Intuition on group homomorphisms So I'm studying for finals now, and came across the idea of homomorphisms again. This is not a new idea for me at all, having seen them in groups, rings, fields ect. However, on reevaluating them I realized suddenly that I really don't understand them on the same level as I thought. While isomorphisms seem to be very natural to think about, I can't visualize what's happening in homomorphisms. At this time, I only have a sketchy idea that the existence of a homomorphism between groups (or between groups and what they're acting on) means that they both somehow combine the same way..

Does someone mind sharing their intuition on the concept? As always, any help is appreciated, thanks
You know that if there is an isomorphism $h:A\to B$ from an algebraic structure (monoid, group, ring, etc.) $A$ onto another algebraic structure $B$ of the same kind, then $B$ is essentially just $A$ ‘in disguise’: the two structures are essentially the same structure. In other words, Isomorphisms are the maps that preserve the structure exactly. 

Homomorphisms preserve some of the structure. (Here some may be all, since every isomorphism is a homomorphism. That is, it’s some in the sense of $\subseteq$, not $\subsetneqq$) They preserve the operations, but they may allow elements that ‘look enough alike’ to be collapsed to a single element. For instance, the usual group homomorphism from $\Bbb Z$ to $\Bbb Z/2\Bbb Z$ (for which you use the notation $\Bbb Z_2$) ‘says’ that all even integers are essentially the same and collapses them all to the $0$ of $\Bbb Z/2\Bbb Z$. Similarly, it ‘says’ that all odd integers are essentially the same and collapses them all to the $1$ of $\Bbb Z/2\Bbb Z$. It wipes out any finer detail than odd versus even. When you learn in grade school that even $+$ even $=$ even, odd $+$ even $=$ odd, and so on, you’re essentially doing the same thing.)

The kernel of the homomorphism is a measure of how much detail is wiped out: the bigger the kernel, the more detail is lost. In the example of the last paragraph, the kernel is the entire set of even integers: the fact that all even integers are in the kernel says that they’re all being seen as somehow ‘the same’, and even more specifically, ‘the same’ as $0$. An isomorphism has a trivial kernel: the only thing that it sees as looking like $0$ is $0$ itself, and no detail is lost.

Another way to put it is that a homomorphic image of an algebraic structure is a kind of approximation to that structure. If the homomorphism is an isomorphism, it’s a perfect approximation; otherwise, it’s more or less crude approximation. As the kernel of the homomorphism gets bigger, the crudeness of the approximation increases. In the case of groups, if the kernel is the whole group, then the homomorphic image is the trivial group, and all detail is lost: all that’s left is the fact that we started with a group.
This answer will be painfully similar to the one here , but in any case the details might be of help.
Say we have groups $ (G, *) $ and $ (G', \cdot) $ and a map $ \phi : G \rightarrow G' $. We can ask ourselves the following question : Under what kinds of $ \phi $ do we have that $ (\phi(G) , \cdot ) $ is a group, and that &quot; relations in $ G $ give corresponding relations of images in $\phi(G)$ &quot; i.e. that $ ( x*y = z \text{ in group } G ) \implies ( \phi(x) \cdot \phi(y) = \phi(z) \text{ in group } \phi(G) ) $ ?
Proceeding as in the answer linked above, we see maps satisfying $ \phi(x*y) = \phi(x)\cdot\phi(y) $ are precisely the ones we were looking for. Such maps are conventionally called &quot;group homomorphisms&quot;.