Intuition on group homomorphisms

Question

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

Answer 1

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.

Answer 2

As has been said, group homomorphisms are maps between groups which respect structure.

This is easy to picture in the context of isomorphisms between groups. The size of the two groups is the same (even if they are infinite), and their structure behaves the same (even if the underlying sets and binary operations are different). So finding an isomorphism is basically like translating a passage between two languages.

But it's not so easy to picture when the maps aren't bijective.

Sometimes you'll be embedding a group into a bigger group, in other words constructing an injective homomorphism that isn't surjective. That's not too hard to picture though, it just looks like this.

You see this is just an isomorphism to a subgroup of the target group. Injective homomorphisms don't need to be unique in this case, for example lots of groups have many subgroups isomorphic to $C_2$, so you could embed $C_2$ in a bunch of different places through different injective homomorphisms.

Surjective homomorphisms (which aren't injective) are a bit of a different story. Here's what they look like. You might recall the first isomorphism theorem, that if $\varphi:G\rightarrow H$ is a homomorphism, then there is another homomorphism $\theta:G\rightarrow G/\mbox{Ker}\varphi$ and an isomorphism $\mu:G/\mbox{Ker}\varphi\rightarrow \varphi[G]$ so that $\varphi=\mu\theta$. (In the case pictured above, $\varphi$ is surjective, so $\varphi[G]=H$.) This is a complicated looking theorem but it is actually quite well illustrated above. Intuitively, it means you can divide the group up into those colored blocks which behave collectively the same way in $G$ as their images do in $H$. Remember that all homomorphisms can be viewed this way, and that in fact an equivalent formulation of a homomorphism is that $\varphi:G\rightarrow H$ is a homomorphism if and only if $\mbox{Ker}\varphi$ is a normal subgroup of $G$. Even injective homomorphisms can be thought of like this, but by definition the kernel of an injective homomorphism is trivial, so the blocks are just one element. It is important to understand this "contracting" of blocks of elements to understand homomorphisms.

Now, there are homomorphisms that are neither injective nor surjective, but they can be pictured easily by combining the above two diagrams. Colored blocks on the left will correspond to preimages of colored blocks on the right as in the second diagram, but like the first diagram they won't cover the whole target group (i.e. there are still some grey blocks that never get mapped to).

So having covered group homomorphisms in detail, it is important to realize that homomorphisms in other algebraic structures behave in exactly the same way. Module, ring, and field homomorphisms all obey the same isomorphism theorems. Though it is a little harder to visualize with more than one operation (and hence more than one multiplication table), they all behave like this, and obey the same "contraction" principle.

Answer 3

Homomorphisms are really similar to isomorphisms except that they need not be injective or surjective. If not injective, different elements may be identified. If not surjective, the image is a proper subset of the codomain. That is all there is to the matter, really. The first fundamental theorem for homomorphisms says it all: that the quotient of a group by the kernel is isomorphic to the image.

Answer 4

Homomorphism simply means a map which respects the structure. Be it a group, a ring, or a partially ordered set.

Homomorphism means that we have a function between two structures of the same type and the map respects the structure. How many homomorphisms (and whether or not there are surjective, injective, or bijective) tells us how comparable the structures are.

If the only homomorphism between two structures is trivial, it tells us the structures are very different.

Answer 5

I am teaching myself some category theory, and I found it very useful in clarifying concepts like group homomorphisms. To get some idea, let's first look at Structures, so it's just a collection of Things with some special Structures.

For instance you can look at a structure called Graph, where the things are dots and the structures are how dots are connected by arrows, which can be thought of as maps between dots, \begin{equation} A\xrightarrow{f_{ab}}B \end{equation} says there is one arrow $f_{ab}$ between the dots $A$ and $B$.

Each graph is just a collection of such maps.

Given two graphs \begin{equation} G_1=\{A_1\xrightarrow{f^{1}_{ab}} B_1\} \end{equation} and \begin{equation} G_2=\{A_2\xrightarrow{f^{2}_{ab}} B_2\}, \end{equation} we want to construct some transform $\Phi$ from $G_1$ to $G_2$ that respect the graph structure, and to define such a $\Phi$ we must specify where the dots go as well as where the arrows go.

