# Can a homomorphism be 1 to n?

1. Oct 26, 2015

### davidbenari

My professor says a homomorphism is an association between groups such that multiplication tables are preserved (he says nothing else, this is his definition).

Imagine two groups $G$ and $H$ with $|G|=2$ and $|H|=8$. In his way of seeing things, there could be a homomorphism $\mu$ that associates elements of G to collections of elements of $H$. The collections must satisfy the same multiplication table.

But I don't agree with such a definition. The definition math textbooks give is that of a mapping $\mu: G \to H$ which follows the multiplicative identity $\mu(gk)=\mu(g) \mu(k)$. His definition is not a mapping at all, since mapping can't be 1 to n, and even if you relax that condition, a homomorphism maps to an element not to a collection.

So where is he getting these notions from? Anyone agrees with him? Where can I read a definition which is OK with his idea? If he's wrong, could you tell me why do you think so?

edit: My professor has been wrong before, so don't think I'm taking false words out of his mouth. That said, if you could give me a sketch on important points on how he's wrong it could be helpful. I've corrected him before and its getting on my nerves that I'm being taught material that's bogus. The other thing though, is that this professor is an important researcher, so he's definitely not dumb; so maybe something complicated is going on.

Last edited: Oct 26, 2015
2. Oct 26, 2015

### PeroK

I'd suggest you're getting confused between the definition of a homomorphism and the first group isomorphism theorem. See this link:

https://en.wikipedia.org/wiki/Isomorphism_theorem

3. Oct 26, 2015

### davidbenari

Hmm. I see don't see the relevance of that theorem here. What motivated your comment?

Thanks.

4. Oct 26, 2015

### PeroK

If you have a homomorphism $f: G \rightarrow H$ then the image of $f$ is isomorphic to the quotient group $G/ker(f)$

A quotient group is a group of equivalence classes of sets of elements. So, effectivey you have a mapping from collections of elements of G into H.

So, I inferred that this was the source of your confusion.

5. Oct 26, 2015

### davidbenari

Well, that isomorphism would be a different mapping altogether than my homomorphism, wouldn't it? I'm gonna think of a good example of what my professor regards as a homomorphism and post it tonight; because I think an example works best here. Thanks for the help meanwhile.

6. Oct 26, 2015

### davidbenari

Okay so consider the homomorphism $f:C12 \to C3$.

Consider C12 as composed of small 30º rotations. C3 as composed of small 120º rotations.

Map al 90º rotations to the identity element in C3. All rotations of the type (60+n90)º map to 240º rotations in C3. All rotations of the type (30+n90)º map to 120º rotations in C3.

As such the mapping direction is $C12 \to C3$ and seems perfectly sound. BUT...

My professor considers $C3\to C12$ a homomorphism as well! Mapping for example, the identity in C3 to the three 90º rotations in C12. This makes no sense! This is not a mapping at all! Its not even mapping you to an element of a group. I can post the image of a color map so that this homomorphism is more clear if you'd like to.

Thanks.

7. Oct 27, 2015

### willem2

Your prof is mapping the elements of C3 to cosets of C12. If C3 is generated by g and C12 by h.
I -> {I, h^3, h^6, h^9}
g -> {h, h^4, h^7, h^10}
g^2 -> {h^2, h^5, h^8, h^11}

8. Oct 27, 2015

### davidbenari

But this satisfy the homomorphism definition does it? I mean, it has to map an element to an element; not an element to a set. Plus, how can we say how coset multiplication behaves ?

9. Oct 27, 2015

### willem2

- a normal subgroup?
- a factor group?

10. Oct 27, 2015

### davidbenari

Yes. But his definition isn't restricted to these kind of groups though. Even if we can find a way for this "definition" to be internally consistent, my question asks if this is compatible with the definition of a group homomorphism. I think the fact that homomorphisms map elements to elements is crucial to its definition and is being violated here.

Thanks.

