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Quotient by the kernel, and you get the image: the four checks that prove it, the counting form, and the backwards use that names an unfamiliar quotient without any coset arithmetic.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state and assemble the proof of the first isomorphism theorem, explain why the well-definedness check comes first, use its counting form to relate the sizes of a group, a kernel and an image, and identify an unfamiliar quotient by producing a surjection whose kernel is the subgroup being collapsed.
A homomorphism comes with a kernel and an image, and a normal subgroup comes with a quotient. The kernel is normal, so it has a quotient too. Three objects — the quotient by the kernel, the image, and the group itself — and one theorem relating them, which is the last structural result of the unit.
The first isomorphism theorem: for a homomorphism $\phi : G \to H$, $G/\ker\phi \cong \operatorname{im}\phi$. The isomorphism is canonical — it is built from $\phi$ and involves no choices. A map is said to factor through a quotient when it can be written as the projection followed by another map, which is what the theorem says every homomorphism does.
> First isomorphism theorem. For any homomorphism $\phi : G \to H$, > $$G/\ker\phi \cong \operatorname{im}\phi,$$ > by the map sending $g\ker\phi$ to $\phi(g)$.
The proof is four checks on the obvious map, in the only order that works.
Check 1 is the one that is easy to skip and impossible to do without: until it is done there is no map to discuss.
The picture the theorem gives is that every homomorphism is a projection followed by an inclusion: collapse the kernel, then include the result into the target. The interesting part of any homomorphism is what it collapses, and the theorem says that is the only interesting part.
Counting form. $|G| = |\ker\phi| \cdot |\operatorname{im}\phi|$, since the quotient and the image have the same size.
How it is actually used: backwards. To identify an unfamiliar quotient $G/N$, find a surjective homomorphism $G \to K$ with kernel exactly $N$; then $G/N \cong K$. No cosets are multiplied and no representatives are chosen. Building the map is the whole of the work.
Another way: picture
Imagine photographing a three-dimensional object. The photograph loses the depth — that is the kernel — and what is left is a flat image. The theorem says the photograph is completely determined by which points of the object are indistinguishable in it: identify each group of indistinguishable points, and what you have built is exactly the photograph. Nothing about the camera beyond what it merges matters.
Another way: steps
To identify a quotient $G/N$: 1. Look for a familiar group $K$ that the elements of $G$ might map onto. 2. Write down a map $G \to K$ and check it respects the operations. 3. Check it is onto $K$. 4. Check its kernel is exactly $N$ — both inclusions. 5. Conclude $G/N \cong K$, and stop.
Four uses, each of which would be laborious by coset arithmetic and is immediate this way.
$\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}_n$. The map is reduction modulo $n$: onto, and its kernel is the multiples of $n$. So the definition of modular arithmetic as a quotient agrees with the definition as remainders.
$S_n/A_n \cong \mathbb{Z}_2$. The sign homomorphism is onto $\{1, -1\}$ with kernel $A_n$.
$GL_n(\mathbb{R})/SL_n(\mathbb{R}) \cong \mathbb{R}^{\times}$. The determinant is onto and has kernel the determinant-one matrices. Note what this says: as far as the determinant can see, two invertible matrices differ only by their determinants.
$\mathbb{R}/\mathbb{Z} \cong$ the circle. The map $t \mapsto e^{2\pi i t}$ is onto the unit circle and its kernel is exactly the integers. So the reals with the integers forgotten is a circle — which is the precise version of the intuition that a number line wrapped round on itself at every integer becomes a loop.
There are two further isomorphism theorems, and both are proved by applying the first one to a cleverly chosen map. They belong to Abstract algebra II, but the shape is worth knowing: the second relates $HN/N$ to $H/(H \cap N)$, and the third says that quotienting twice is quotienting once by the bigger subgroup, $(G/N)/(M/N) \cong G/M$. Neither introduces a new idea; both are this theorem, pointed somewhere else.
Matching the quotient with the target. $G/\ker\phi \cong \operatorname{im}\phi$. Only when $\phi$ is onto is the image the target, and most homomorphisms are not onto.
Skipping the well-definedness check. It is the first check and the only one that can fail for a reason other than carelessness. A map defined on cosets by a representative is not a map until it has been done.
Checking the kernel is normal. No need: a kernel always is. Time spent on it is wasted.
Using the theorem forwards when backwards is wanted. If the question is what is $G/N$, the move is to build a surjection with kernel $N$, not to start listing cosets.
Thinking two groups of the same size must be isomorphic. The counting form is weaker than the theorem. The theorem gives an isomorphism; equal sizes alone never do.
What is $\mathbb{Z}_{12}/\{0, 4, 8\}$? Rather than multiplying cosets, look for a surjection out of $\mathbb{Z}_{12}$ whose kernel is $\{0, 4, 8\}$.
Backwards, from the start.
Try $\phi(x) = 3x$ into $\mathbb{Z}_{12}$. Its kernel is $\{x : 3x \equiv 0\} = \{0, 4, 8\}$ — exactly right — and its image is $\{0, 3, 6, 9\}$, cyclic of order $4$.
One map, both conditions.
So the quotient is isomorphic to $\mathbb{Z}_4$. The same answer as multiplying out the table, with none of the arithmetic.
Identified.
$\phi : \mathbb{Z}_6 \to \mathbb{Z}_{12}$ with $\phi(x) = 2x$. Kernel: $2x \equiv 0 \pmod{12}$ with $x < 6$ gives only $x = 0$.
Trivial kernel.
So $\mathbb{Z}_6/\{0\} \cong \operatorname{im}\phi$, and the image is $\{0, 2, 4, 6, 8, 10\}$, of order $6$.
The theorem, applied.
The quotient is $\mathbb{Z}_6$ itself, matched with the image and not with $\mathbb{Z}_{12}$, which has twice as many elements.
Image, not target.
Look for a surjection out of the reals whose kernel is exactly the integers.
Backwards.
The map $t \mapsto e^{2\pi i t}$ sends sums to products, is onto the unit circle, and gives $1$ exactly when $t$ is an integer.
A homomorphism, onto, with the right kernel.
So the quotient is the unit circle under multiplication. The real line with the integers forgotten is a circle of circumference one.
Build the proof that the quotient of a group by the kernel of a homomorphism is isomorphic to its image.
This task has no paper form; do it on a device.
Each row gives a surjective homomorphism. Match it to the quotient the theorem identifies.
| a group with two elements | the integers modulo $n$ | the non-zero reals under multiplication | the unit circle | the trivial group | |
|---|---|---|---|---|---|
| Reduction modulo $n$, from the integers | |||||
| The sign of a permutation | |||||
| The determinant, on the invertible matrices | |||||
| The map sending $t$ to $e^{2\pi i t}$, from the reals |
A homomorphism has a source group with $15$ elements and a kernel with $3$. How many elements does its image have?
Answer:
Let $\phi$ send $x$ to $2x$ in the integers modulo $6$. Fill in the three sizes the first isomorphism theorem relates.
| How many elements | |
|---|---|
| The kernel | |
| The quotient by the kernel | |
| The image |
Select every statement about the first isomorphism theorem that is true.
This task has no paper form; do it on a device.
Given the sign of a permutation, with kernel the even permutations, what is the quotient by that kernel isomorphic to?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A homomorphism has a source group with $12$ elements and a kernel with $2$. How many elements does its image have?
Answer:
You can prove the first isomorphism theorem and use it backwards to name a quotient. Say in your own words why the quotient matches the image rather than the target. Next: rings, where a second operation arrives and the same story is told again.
9. Your turn: what is the quotient of the reals under addition by the subgroup of integers?, step 3