S, ~) is a group.
(d) Is the converse of (c) true? Either give a proof or a counterexample.
[The converse of the statement “if P then Q” is “if Q then P ”.]
Solution (a) They are transposes of one another, since the equation x∗y = y ~x
shows that the (x, y) entry of the Cayley table for ∗ is equal to the (y, x) entry of
the Cayley for ~.
(b) We have
∗ is commutative ⇐⇒
⇐⇒
⇐⇒
⇐⇒
⇐⇒
⇐⇒
x ∗ y = y ∗ x for all x, y ∈ S
x ∗ y = x ~ y for all x, y ∈ S
∗=~
x ∗ y = x ~ y for all x, y ∈ S
y ~ x = x ~ y for all x, y ∈ S
~ is commutative.
(c) Suppose that (S, ∗) is group.
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• If x, y, z ∈ S then, by the associativity of ∗, we have
x ~ (y ~ z) = (z ∗ y) ∗ x = z ∗ (y ∗ x) = (x ~ y) ~ z.
So ~ is associative.
• If e is the identity element for ∗, then x ~ e = e ∗ x = x = x ∗ e = e ~ x for
every x ∈ S, so e is the identity element for ~.
• If x ∈ S and y is the inverse of x with respect to ∗, then x ~ y = y ∗ x = e =
x ∗ y = y ~ x, so y is also the inverse of x with respect to ~. So every x ∈ S
is invertible with respect to ~.
Hence (S, ~) is a group.
(d) The converse of (c) is the statement: if (S, ~) is a group, then (S, ∗) is a group.
~ is a group. But,
This is true. Indeed, if (S, ~) is a group, then by (c), (S, )
~ ∗. So (S, ∗) = (S, )
~ is a group.
clearly, =
5. Given a group (G, ∗) and x, y ∈ G, consider the following two mappings:
Lx : G → G, Lx (z) = x ∗ z
and Ry : G → G, Ry (z) = z ∗ y
for z ∈ G.
(a) Prove that Lx ◦ Ry = Ry ◦ Lx .
(b) Prove that (G, ∗) is abelian if and only if Lx = Rx for every x ∈ X.
(c) Prove that the following statements are equivalent:
(i). x = y
(ii). Lx = Ly
(iii). Rx = Ry
[This means: show that any of these three statements is true if and only if
any of the other statements is true. One way to do this is to prove that
(i) ⇐⇒ (ii), and that (i) ⇐⇒ (iii).]
6. Which of the following is a group operation on M (2, R), the set of 2 × 2 matrices
with real entries? As always, you should prove that your answers are correct.
(a) matrix addition
(b) matrix multiplication
(c) the operation ∗ given by A ∗ B = AB − BA for matrices A and B
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Solution (a) This is a group operation. Indeed
• M (2, R) is closed under matrix addition, since A, B ∈ M (2, R) =⇒ A + B ∈
M (2, R). Hence matrix addition is an operation on M (2, R).
• For A, B, C ∈ M (2, R) we have (A + B) + C = A + (B + C). So addition is
associative.
• Writing 0 = ( 00 00 ) for the zero matrix in M (2, R), we have A + 0 = 0 + A = A
for all A ∈ M (2, R). Hence 0 is the identity element for (M (2, R), +).
• If A ∈ M (2, R) then −A ∈ M (2, R), and A + (−A) = (−A) + A = 0. So every
A ∈ M (2, R) is invertible with respect to addition.
(b) Since the zero matrix is not invertible with respect to matrix multiplication,
this is not an operation on M (2, R).
(c) This operation is not associative on M (2, R), by the last exercise on Homework
sheet 1. So it is not a group operation.
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