Question: Consider the group homomorphism f : Z × Z Z × Z with the property that

Question

Question: Consider the group homomorphism f : Z × Z Z × Z with the property that f((1, 0)) = (6, 12) and f((0, 1)) = (2, 4).

Please explain answer it doesn't make sense to me.
(a) Determine the image of f. [How do I get image??]
Im(f) = f(Z × Z) =< (2, 4) >= {(2k, 4k)|k Z}
(b) Determine the kernel of f. [How do I get Kernel, I do not understand kernel??]
Ker(f) = {(x, y)|x, y Z, 6x + 2y = 0} =< (1, 3) >= {(k, 3k)|k Z}

(c) Consider the factor group Z × Z/f(Z × Z). Does this group contain any elements of order 5? Does it contain any elements of order 2? Does it contain any elements of infinite order? Is it cyclic?

Z × Z/f (Z × Z) = Z × Z/ < (2, 4) >.

If (x, y) is an element of order 5 in Z×Z/f(Z×Z) then (5x, 5y) {(2k, 4k)|k Z},that is 5x=2k and 5y=4k forsome kZ. This implies 2|x and 4|y, and therefore (x, y) < (2, 4) >, that is (x, y) is the identity in Z × Z/f (Z × Z).

(1, 2) is an element of order 2 in Z × Z/ < (2, 4) >.

(1, 0) is an element of infinite order in Z × Z/ < (2, 4) >. In particular, this implies that Z × Z/ < (2, 4) > is infinite.

Z × Z/ < (2,4) > is not cyclic. Indeed, assume Z × Z/ < (2,4) > is cyclic, then since Z×Z/ < (2,4) > is infinite, we have Z×Z/ < (2,4) >= Z. We notice that Z does not have elements of order 2, while Z × Z/ < (2, 4) > has such an element.


Please explain the solution. I do not understand the solution.

Explanation / Answer

(a) f : Z × Z Z × Z be a group homomorphism with the property that f((1, 0)) = (6, 12) and f((0, 1)) = (2, 4).

f((1, 1)) = f((1, 0) + (0, 1))

= f((1, 0)) + f((0, 1)) by operation preserving property of homomorphism f

= (6, 12) + (2, 4)

= (6+2, 12+4)

= (8, 16)

Since Z × Z be a cyclic group generated by (1, 1), i.e. Z × Z = < (1,1) >

The imgae of f, i.e. f(Z × Z) must be a cyclic group generated by f((1, 1))

Therefore, f(Z × Z) = < f((1, 1)) > = < (8, 16) >

Answer

(b) Kernel f = {(x,y) | x, y Z; f((x,y)) = (0,0)}

Let, x, y Z & x > y

(x, y) = (x - y, 0) + (y, y) = (x-y)(1, 0) + y(1, 1)

f((x, y)) = (x-y)f((1, 0)) + yf((1, 1) by operation preserving property of homomorphism f

= (x - y)(6, 12) + y(8, 16)

= (6x + 2y, 12x + 4y)

Therefore, f((x, y)) = (0, 0)

Implies, (6x + 2y, 12x + 4y) = (0, 0)

Implies, 6x + 2y = 0 = 12x + 4y

Implies, 6x + 2y = 0

Implies, 3x + y = 0

Implies, y = - 3x

General solution: k(1, - 3) for k Z

Therefore, Kernel f = {(x,y) | x, y Z; 3x + y = 0} = < (1, - 3) >

Answer

(c) Since f(Z × Z) = < (8, 16) > is a cyclic subgroup of Z × Z, the factor group Z × Z/f(Z × Z) is defined.

This is because any cyclic subgroup is a normal subgroup.

Z × Z/f(Z × Z) = {(x,y) + f(Z × Z) | x, y Z} = {(x,y) + < (8,16) > | x, y Z}

If (x,y) + < (8,16) > is an element of order 5 in Z×Z/f(Z×Z) then, 5x = 8k and 5y = 16k for some k Z.

This implies, 8|x and 16|y as gcd(5,8) = 1, gcd(5,16) = 1

Therefore, (x,y) < (8,16) >

This implies, (x,y) + < (8,16) > is the identity in Z × Z/f (Z × Z)

(4, 8) is an element of order 2 in Z × Z/ < (8,16) >

(1, 0) is an element of infinite order in Z × Z/ < (8,16) >.

Z × Z/ < (8,16) > is not cyclic as it does not have a generator.

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