Last edited: Oct 27, 2015
11. Oct 28, 2015

### willem2

A factor group is a group whose elements are cosets. And you can do this if a subgroup H of G is normal (the coset gH is always equal to Hg} because in that case the coset of the result of the multiplication of two elements of G only depends on the cosets of the elements of G.
C12 is abelian, so all its subgroups are normal.

12. Oct 28, 2015

### davidbenari

But this wouldn't be the homomorphism $\mu : C3 \to C12$ namely, it would be the $\mu' : C3 \to U$ defined by

I -> {I, h^3, h^6, h^9}=A
g -> {h, h^4, h^7, h^10} =B
g^2 -> {h^2, h^5, h^8, h^11}=C

with each element of U being the sets above, and with some strange binary operation which makes sense superficially but isn't quite easy to define; especially since the message it tries to convey is not that of multiplying sets (instead it is saying that e.g. given any element in A and multiplied by an element in B it will map you to some element in C and likewise for all other combinations)...

Do you agree with me that this isn't the map $C3 \to C12$? A map $C3 \to C12$ would map e.g. I->h^3 g^2 -> h^2 (I'm not intending to make this shallow example a homomorphic map in any way).

Probably I'm not understanding you completely, but how do you object the statement that homomorphisms map elements to elements specifically? And cannot map elements to sets?

Thanks.

13. Oct 29, 2015

### willem2

It seems to me this mapping is probably what your professor meant, even if it isn't a homomorphism from C3 to C12, but to the factor group C12/C4.
The idea of a factor group and a mapping to cosets should become important soon.
A homomorhpism from C3 to C12 could only have a subgroup of C12 as range, since a one to many relation doesn't make sense.

14. Oct 29, 2015

### HallsofIvy

To answer the question in the title, a homomorphism must first be a function so not "1 to n".
However, a homomorphism, unlike an isomorphism, does not have to be "onto". It can map one group to a subgroup of another group.

Since 2 is a prime divisor of 8, any group of order 8 has a subgroup of order 2 so an element of order 2. An obvious homomorphism from a group of order 2 to a group of order 8 is to map identity to identity and to map the non-identity to an element of order 2. Similarly, since 3 is a prime divisor of 12, any group of order 12 has a subgroup of order 3. Map that each member of C3 to that subgroup.

15. Oct 29, 2015

### davidbenari

This is a good idea, but I don't think its the same map as the map $\mu C3 \to U$ explicitly given by

I -> {g^0,g^3,g^6,g^9}= $\mathbf{I}$
h^1->{g^1,g^4,g^7,g^10}=B
h^2->{g^2,g^5,g^8,g^11}=C
and where $U$ is a group given by disjoint sets of $C12$ , where the binary operation in $U$ is supposedly representing that

"given any element in one of the sets above, multiplied by another element of another set, the result would map you to another element in a set that would be the one which you would have been mapped to if you had done the operations in C3 all along" (Sorry for that messy sentence).

In other words if you have $g^i g^j$ then you would be mapped to the set $\mu^{-1}(g^i)\mu^{-1}(g^j)$ where $\mu^{-1}$ simply recognizes the set to which $g^i$ belongs to and maps it back to C3. This is the professor's example, so please stick with it.

ASIDE---

I went to talk with the professor today and it seems he wasn't too familiar with the definition that talks about $\mu(gh)=\mu(g)\mu(h)$; this baffles me since he's the guy supposed to be teaching me and this isn't an obscure notion at all.... He ended up admitting I was right though, but I'm not sure I managed to stick my message well inside his brain.

I suppose there's a way to formalize his idea of the map $\mu : C3 \to U$ as a homomorphism. Are there popular examples of this kind of map? It seems like a useful kind of map. One that could reduce the complexity of your calculations if you go to a smaller group.

It is worth mentioning again that this professor is a renowned researcher of optics and I highly doubt he's dumb; there has to be some kind of motivation behind his idea. But I guess all this might be due to the fact that he's a physicist and not a mathematician, and probably never saw a course on abstract algebra...

Thanks, and I hope I'm not frustrating you guys with unnecessary semantics, but I would really like to formalize these ideas.