Commutator subgroup help?!

Let G={(a,b) : a,b E Z} Define * on G by (a,b)*(c,d) = (a + c(-1)^b , b+d)

a) Show G is non abelian group
b) Show G'(commutator subgroup) = <(2,0)>

a) Since * is defined in terms of operations closed in Z, G is closed under *.
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Identity:
First, we need to find (c, d) such that
(a,b) * (c,d) = (a + c(-1)^b , b+d) = (a, b) for all (a, b) in G
==> a + c(-1)^b = a and b + d = b.
==> c = 0 and d = 0.

Hence, the identity on G is (0, 0).
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Now, we can find the inverse of (a, b):
We want to now find (c, d) such that (a,b) * (c,d) = (a + c(-1)^b , b+d) = (0, 0)
==> a + c(-1)^b = 0 and b + d = 0
==> c = a(-1)^(b+1) and d = -b.

So, (a, b)^(-1) = (a(-1)^(b+1), -b).
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Associativity:
For all (a,b), (c,d), (e,f) in G:

[(a, b) * (c, d)] * (e, f)
= (a + c(-1)^b , b+d) * (e, f)
= (a + c(-1)^b + e(-1)^b, (b+d)+f)

(a, b) * [(c, d) * (e, f)]
= (a, b) * (c + e(-1)^b, d + f)
= (a + (c + e)(-1)^b, b + (d + f)).

Now, it's easy to see that [(a, b) * (c, d)] * (e, f) = (a, b) * [(c, d) * (e, f)].
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Lack of Commutativity:
An example will suffice.

(1, 1) * (2, 2) = (1 + 2(-1)^1, 1 + 2) = (-1, 3),
but (2, 2) * (1, 1) = (2 + 1(-1)^2, 2 + 1) = (3, 3).

Hence, G is not an abelian group.
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b) G' is generated by all elements of the form (a, b)^(-1) * (c, d)^(-1) * (a, b) * (c, d).

Note that (a, b)^(-1) * (c, d)^(-1) * (a, b) * (c, d)
= (a(-1)^(b+1), -b) * (c(-1)^(d+1), -d) * (a, b) * (c, d)
= [(a(-1)^(b+1), -b) * (c(-1)^(d+1), -d)] * [(a, b) * (c, d)]
= (a(-1)^(b+1) + c(-1)^(d+1) (-1)^(-b), -b + -d) * (a + c(-1)^b , b+d)
= (a(-1)^(b+1) + c(-1)^(b+d+1), -b - d) * (a + c(-1)^b , b+d), since (-1)^2 = 1
= ([a(-1)^(b+1) + c(-1)^(b+d+1)] + [a + c(-1)^b] (-1)^(-b-d), [-b - d] + [b + d])
= ([a(-1)^(b+1) + c(-1)^(b+d+1)] + [a + c(-1)^b] (-1)^(b+d), 0)
= (a(-1)^(b+1) + c(-1)^(b+d+1) + a(-1)^(b+d) + c(-1)^d, 0)
= (a(-1)^b (-1 + (-1)^d) + c(-1)^d ((-1)^(b+1) + 1), 0)

Note that each of a(-1)^b (-1 + (-1)^d) and c(-1)^d ((-1)^(b+1) + 1) are 0 or even,
depending on whether (-1)^k + 1 = 0 or 2.

Hence, G' is contained in <(2, 0)> = {(2k, 0) : k in Z},
[Note that (2, 0)^k is obtained from composing (2, 0) with itself via * k times for k > 0.
For k < 0, use the inverse of (2, 0)^(-k).]

To show that G' has (2, 0) in it, observe that letting (a, b) = (1, 1), (c, d) = (0, 1)
yields a commutator of (2, 0).

I hope this helps!