How do I show a mapping is a homomorphism?

Question

I don't want to make this question too broad, or non-specific. I'll will discuss a simple situation so we can all share a common context, but my question is less about this particular group, and more about the strategy of showing that a particular mapping you're interested in is a valid homomorphism.

Let $G$ be a group, and let a non-identity $(\ne 1)$ element $a \in G$ be such that $a^2=1$. Then $K:=\left\{ 1, a \right\} \lt G$. I don't see right away that $K \lhd G$, so I want to find a homomorphism with kernel $K$. If we were to show that $K$ is normal, we would need $gK = Kg$ $\forall g \in G$, but since $g1 = 1g$, this would imply we need $ga = ag$ $\forall g \in G$. (I point this out because I'll refer back to it later. I'm willing to hear arguments about $K$ being normal, but that's not the point of this question.)

Defining a mapping $\phi:G \rightarrow G$ such that $\phi(a)=1$ (and necessarily $\phi(1) = 1$) would certainly put $a$ in the kernel if $\phi$ is indeed a homomorphism as we'd like. But how would I clearly show that this is a homomorphism, with exactly $K$ as its kernel? Surely I can't just note that $\phi(ag) = \phi(a)\phi(g) = \phi(g) = \phi(g)\phi(a)=\phi(ga)$. This would depend on $\phi$ being a homomorphism, and would be an example of assuming the conclusion. If $\phi$ were a homomorphism and well defined, this would show ... well, I guess it would not show that $ag = ga$, but it would show that $ag$ and $ga$ are both members of the same equivalence class in the quotient group formed by modding $K$ out of $G$.

I freely admit that my background is strong in analysis and weak in algebra. When I watch an algebraist at work, everything looks quick and simple and leaves the impression that I'm watching magic, as if we could prove anything we want to, which clearly isn't so.

So I suppose my clear, narrowly focussed question is this: What is the list of criteria that one must show to prove that a particular mapping is a well defined homomorphism, and as an example how would they be used to show that this is (or is not) a homomorphism?

Answer 1

I will share here what I've learned in my effort to understanding how to show one has a well defined group homomorphism. I invite the community to edit this answer if they can remove errors or increase clarity. Also, if anyone can address my question about the strategy of going about doing this (good approaches that make it easier), please chime in.

You cannot have a well defined homomorphism, without first having a well defined mapping. The "error" in my question above, was that I thought I had a well defined mapping and was struggling to show it was a well defined homomorphism.

1. First you show you have a well defined mapping,
2. Then you show your mapping is a homomorphism.
   • This will be a well defined homomorphism. There is no distinction between a well defined mapping that is a homomorphism, and a well defined homomorphism.

Well defined mapping

There are 3 things one must have to show a mapping is well defined. Suppose you want to show $f:A \to B$ is a well defined mapping

1. $\forall a \in A, f(a)$ must be unambiguous. That is, it must be clear what $f(a)$ is.
   • If there's going to be a hope of being a group homomorphism, $f(a)$ must be defined, but to be a well defined mapping, at a minimum it must be unambiguous. In real analysis, when $f(x)$ is undefined, the typical way to treat this is to exclude $x$ from the domain, so that the function is defined for everything in its domain. However, even when this approach isn't taken, to say that $x$ is in the domain, but that $f(x)$ is undefined is not ambiguous.
2. $f(a) \in B$. If $f$ maps some elements of $A$ to $B$, and some elsewhere, then it's not a mapping from $A$ to $B$.
   • This is why elements for which $f(a)$ are not defined, are typically excluded from the domain. There is no "undefined" element in $B$. For homomorphisms, it it typically more of a concern to show that $f(a)$ is unambiguous than to show that all elements are mapped inside the range. If your "formula" maps one element of $A$ to an element of $B$, odds are you specified it so that it's taking all elements there.
3. $b_1 = b_2 \implies f(b_1) = f(b_2)$. No element may be mapped to two distinct places.
   • This is equivalent in real analysis to the vertical line test. One typically starts with two (apparently) distinct elements (differing in notation, not value) in the domain, applies the mapping to them, and then shows that if they are the same element then the mappings are the same.

