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What a homomorphism throws away and what it reaches: the kernel is always normal, injectivity is exactly a trivial kernel, and the two sizes multiply to the size of the source.
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 compute the kernel and image of a homomorphism, prove that a kernel is always a normal subgroup, use the size relation between source, kernel and image, decide injectivity by looking at the kernel alone, and explain why normal subgroups and kernels are the same class of objects.
A homomorphism need not be injective, and the determinant was the case in point: everything of determinant one went to the same place. That collapsing is not a defect to be apologised for — it is the useful part, because it is how a complicated group gets replaced by a simpler one. This lesson measures it.
The kernel of $\phi : G \to H$ is $\ker\phi = \{g \in G : \phi(g) = e'\}$, the elements sent to the identity of $H$. The image is $\operatorname{im}\phi = \{\phi(g) : g \in G\}$. A kernel is trivial when it is $\{e\}$. The set of elements sharing one image is sometimes called a fibre; each is a coset of the kernel.
The kernel is what the map sends to the identity, and the image is what it reaches. Three facts.
The kernel is a normal subgroup of $G$. A subgroup by the one-step test, since $\phi(ab^{-1}) = \phi(a)\phi(b)^{-1} = e'$; and normal because
$$\phi(gng^{-1}) = \phi(g)\phi(n)\phi(g)^{-1} = \phi(g)e'\phi(g)^{-1} = e'.$$
One line, and it is the reason normal subgroups were worth defining. The converse holds too — every normal subgroup is a kernel — so the two notions pick out exactly the same subgroups.
Injective exactly when the kernel is trivial. $\phi(x) = \phi(y)$ says $\phi(x^{-1}y) = e'$, so the elements with the same image as $x$ are precisely the coset $x\ker\phi$. If the kernel is $\{e\}$ that coset is $\{x\}$ and the map is injective; otherwise every image is hit $|\ker\phi|$ times.
The sizes multiply. For a finite $G$,
$$|G| = |\ker\phi| \cdot |\operatorname{im}\phi|,$$
because $G$ is partitioned into $|\operatorname{im}\phi|$ cosets of the kernel, each of size $|\ker\phi|$. It is Lagrange, applied to the kernel, with the index identified as the size of the image — and it is the first isomorphism theorem in counting form, two lessons early.
So a homomorphism has exactly two measurements: how much it throws away, and how much it reaches. The larger the first, the smaller the second.
Another way: picture
Picture the source group as a stack of trays, all the same size, and the homomorphism as a hand pressing the stack flat onto the target. Each tray is a coset of the kernel and lands on a single point of the image; the kernel itself is the tray containing the identity. A tall stack of small trays means a big kernel and a small image; a single layer means an injective map. Multiplying the height of the stack by the area it covers gives the whole group, every time.
Another way: steps
Given a homomorphism: 1. Kernel: solve $\phi(x) = e'$ in the source. 2. Image: find what values $\phi$ actually takes; it is a subgroup of the target. 3. Check $|\ker\phi| \cdot |\operatorname{im}\phi| = |G|$; a mismatch means one of them is wrong. 4. Injective exactly when the kernel is trivial; onto exactly when the image is the whole target.
The kernel of a homomorphism is normal. The striking half of the story is the converse: given any normal subgroup $N \trianglelefteq G$, there is a homomorphism whose kernel is exactly $N$.
The map is the one onto the quotient, $\pi : G \to G/N$ with $\pi(g) = gN$, and it is built in the next lesson. Its kernel is the set of $g$ with $gN = N$, which is precisely $N$. So normal subgroup and kernel are two descriptions of one class of objects, and which description is more useful depends on the question.
This is worth pausing on, because it explains the shape of unit 4. Normality looked like a technical condition on cosets; it turns out to be the exact answer to the question which subgroups can be collapsed? Collapsing $N$ means treating two elements as equal when they differ by an element of $N$, and that is consistent with the operation precisely when $N$ is normal.
The measurement is where the usefulness comes from. Consider $\det : GL_2(\mathbb{R}) \to \mathbb{R}^{\times}$: the kernel is $SL_2(\mathbb{R})$, a large and complicated group, and the image is the whole of $\mathbb{R}^{\times}$, which is simple to understand. The homomorphism has separated a complicated group into a complicated part that it ignores and a simple part that it sees. Doing that deliberately — choosing what to ignore — is most of what algebra is for.
