Question about Z12≃Z4⊕Z3 generators and order I apologize in advance for my messy language and questions; I've only been studying group theory for a month and thus these concepts aren't clearly locked in yet; hence my questions. :)
Take Z12≃Z4⊕Z3.   
Call the generator of Z12 "a", Z4 "b", and Z3 "c".  Thus:
Z12 = aaaaaaaaaaaa
Z4 = bbbb
Z3 = ccc
Q1):  Am I using the term "generator" correctly as in "a single element whose repeated application generates the group"?
However, the automorphic generators of Z12 are {1,5,7,11}.
Q2)  Is "a" a generator since repeated applications of it produce Z12 or is the set {1,5,7,11} the generators since they are coprime to Z12?  How do I reconcile these two usages of the term "generator"?
Since Z4 and Z3 are characteristic subgroups of Z12, then:
b = aaa
c = aaaa
if we denote a= 1, then b = 3 and c = 4.
Q3) Is that construction correct?  If so, what is the group theory concept behind building a "child" generator from a repeated application of the "parent" generator?  In other words, is there a formal group theoretical concept that "ranks" the generators such that "a" is rank 1, "b" is rank 3, and "c" is rank 4?
Finally, I want to find a way to combine both b and c generators such that I get a new generator that would have rank 7.  I don't think the direct product works since I read that as a cartesian product (a,b) =ab = 12.  Instead I want a summation, a direct sum, where [a,b] = a+b = 7.  It seems like a direct sum is involved but I thought that direct products and direct sums where the same thing for abelian groups
Q4) I suspect this may be non-sense, but is there a way to combine Z3 and Z4 or interpret their direct product or sum such that we get a group of order 7 since there are 3 elements in Z3 and 4 elements in Z4??
 A: $\mathbb{Z}/12$ is cyclic, and it is generated by any element prime to $12$, which as you have noted is $\{1,5,7,11\}$. So any of these elements could be the $a$ that you selected as your generator. For example, if you selected $5$, you would get
$$ 5, 5+5 = 10, 5+5+5 = 15 = 3, 5+5+5+5 = 20 = 8,$$
and so forth. You will see that you get all twelve elements of $\mathbb{Z}/12$ in sequence.
Now, if $a$ is any generator, then $12\cdot a = 0$, so that $3\cdot (4\cdot a) = 0$. That means that $4\cdot a$ generates a subgroup of order $3$ ($\{4\cdot a, 2\cdot (4\cdot a), 3\cdot (4\cdot a) = 0\}$). So this subgroup is isomorphic to $\mathbb{Z}/3$. Thus you can select $4\cdot a$, where $a$ is any generator, for your $b$. Similarly, $3\cdot a$ generates a subgroup of order $4$, which is isomorphic to $\mathbb{Z}/4$.
In general, if $G$ is cyclic of order $n$, with generator $g$, then $G$ has a (unique) subgroup of order $d$ for an divisor $d$ of $n$, and a generator of that subgroup is $g^{n/d}$ (or, if you wish to write your group additively, $\frac{n}{d}g$).
A group of order $12$ cannot have a subgroup of order $7$, since by Lagrange's Theorem the order of any subgroup divides the order of the group.
Finally, in general if $a, b\in \mathbb{Z}/12$, then the order of the subgroup generated by $ab$ is the least common multiple of $a$ and $b$.