For the purpose of memory, the simplest way I can compact those 3 rules into notation (encapsulating the convention that $f$ must be defined for every element in the domain) is $$f:A \to B \text{ is well defined} \iff \forall a \in A, f(a)\in \{b\} \subseteq B.$$

For every $a$, $f(a)$ is unambiguous (it's $b$), it takes a single value ($\{b\}$ is a singleton), and that value is in the range $B$.

Homomorphism

Showing that a well defined mapping is a homomorphism, is a single step application of the definition. Suppose you have two groups $G$ and $H$ and a well defined mapping $\phi : G \to H$, whose domain is all of $G$ (greek letters are preferred for homomorphisms).

1. $\phi(g_1 g_2) = \phi(g_1)\phi(g_2)$
   • This is not taken as a given, it's what you're trying to show. In other words, you cannot compute $\phi(g_1)$ and $\phi(g_2)$ and say, well I want a homomorphism, so I'll just define $\phi(g_1 g_2)$ to be the product of those two. To get to this point, you must already have a well defined mapping, so $\phi(g_1 g_2)$ already has a specified value. You must compute all 3 mappings and show the equivalence.
   • In creating your well defined mapping, you can set $f(a_1 a_2)$ to be what it needs to be for this to work out, but you still have to show all 3 steps of the well-defined mapping. In particular, you cannot be mapping $f(a_1 a_2)$ to two distinct elements of the range (step 3). Since in a group, there are many different ways to get the same product, if $a_1 a_2 = a_3 a_4$, then you must have $f(a_1 a_2) = f(a_3 a_4)$. If you take this approach, then once you've shown your mapping is well defined, this step of showing it is a homomorphism should be trivial.

Example

In the example in the question above, I thought I had a well defined mapping and was failing to see how to show this last step of showing it is a homomorphism (because I didn't know - or care - where the mapping took abstract elements I wasn't interested in). This itself showed that I'd failed the first step of showing the mapping was well defined. I knew where I wanted it to map a particular subgroup, but it was ambiguous where it took everything else. There were potentially MANY mappings that would map my subgroup to the identity in the range (and thus place the subgroup in the kernel if I could show it were a homomorphism), but for all we know, they could have mapped every other element to the same element. Even if I insisted criteria 2 and 3 of well defined held, and it respected the group operations (which would show it's a homomorphism), where the mapping took elements was still ambiguous. There may be several homomorphisms that accomplished such a thing.

There are many times in mathematics where we can define a portion of a mapping, and then claim there is a unique extension of that to a mapping over the entire domain, but identifying a subgroup that one would like to lie in the kernel of a homomorphism, and then seeking the homomorphism that does this is not valid.

Answer 2

I will assume that you are refering to group homomorphism. By definition, a group homomorphism from $(G, \star)$ to $(H, \cdot)$ is a map $\phi : G \to H$ such that for all $x,y \in G$, we have $$\phi(x \star y) = \phi(x) \cdot \phi(y).$$

Let's take your example. In this case, the group operations are the same, namely multiplication. We have $K = \{1,a\} < G$, where $a^2 = 1$, that is, $a$ has order 2 in $G$. There are multiple ways to show that a subgroup $H$ is normal. The easiest one, in my opinion, is to go with the definition and check that for all $g \in G$, $gHg^{-1} \subseteq H$. However, keep in mind that this really depends on the structure of your group $G$. Not all subgroups of order 2 are normal. For instance, $H = \langle (1 \ 2) \rangle < S_4$ is not normal since $(1 \ 2 \ 3 \ 4) ( 1 \ 2) (1 \ 4 \ 3 \ 2) = (2 \ 3) \notin H$.

Finding a direct group homomorphism with kernel precisely $K$ is usually not the best approach to show that $K$ is normal. Surely, this will give you what you want by the First Isomorphism Theorem for Groups, but it is rather tedious and very inefficient to do so for groups of larger order.

One usually works backward: define a map between groups, then show it is a well-defined homomorphism and finally realize its kernel as a normal subgroup.