One more consequence, used later: a homomorphism from a simple group is either injective or trivial, because its kernel is normal and a simple group has only two normal subgroups. So simple groups cannot be collapsed at all, which is why they are called simple and why classifying them was worth forty years.
Putting the kernel in the target. The kernel is a subset of the source: the things sent to the identity. The image is the subset of the target.
Thinking a trivial kernel means a trivial map. It means the opposite — nothing is collapsed, so the map is injective. The map with the largest kernel is the one sending everything to the identity.
Expecting the image to be the whole target. It need not be; it is a subgroup. Saying a map is onto its image is a tautology, not a result.
Forgetting that the fibres are cosets. Every image value is hit exactly $|\ker\phi|$ times, never some other number, which is why the sizes multiply.
Assuming a normal subgroup needs a homomorphism to be found first. It does not, but one always exists — that is the theorem, and it is what makes the two notions interchangeable.
$\operatorname{sgn} : S_4 \to \{1, -1\}$, a group of order $24$ mapping onto a group of order $2$.
Source and target.
The kernel is the even permutations, $A_4$, of order $12$; the image is all of $\{1, -1\}$, of order $2$.
Kernel and image.
Check: $12 \times 2 = 24$. And the kernel is normal, which reproves that the alternating group is normal without mentioning index two.
The counting relation, and normality for free.
$\phi : \mathbb{Z}_3 \to \mathbb{Z}_6$ with $\phi(x) = 2x$. Check: $\phi(x + y) = 2(x+y) = \phi(x) + \phi(y)$.
A homomorphism between different groups.
Kernel: $2x \equiv 0 \pmod 6$ with $x$ in $\{0, 1, 2\}$ gives only $x = 0$. So the kernel is trivial and $\phi$ is injective.
Nothing collapses.
Image: $\{0, 2, 4\}$, of order $3$. And $1 \times 3 = 3 = |\mathbb{Z}_3|$, as required. An injective map into a bigger group — an embedding, not an isomorphism.
Injective, not onto.
Kernel: $3x \equiv 0 \pmod 9$ means $3$ divides $x$, so the kernel is $\{0, 3, 6\}$, of size $3$.
Solve for the identity.
Image: the multiples of $3$ modulo $9$, which is $\{0, 3, 6\}$ again, also of size $3$.
The values taken.
Check $3 \times 3 = 9$. Here the kernel and the image happen to be the same subgroup, which is a coincidence of this map rather than a general fact.
Let $\phi$ send $x$ to $3x$ in the integers modulo $9$. Fill in the size of its kernel, the size of its image, and the two multiplied together.
| How many | |
|---|---|
| Elements of the kernel | |
| Elements of the image | |
| The two multiplied together |
A homomorphism has a source group with $18$ elements and an image with $6$. How many elements does its kernel have?
Answer:
Build the proof that the kernel of a homomorphism is a normal subgroup.
This task has no paper form; do it on a device.
Match each homomorphism to its kernel.
| the even permutations | the matrices of determinant one | the multiples of $n$ | the whole group | the identity alone | |
|---|---|---|---|---|---|
| The determinant, on the invertible matrices | |||||
| The sign of a permutation | |||||
| Reduction modulo $n$, on the integers | |||||
| The map sending every element to the identity |
Select every statement about the kernel of a homomorphism that is true.
This task has no paper form; do it on a device.
Let $\phi$ send $x$ to $2x$ in the integers modulo $8$. How many elements does its kernel have?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Let $\phi$ send $x$ to $2x$ in the integers modulo $6$. Fill in the size of its kernel, the size of its image, and the two multiplied together.
| How many | |
|---|---|
| Elements of the kernel | |
| Elements of the image | |
| The two multiplied together |
You can find a kernel and an image and use the fact that their sizes multiply to the size of the source. Say in your own words why a kernel is always normal, and why a trivial kernel means an injective map. Next: quotient groups, which are what collapsing a kernel actually produces.
9. Your turn: find the kernel and image of $\phi(x) = 3x$ on the integers modulo $9$., step 3