MATH 113: ABSTRACT ALGEBRA
SOLUTIONS TO PRACTICE PROBLEMS FOR MIDTERM 1
1. Show that if (G, ·) is a group of order 9, then G is abelian.
Using material we have not yet covered (namely, Lagrange’s Theorem and
the class equation), this problem is not so difficult. Just using the
material up through Section 9, it is very complicated. No question of
this kind will appear on the this test.
2. Let (G, ·) be a group and X any set. Let F be the set of functions with domain
X and range G. Define a binary operation ∗ on F by (f ∗ g)(x) := f (x) · g(x). Is
(F, ∗) a group? If so, prove that it is. If not, give an axiom which is violated and
prove that this is so.
Yes, (F, ∗) is a group.
Proof:
identity The identity element is the function I : X → G which is identically equal
to the identity element, e, of G. Indeed, for any f ∈ F and any x ∈ X we
have (I ∗ f )(x) = I(x) · f (x) = e · f (x) = f (x). Hence, I ∗ f = f .
inverse Let f ∈ F be any element of F . Let g : X → G be defined by g(x) :=
(f (x))−1 . Then for any x ∈ X we have (g ∗ f )(x) = g(x) · f (x) = (f (x))−1 ·
f (x) = e = I(x). Hence, g ∗ f = I so that g is a left-inverse of f .
associativity Let f , g, and h be elements of F . For any x ∈ X we have f ∗ (g ∗ h)(x) =
f (x)·(g ∗h)(x) = f (x)·(g(x)·h(x)) = (f (x)·g(x))·h(x) = (f ∗g)(x)·h(x) =
(f ∗ g) ∗ h(x). Hence, f ∗ (g ∗ h) = (f ∗ g) ∗ h.
1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8
3. Let σ = and τ =
8 4 3 2 7 6 1 5 3 5 2 7 8 1 6 4
a. Write τ as a product of cycles.
Solution: τ = (1, 3, 2, 5, 8, 4, 7, 6) is already a cycle.
b. Write σ as a product of transpostions.
Solution: σ = (1, 7)(1, 5)(1, 8)(2, 4)
c. Compute στ and τ σ.
1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8
Solution: στ = and τ σ =
3 7 4 1 5 8 6 2 4 7 2 5 6 1 3 8
d. What is the order of σ? of στ ?
Solution: The order of σ = (1, 8, 5, 7)(2, 4) is four while the order of
στ = (1, 3, 4)(2, 7, 6, 8) is twelve.
4. How many generators does the group Z225 have?
Date: 27 September 2007.
1
SOLUTIONS TO PRACTICE PROBLEMS FOR MIDTERM 1
1. Show that if (G, ·) is a group of order 9, then G is abelian.
Using material we have not yet covered (namely, Lagrange’s Theorem and
the class equation), this problem is not so difficult. Just using the
material up through Section 9, it is very complicated. No question of
this kind will appear on the this test.
2. Let (G, ·) be a group and X any set. Let F be the set of functions with domain
X and range G. Define a binary operation ∗ on F by (f ∗ g)(x) := f (x) · g(x). Is
(F, ∗) a group? If so, prove that it is. If not, give an axiom which is violated and
prove that this is so.
Yes, (F, ∗) is a group.
Proof:
identity The identity element is the function I : X → G which is identically equal
to the identity element, e, of G. Indeed, for any f ∈ F and any x ∈ X we
have (I ∗ f )(x) = I(x) · f (x) = e · f (x) = f (x). Hence, I ∗ f = f .
inverse Let f ∈ F be any element of F . Let g : X → G be defined by g(x) :=
(f (x))−1 . Then for any x ∈ X we have (g ∗ f )(x) = g(x) · f (x) = (f (x))−1 ·
f (x) = e = I(x). Hence, g ∗ f = I so that g is a left-inverse of f .
associativity Let f , g, and h be elements of F . For any x ∈ X we have f ∗ (g ∗ h)(x) =
f (x)·(g ∗h)(x) = f (x)·(g(x)·h(x)) = (f (x)·g(x))·h(x) = (f ∗g)(x)·h(x) =
(f ∗ g) ∗ h(x). Hence, f ∗ (g ∗ h) = (f ∗ g) ∗ h.
1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8
3. Let σ = and τ =
8 4 3 2 7 6 1 5 3 5 2 7 8 1 6 4
a. Write τ as a product of cycles.
Solution: τ = (1, 3, 2, 5, 8, 4, 7, 6) is already a cycle.
b. Write σ as a product of transpostions.
Solution: σ = (1, 7)(1, 5)(1, 8)(2, 4)
c. Compute στ and τ σ.
1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8
Solution: στ = and τ σ =
3 7 4 1 5 8 6 2 4 7 2 5 6 1 3 8
d. What is the order of σ? of στ ?
Solution: The order of σ = (1, 8, 5, 7)(2, 4) is four while the order of
στ = (1, 3, 4)(2, 7, 6, 8) is twelve.
4. How many generators does the group Z225 have?
Date: 27 September 2007.
1
