Abstract Algebra Solutions

This document presents solutions to selected problems from Chapters 5, 6, and 7 of an abstract algebra textbook. Chapter 5 solutions cover even permutations and the order of permutations. Chapter 6 focuses on isomorphisms and automorphisms, demonstrating the abelian nature of groups under specific conditions. Chapter 7 solutions delve into cosets, Lagrange's theorem, Fermat's Little Theorem, and the properties of group centers. The solutions are detailed and provide step-by-step explanations, making them a valuable resource for students studying abstract algebra.

Chapter 5
Solution 10 :
We know that
3 –cycle (abc) = (ab) (ac) is even
And, 5 – cycle (abcxy) = (ab) (ac) (ax) (ay) is also even.
Considering this, we can say (123) ϵ A8 and (45678) ϵ A8.
Therefore, α = (123) (45678) ϵ A8.
Now, according to the Ruffini’s theorem,
| α | = 15, since (123) and (45678) are disjoints.
Solution 26:
We can write any permutation as a product of 2 cycles
Therefore, let α = α1 α2 …… αk and  = ❑1 ❑2 … ..❑l where α i∧❑j is two-cycle.
It can also be written as α−1 = αk … ..α 2 α1and ❑−1 = ❑l …..❑2 ❑1
Now, -1-1 = (α 1 α2 … … αk ¿ (❑1 ❑2 … ..❑l ¿ (αk … ..α2 α1) (❑l …..❑2 ❑1 ¿
This is the product of 2(k + l) 2-cycles.
Therefore, -1-1 is an even permutation.
Chapter 6
Solution 12.
Considering α as an isomorphism, we can say that any two element a, b belongs to G.
a−1 b−1 = ab−1 = (ab) = (a) (b) = b−1 a−1
Therefore, ab = (a ¿¿−1)−1 ¿ ¿ = (b−1 a−1)−1 = (a−1 b−1)−1 = (b−1)−1 (a−1 )−1 = ba

Hence, G is abelian.
Now, considering that G is abelian and  being the inverse function of ,  : G → G is a
bijective function
And, also,
 (ab) =  (ba) = (ba)−1 = a−1 b−1 =  (a) (b)
It suggest that it has operation preserving property. Hence, G is an automorphism.
Solution 38:
Let R : G → H be the defined map by R(a + b √2) = [ a 2 b
b a ]
It is evident that R is a bijection between G and H.
Let us prove that the group structure is preserved,
0 is the identity element of G and it can be represented as 0 + 0 √2
Now,
R (0) = [ 0 0
0 0 ]
The identity of H.
Now,
(a+ b √ 2)−1 = -a - b √2
And R (-a - b √2) = [ −a −2 b
−b −a ]
= [a 2b
b a ]−1
Now, for a1,a2,b1,b2 ∈ Q we have
R ((a1 + b1 √ 2) + (a2 + b2 √ 2)) = R( ( a1+a2 ) + (b1+b2 ¿ √ 2)
= [ a1+a2 2(b1+b2)
b1+b2 a1+ a2
]

= [ a1 2b1
b1 a1
] + [ a2 2 b2
b2 a2
]
= R (a1 + b1 √2) + R (a2 + b2 √2)
Therefore, R is an isomorphism.
Moving to the multiplication,
(a1 + b1 √ 2) ( a2 + b2 √ 2) = a1 a2+ ( a1 b2 + b1 a2 ¿ √ 2 + 2b1 b2
= (a1 a2 + 2b1 b2 ¿ + (a1 b2 + b1 a2 ¿ √ 2
Now, considering
R (a1 +b1 √2). R (a2 +b2 √ 2) = [ a1 2b1
b1 a1
] [ a2 2 b2
b2 a2
]
= [ a1 a2+2 b1 b2 2a1 b2+2 b1 a2
b1 a2 +b2 a1 2 b1 b2 +a1 a2
]
= R(( a1 a2 + 2b1 b2 ¿ + (a1 b2 + b1 a2 ¿ √ 2)
= R (a1 +b1 √2) (a2 +b2 √ 2)
Therefore, B preserves the multiplication
Solution 52:
|Inn(G)| = 1 if and only if G is Abelian
Chapter 7
Solution 6.
Suppose that a has order 15. Find all of the left cosets of a5 in a.
We know that,
|a5| = 3
Also, |a| = 5
By Lagrange theorem, |a : a5| = 5
Hence, there are 5 cosets and they are:

e a5 = e, a5, a10
a a5 = a, a6, a11
a2 a5 = e, a7, a12
a3 a5 = e, a8, a13
a4 a5 = e, a9, a14
Solution 24.
According to the FML,
a p = a mod p
Or, a p mod p = a mod p
Both sides multiplying a−1
We have,
a p a−1= a a−1
Or, a p−1 = 1
Or, a p−1 = 1 (mod p)
Substituting power both sides:
(a p−1)m = 1m
Upon multiplyingak−1 on both sides, we have
(a p−1)m ak−1 = 1m ak−1
Or, am ( p−1 ) + (k−1 ) = ak−1
Or, m(p-1) + (K-1) = (K-1)
According to the definition, x mod(p-1) if x = q(p-1) = k -1
Therefore,
m(p-1) + (k-1) = (k-1)
Or, (k-1) mod (p-1) = (k-1)
As per our hypothesis,
ak=a mod p
ak a−1=a mod p . a−1

ak−1 = 1 mod p
Substituting the value of ak−1, we have
amod( p−1)+(k−1) = 1 mod p = a0 modp = a0 mod( p−1) modp
Therefore, (k-1) mod p = 0 mod p
According to the definition, it means
m(p-1) = k-1
It can also be said as k-1 divides p-1
Solution 38.
Let us assume by contradiction, |G:Z(G)| = p and p is prime.
Also, |Z(G)| = k, where k is a positive integer.
Therefore, |G| = pk. Take g ∈ G\ Z(G)
Consider the situation:
C(g), where |C(g):Z(G)| = y, xy =p and |G:C(g)| = x
This can only be true if either
1) x = p and y = 1
2) x = 1 and y = p
Consider 1st situation: |G: C(g)| = p and |C(g): Z(G)| = 1
Which means, Z(g) = C(g) and g ∈ C(g) = Z(G)
Hence, g ∈ Z(G) and this is contradiction
Consider 2nd Situation: |G:C(g)| = 1 and |C(g): Z(G)| = p
Which means, G = C(g) Or g◦x = x◦g, ∀x ∈ G, the centre definition
Hence, g ∈ Z(G) and this is contradiction
Thus it cannot be prime.