Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$\DeclareMathOperator{\Ker}{Ker}$ Hopefully this question won't be too vague, naive, or have incorrect information.

I am taking an algebra class (using Robert Ash's algebra text) and we have finished the first groups theory sections and I have been looking it over, especially the isomorphism theorems and kernels/normal subgroups. It just so happens that I started reading a paper a little before I started looking over the group which defines kernels of a function in a different way than it is defined in group theory.

First we will define the kernel in the paper. Let $f:S \to U$ be a function and $\Ker f$ will be the partition induced on $S$ by the equivelence relation $\sim$ defined by $a \sim b \iff f(a)=f(b)$. Unless I am mistaken, $\Ker f$ where $f$ is a group homomorphism $G \to H$ is the set $G/\ker f$ (I am pretty sure you can actually use any element of $\Ker f$ as the quotient). In my mind, $\Ker$ seems like a more natural concept to come up with and "more directly" explains the first isomorphism theorem, at least at first blush. By "more directly" I guess I mean you don't have to pass through defining normal subgroups, the typical group kernel, or even cosets.

Is there a reason why the concept of $\ker$ seems to be favored more than $\Ker$?

In an attempt to think of why I came up with a few potential things, but not sure how valid they are. One is that you don't need to develop cosets for this and cosets are more useful than just objects for quotient group, but I suspect from the concept of $\Ker$ one could better motivate looking at cosets (although I am not sure how). Another thought I have is that maybe notationally and conceptually it becomes more useful to think of $G/N$, especially when you are given a normal subgroup or need to remove/preserve some properties. Then again I feel like $\Ker$ could motivate those concepts, so I am not sure if these are great arguments against $\Ker$.

share|improve this question
    
This is how quotient maps are defined in, for example, semigroup theory. You have to ensure they the left and right multiplication preserves the structure of the classes, then you are done. –  user1729 Sep 13 '13 at 9:32
add comment

3 Answers 3

up vote 4 down vote accepted

Some authors do define quotient groups in this way (e.g. see Jacobson's Basic Algebra I, where he defines "congruence relations" as equivalence relations which respect the group operation). In fact, this "congruence relation" definition is the right generalization to quotients of other structures, e.g., quotient semigroups.

I think the main reason for doing it the usual way for groups, however, is that constructing quotient groups is much easier when it is done the usual way. Checking that a subgroup is normal is relatively straightforward (assuming you can do computations in the group). If defined with equivalence relations, you'd need to construct an entire equivalence relation each time to construct a quotient group (until you eventually realize the concept of normal subgroup).

share|improve this answer
    
I grew up with normal subgroups. It was very healthy for my mathematical mind to 'discover' the concept of 'congruence relation'. My look grew wider and 'quotients' became more natural and general. Looking backwards I would have preferred a start with congruences followed by the observation that by groups the equivalence class that contains the identity determines the whole equivalence (wich explains why normal subgroups are exploited there). After that observation it is off course fruitful to go on concentrating on that special equivalence class, calling it 'normal subgroup'. –  drhab Sep 13 '13 at 8:31
1  
I think that you’ve missed one important reason: at least in most U.S. programs the students in a first abstract algebra course have already had a semester of linear algebra and are likely to have been introduced to the kernel of a linear transformation (though they may only know it as the null space). It makes good pædagogical sense to take advantage of this. For that matter, those who have seen some modular arithmetic are generally accustomed to seeing the $0$ class as the ‘special’ one: $m\mid n$ is seen as ‘nicer’ than $n\equiv 2\pmod{m}$, say. –  Brian M. Scott Sep 13 '13 at 9:05
    
@BrianM.Scott . I think you meant to write $n\equiv0\text{ (mod m)}$. I was raised in the Netherlands. Maybe that plays a part. I understand the pedagogical reason, but purely thinking mathematically I stay with my view: first congruences. –  drhab Sep 13 '13 at 10:04
    
@drhab: No, I meant what I wrote: I was contrasting the ‘nice’ $0$ class the secondary, less ‘nice’ ones like remainder $2$. –  Brian M. Scott Sep 13 '13 at 10:06
1  
@BrianM.Scott .I understand now. My reading was uncarefully and I thought that 'nicer' involved only the notation. In fact I missed the essence of what you said there and was too eager to answer. –  drhab Sep 13 '13 at 10:13
show 1 more comment

I teach a first course of Algebra to Maths majors, and my first approach to quotient structures (groups, rings, etc.) is exactly the same. Only later in the course I show that all equivalence relations that are compatible with the operations are congruences modulo a normal subgroup, an ideal, etc.

It is certainly a matter of opinion, but the first examples I deal with are cyclic groups and simple extensions of fields, and this approach seems to me perfectly adequate, because the students can see immediately how these are related to congruences they are already familiar with.

Clearly the normal subgroups, ideals, etc. stuff will come handy later!

share|improve this answer
add comment

Your explanation seems a bit muddled to me.

The reason we have a Normal subgroup - rather than any subgroup - as the kernel of a homomorphism $f:G\to K$ is (one interpretation) that Normal subgroups are what you need to define an equivalence relation on $G$ which carries the same group structure on equivalence classes as the group structure of the image of $G$ in $K$.

Your "Ker" seems to have more to do with the image of the homomorphism, which is a different idea from the kernel (which, in equivalence terms, is the equivalence class containing the identity).

You should also note that the cosets are the equivalence classes (in this way of thinking) - so if you are using equivalence classes, you have implicitly defined cosets.

share|improve this answer
1  
Maybe I didn't express myself well, but I know that these are different ideas. Typically the kernel is introduced to create this a nice equivalence relation that respects the operation, but why not start with the equivalence relation? It seems more natural to define and see why the operation is preserved. It seems the typical kernel is used to define the equivalence relation almost exclusively even though just defining the equivalence relation seems more natural(to me). My question is wonder if there is a particular reason for that. –  Paul Plummer Sep 13 '13 at 5:44
    
I agree with you when you are saying that it is better to start with congruences (see my comment above, where I also try to explain why in groups there is a focus on normal subgroups instead of congruences). –  drhab Sep 13 '13 at 8:37
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.