I guess it's clear the natural thing to do is if $A_1,B_1$ are connected by $f_{ab}^1$ in $G_1$, then $\Phi(A_1),\Phi(B_1)$ better get connected by $\Phi(f_{ab}^1)$ in $G_2$. Or in the language of category, the following diagram commutes:

It just says if there is one arrow in the first graph then the image of this arrow is an arrow in the second graph, and this is what we mean by respect the structure.

We now look at Groups as structures. The structure is given by the binary operation, multiplication \begin{equation} G\times G\xrightarrow{m} G \end{equation} so the maps $\Phi$ respect this structure make the following diagram commute

But this precisely means \begin{equation} \Phi(g)\cdot\Phi(h)=g\cdot h, \end{equation} which is the definition for group homomorphisms.

So there is nothing strange about this definition, it just makes sure dots (elements) get mapped to dots (elements) while arrows (multiplication) got mapped to arrows (multiplication). And to get an isomorphism, you just require this $\Phi$ is bijective.

Answer 6

One way of thinking about group homomorphisms is maps that preserve group structure. This seemingly abstract idea is quite simple to comprehend.

If a group $G_1$ is defined with the product operation $*$, then a homomorphism is a map that takes any such product of elements in $G_1$ to another group $G_2$ such that this product is taken to another product in $G_2$ with it's corresponding operation $\#$. Essentially, products are taken to products, inverses to inverses and identity elements to identity elements. This is what is meant by preservation of structure.

However, a Homomorphism need not be bijective like an isomorphism. For example the exponential map from the set of real numbers with the $+$ operation to the set of non -zero real numbers with the multiplication operation is a homomorphism but not an isomorphism since it is not bijective but preserves group structure.

Hope this helps.

Answer 7

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 " relations in $ G $ give corresponding relations of images in $\phi(G)$ " 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 "group homomorphisms".

Answer 8

Sometimes, understanding the utility of a mathematical concept with some "use cases" helps to understand the concept itself. This is what I am trying to do with this answer.

Let us say that you want to study a certain group. One of the ideas is to find homomorphisms from this group to other groups.

One-to-one homomorphisms provide new ways to think about the elements of the group. For example, there is a one to one homomorphism from the set of symmetries of an equilateral triangle (the dihedral group $D_3$) to the set of symmetries of a regular hexagon (the dihedral group $D_6$). This means that the symmetries of an equilateral triangle can be seen as symmetries of a regular hexagon as well.

Many-to-one homomorphisms may help you to “classify” the elements of a group according to their image under the homomorphism, and to understand the internal structure of the group. For example, there is a many-to-one homomorphism from the set of symmetries of an equilateral triangle $D_3$ to the multiplicative group $\{1,-1\}$. The elements of $D_3$ mapped to $1$ are rotations around the center of the triangle (including the identity, which is a rotation of null angle). The elements mapped to $-1$ are reflections about the symmetry axes of the triangle. Then, the homomorphism “classifies” the elements of $D_3$ as either rotations or reflections. Moreover, because the homomorphism preserves the group operation ($\phi (x \circ y) = \phi (x)\phi (y)$), it makes clear that:

• the composition of two rotations is a new rotation $(1·1=1)$,
• the composition of two reflections is a rotation $[(-1)·(-1)=1]$,
• the composition of a reflection and a rotation is a reflection: $[(1)·(-1)=-1]$.

Answer 9

I will just add to and adjust the answer by @Brian, by saying that not only does the kernel tell you how much structure is "wiped out" but, and this is a view I think I remember being espoused in Herstein, if not others, that (and as we all know, and it's a very useful fact, that they are normal subgroups) kernels/normal subgroups are in one-one correspondence with homomorphisms out of a group.

Furthermore, in view of the first isomorphism theorem, the image of a homomorphism is isomorphic to the quotient of the domain group by its kernel.

Homomorphisms have a number of nice properties, they always map the identity to the identity, as a result map an element to an element dividing its order, and, if it's surjective, maps normal subgroups to normal subgroups.

Under certain conditions (known as a split extension) the image of a homomorphism is isomorphic to a subgroup of the domain.