First Isomorphism Theorem.

Question

After proving the theorem, the author proceeds to say:

As a consequence of the theorem:

1- We see that all homomorphic images of G can be determined using G.

I can't see how this follows. That assumes that we know the factor groups which contain the cosets of a normal group which is a kernel of a homomorphism which we are supposed to find in the first place!

2- We know that the number of homomorphic images of a cyclic group G of order n is the number of divisors of n, since there is exactly one subgroup of G (and therefore on factor group of G) for each divisor of n. (Be careful: The number of homomorphisms of a cyclic group of order n need not be the same as the number of divisors of n, since different homomorphisms can have the same image).

But again, that assumes that different homomorphisms can't have the same kernel with different images and thus a factor group can't be isomorphic to different homomorphic images! But that is not correct!

What I'm missing?

Answer 1

By the theorem, any two homomorphisms with the same kernel have isomorphic images.

For the second part, cyclic groups have unique normal subgroups of any order dividing the order of the group.

It's safe to say that the author's comments are to be taken to mean "up to isomorphism". Thus if two homomorphisms differ by an isomorphism, their images are isomorphic. So we regard their images as the same. They are still different homomorphisms though, if the isomorphism isn't the identity. Notice the author said "homomorphic images" are determined. Not that the homomorphisms are determined.

In the second case, he even cautions to distinguish the two (homomorphism and homomorphic image).

Answer 2

There are two approaches you can take to studying a group (or any algebraic structure for that matter):

1. Stare at it intently until you see something interesting about it.

2. See how it interacts with other objects and deduce interesting things about it that way.

The second method tends to be more fruitful. And one of the ways in which a group "interacts with other objects" involves homomorphisms between groups. So you want to study homomorphisms from your group $G$ to other groups.

One issue about this is that to understand homomorphisms from your group $G$ to other groups, you need lots of other groups you can map into. And no matter how many groups you know you can map into, you are never going to exhaust all possible groups. How do you know that there isn't some group $K$ you have not considered yet, and a homomorphism $f\colon G\to K$, such that $f(G)$ will get you some new insight about $G$?

The First Isomorphism Theorem tells you that you don't need to worry about "other" groups: up to an isomorphism (which just amounts to changing the names of elements), any homomorphic image of $G$ can be obtained by finding a normal subgroup $N$ of $G$ and looking at the canonical projection $\pi\colon G\to G/N$. In a sense, it tells you that as far as looking at homomorphic images of $G$ (which you can think of as "shadows" of $G$), you actually can accomplish (2) above by doing (1): Just stare at $G$ intently, figure out what its normal subgroups are, and you know all possible homomorphic images of $G$.

Thus, up to an isomorphism, figuring out a possible image of $G$ can be accomplished just by staring at $G$. No need to consider any group or object outside of $G$ itself.

But in the second part, he is distinguishing between "homomorphic images" and "homomorphisms. So for example, the map $\mathbb{Z}\to\mathbb{Z}$ given by the identity, and the map $\mathbb{Z}\hookrightarrow\mathbb{Q}$ given by the inclusion $n\mapsto\frac{n}{1}$ are different homomorphisms, but have isomorphic image. The "homomorphism" is different, but the "homomorphic image" is the same.

I'll comment that I think the authors are less than crystal clear, though; rather than a passing parenthetical comment, I would prefer them to be explicit about what distinction they are drawing there.