The anatomy of a homomorphism. I am a bit unclear about how to think of homomorphisms.
Hence, I am trying to sketch the anatomy of a homomorphism, if there are any flaws I would be very happy if someone could correct my sketch....
What I know (about homomorphisms ($\phi: G\rightarrow H$))


*

*$\phi$ is map b/w two groups ($G,H$ & let g, h be arbitrary elements in them respectively)

*$\phi(g_1*g_2)=\phi(g_1)\times \phi(g_2)$ (* is the group operation in G, while $\times$ is the group operation in H)

*$\phi$ maps $e_h$ to $e_g$ (these are identity elements of the two groups)

*$\phi$ maps inverses of one group to inverses in second group.

Some Assumptions
$\phi$ maps elements of a cycle of $G$ ($C_g$) to corresponding elements of a cycle of $H$ ($C_h$) [the number of elements in these two may not be the same, as $\phi$ is not necessarily injective and bijective or either one of the two]. 
My sketch of a homomorphism
Let the corresponding cycles of $G$ and $H$ mapped by $\phi$ be $C_g$ and $C_h$. Then, $C_g$ and $C_h$ can be represented as two circular dials marked with elements at equally spaced intervals fixed one each to the parallel shafts shown in the figure below with fixed pointers to read of the value (element) on the dial (They are fixed in such a way that before rotating the shafts, the identity element marked on one dial was read out when the pointer at the other was at its identity):

The group operations in the two groups be unified as a quantized rotation of the corresponding ring where one quanta for either of the dials is the angle subtended by two consecutive elements at the center of the dial (the direction of rotation in the two dials has opposite directions, same applies to the quanta). The gear assembly connecting the two shafts represents the homomorphism $\phi$ and defines the rotational multiplier which depends upon the relative circumferences of the coupled gears. 
[CLARIFICATION: Rotation by one unit meaning from the identity element to the next element corresponds to the group operation which may be multiplication in one group and addition in the other, they may or may not be the same for both, but we still will use rotation to represent both.]
Rotating the dial $C_g$ starting from identity element to an arbitrary element $g$ should give a read out of $\phi (g)$ on the $C_h$-dial, and vice-versa.


I understand that my concepts can be completely wrong, in such a case I request you to kindly give an alternative sketch / physical representation of a homomorphism, or if none exist then please let me know the reason why...Thanks.
 A: I do think this is an interesting model of a homomorphism, in fact I think its a great example. The construction itself does have a flaw, but it's an interesting flaw that reveals the properties of homomorphisms a little bit. So I think this is a very instructive set of examples.
Let me first describe the operation of your machine in a more detail, and then I'll compare your map $\phi$ to a homomorphism. 
The rotation of each gear models an action of a cyclic group on the circle, sometimes referred to as "clock arithmetic" where the clock has $m$ positions marked (or where your gear has $m$ teeth), and the group is cyclic of order $m$, modelled by rotating a clock hand through fractions $i/m$ of a full rotation.
Here's more details. Suppose that the gear on the left has $m$ teeth. The rotations of the gear on the left form a cyclic group of order $m$. Let me choose as generator for this group the clockwise rotation of the gear by just a single tooth (clockwise rotation through $1/m$ of a full rotation). I'll denote this generator $r$. So the elements of the rotation group of the left gear can be written as $\{r,r^2,...,r^{m-1},r^m\}$ where $r^i$ is a clockwise rotation by $i$ teeth. The full rotation $r^m$ is the equal to the identity element of the group, meaning no rotation at all; I'll denote that identity as $r^m=I$. The group operation is simply to do one rotation and then follow it with another, for example $r^i * r^j$ means first do the clockwise rotation through $i$ teeth, and then do the clockwise through $j$ teeth, the result of which is a clockwise rotation through $i+j$ teeth. But you have to reduce $i+j$ modulo $m$:
$$r^i * r^j = \begin{cases}
r^{i+j} & \quad\text{if $i+j \le m$} \\
r^{i+j-m} & \quad\text{if $i+j > m$}
\end{cases}
$$
Suppose similarly that the gear on the right has $n$ teeth, with rotations forming a cyclic group of order $n$. This time, though, I'll choose as generator the counterclockwise rotation by just a single tooth, denoting that as $s$, and so the elements are $\{s,s^2,...,s^{n-1},s^n\}$, with identity $s^n=J$. The group operation is again just addition of exponents, reduced modulo $n$.
Your map $\phi$ is defined by $\phi(s^i)=r^i$ for all $i=1,...,m$: if I rotate the left gear $i$ teeth clockwise, then the effect on the right gear is a rotation of $i$ teeth clockwise.
Now let's look at specific values of $m,n$. 


*

*If $m=12$ and $n=6$ then this map $\phi$ is indeed a homomorphism; it's not hard to check.

*But if, on the other hand $m=6$ and $n=12$ then $\phi$ is not a homomorphism, because it $\phi(I) = \phi(r^6) = s^6 \ne J$, and for a homomorphism it is required that $\phi(I)=J$.

*For an example as asked for in the comments, suppose that that $m,n$ have a common factor. As long as the greatest common factor is $<n$ then $m$ will not be a multiple of $n$ and the argument below can be applied to prove that $\phi$ is not a homomorphism. For a specific example take $m=6$ and $n=4$, and we have $\phi(I)=\phi(r^6)=s^6=s^2 \ne J$. 


There is a completely general statement one can make: for your map to be a homomorphism, it is necessary and sufficient that $m$ be a multiple of $n$.
To prove necessity, suppose $\phi$ is a homomorphism. It follows that
$$\phi(r^m) = s^m = J
$$
which means that $m$ full rotations of the right gear is the same as no rotation at all, which implies that $m$ is a multiple of $n$.
The proof of sufficiency is a good exercise in group theory: assuming that $m$ is indeed a multiple of $n$, prove that $\phi$ is a homomorphism.

Let me add some final remarks. Your model is designed to work only for a specific class of groups, namely the finite cyclic groups (and as said it produces a homomorphism if and only if the number of teeth on the left gear is a multiple of the number of teeth on the right gear). However, there is another completely general description of a model similar to yours, which is known as Cayley's Theorem, and which says that every group of order $n$ may be represented as a subgroup of the group of permutations of a set of size $n$; your example of rotating a gear with $n$ teeth is mathematically the same as Cayley's representation of a cyclic group of order $n$.
