Department of Mathematical Sciences
Instructor: Daiva Pucinskaite
Modern Algebra
March 23, 2017
Homework (Classification of finite groups up to isomorphism)
Recall: Lagrange’s theorem. Let G be a finite group of order n, and H a subgroup of order m. Then m divides n.
If x ∈ G, then the subgroup hxi = {e, x, x2 , . . . , xord(x)−1 } of G has ord(x) elements. By the Lagrange’s theorem
ord(x) divides |G|.
I. Let hG, •i be a group where every non-identity element has order 2. The identity
element we denote by e.
(Notice, that in this case g −1 = g for all g ∈ G, also for e.)
1. Show: G is abelian.
(In other words: If x2 = e for all x ∈ G, then a • b = b • a for all a, b ∈ G).
2. Show: If g, g̃ ∈ G, such that g 6= e, g̃ 6= e, and g 6= g̃, then
g • g̃ 6∈ {e, g, g̃}.
3. Using 1. and 2. show:
If g, g̃ ∈ G, such that g 6= e, g̃ 6= e, and g 6= g̃, then the subset
H = {e, g, g̃, g • g̃} is a subgroup of G.
4. Using 3. and Lagrange’s theorem show:
If |G| = n, and there are g, g̃ ∈ G, such that g 6= e, g̃ 6= e, and g 6= g̃, then
n = 4 · a , for some
a ∈ N.
(i.e. 4 is a divisor of n.)
II. Using I.4. show:
Let hG, •i be a group such that |G| = n where
n ∈ {3, 5, 6, 7, 9, 10, 14, 15, 22, 26, 111, 2017, 4 225, 1 234 567}.
Then there exist an element g ∈ G such that g 6= e and ord(g) 6= 2.
III. Let hG, •i be a group such that |G| = n, and G ∼
= Zn . Show:
1.
2.
3.
4.
5.
If n = 6, then there exists an element a ∈ G, such that ord(a) = 3.
If n = 10, then there exists an element a ∈ G, such that ord(a) = 5.
If n = 14, then there exists an element a ∈ G, such that ord(a) = 7.
If n = 26, then there exists an element a ∈ G, such that ord(a) = 13.
If n = 2 · p, where p is a prime number, then there exists an element a ∈ G, such that
ord(a) = p.
IV. Let hG, •i be a group such that |G| = n, and G 6∼
= Zn . Show:
(a) Show: If g ∈ G, then ord(g) 6= n.
(b) Using II. and (a) conclude:
If n = 6, then there exists an element a ∈ G, such that ord(a) = 3.
If n = 10, then there exists an element a ∈ G, such that ord(a) = 5.
If n = 14, then there exists an element a ∈ G, such that ord(a) = 7.
If n = 26, then there exists an element a ∈ G, such that ord(a) = 13.
If n = 2 · p, where p is a prime number, and p 6= 2, then there exists an element
a ∈ G, such that ord(a) = p.
V. Let hG, •i be a group such that |G| = 10, and G ∼
6= Z10 . Let a be an element in G, such
that ord(a) = 5. Show:
1.
2.
3.
4.
5.
If ξ 6∈ hai = {e, a, a2 , a3 , a4 }, then ord(ξ) = 2.
VI. Let hG, •i be a group such that |G| = 10, and G ∼
6= Z10 . Let a be an element in G, such
that ord(a) = 5.
1. Show: If b 6∈ hai = {e, a, a2 , a3 , a4 }, then a • b 6∈ hai.
2. Show: If b ∈
6 hai, then a • b = b • a4 .
(Use 1. and V. )
3. Show: If b 6∈ hai, then G = {e, a, a2 , a3 , a4 , b, b • a, b • a2 , b • a3 , b • a4 }.
4. Let b 6∈ hai, compute the multiplication table of G:
•
e
a
a2
a3
a4
b
b•a
b • a2
b • a3
b • a4
e
a
a2
a3
2
a4
b
b • a b • a2 b • a3 b • a4