2.3#12a. Z_2 x Z_2 has no element of order 4.
b,c. Consider the map A x Z -> Z by forgetting the first element. If we had a generator of A x Z, it would map to a generator of Z. We know the two generators of Z: 1 and -1. Let's assume we map to 1 by replacing our purported generator by its inverse. Now our generator looks like (a,1).

Let b in A, but b not a. (This uses A not {1}; we only care about Z_2 and Z.) This generator (a,1) can't generate (b,1), since its nth power only has second entry 1 if n=1.

7.1#2. First let's note that (-1)v = -v, i.e. v + (-1)v = 1v + (-1)v = (1 + -1)v = 0v = 0. (For that matter, why is 0v=0? Because 0v = (0+0)v = 0v + 0v, and we can subtract 0v from both sides.) Likewise, v(-1) = -v using distributivity on the other side.

Then if uv=1, (-u)(-v) = u(-1)(-1)v = uv = 1.

7.1#3. If u is a unit in S, then us = 1 for some s in S, so us = 1 for that same s but considered in R. Whereas if 2 is a unit in the rationals Q (it is), it doesn't tell you that 2 is a unit in the integers Z.

7.1#11. If x^2=1, then (x-1)(x+1)=0. If we're in a domain, then there are no zero divisors, so either x-1=0 or x+1=0.

7.3#7. Checking that it's a homomorphism is just matrix multiplication. Ontoness is totally stupid. The kernel is the strictly upper triangular matrices.

7.3#8. No a or d, yes b and c. (a,b)*(1,0) = (a,0) which sinks a & d.

7.3#10a. Yes, it's (3,x).
b,d,f. No. It contains 1, but not x*1.
c. Yes, it's (x^3).
e. Yes, it's (x-1).