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Maps that respect the operation: what the single defining equation forces, what an isomorphism carries across, and how to show two groups are not the same group.
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 check whether a map is a homomorphism, derive from the defining equation that the identity and inverses are preserved and that the image is a subgroup, say what an isomorphism carries across and what it does not, show two groups are not isomorphic by comparing element orders, and count the homomorphisms between two cyclic groups.
Every group so far has been studied on its own. Twice already two of them turned out to be the same group in different clothing — the fourth roots of unity and the integers modulo four, the residues modulo three and a subgroup of the units modulo seven. Saying exactly what the same group means, and what a weaker comparison looks like, is the business of this lesson.
A homomorphism $\phi : G \to H$ satisfies $\phi(ab) = \phi(a)\phi(b)$ for all $a, b$. It is an isomorphism when it is also a bijection, and then $G \cong H$. A homomorphism from a group to itself is an endomorphism, and a bijective one an automorphism. The image $\operatorname{im}\phi$ is the set of values $\phi$ takes, a subgroup of $H$.
A homomorphism is a map $\phi : G \to H$ with
$$\phi(ab) = \phi(a)\phi(b) \quad \text{for all } a, b \in G,$$
where the product on the left uses $G$'s operation and the one on the right uses $H$'s. That is the entire definition. Four consequences come out of it immediately, each by applying the equation to a chosen product.
An isomorphism is a bijective homomorphism. Then $G$ and $H$ are the same group with different labels: every structural statement true of one is true of the other. Being abelian, being cyclic, the number of elements of each order, the number of subgroups — all transfer. What does not transfer is anything about the labels, the notation or what the elements are made of.
That gives the standard way of showing two groups are not isomorphic: find a structural property one has and the other does not. Counting elements of each order usually settles it in a line.
Out of a cyclic group, there is not much choice. Since every element is a power of the generator, $\phi$ is determined by $\phi(g)$, and the only constraint is that the order of $\phi(g)$ divides the order of $g$. Counting homomorphisms $\mathbb{Z}_n \to \mathbb{Z}_m$ is therefore counting suitable images: there are $\gcd(n, m)$.
An algebraic fact is about the operation, not about what the elements happen to be. The same group turns up as rotations of a square, as residues under addition and as matrices, and a theorem proved once from the axioms holds in all of them at once.
Another way: picture
Think of a homomorphism as a translation between two languages that gets the grammar right. It need not be word for word: several words of one language may translate to the same word of the other, and some words of the target may never be used. What it must never do is scramble the grammar — if you combine two things and then translate, you must get the same as translating and then combining. An isomorphism is a translation that is also a perfect dictionary, one word each way.
Another way: steps
To check a map is a homomorphism: 1. Identify the operation on each side; they are usually different. 2. Compute $\phi(ab)$ and $\phi(a)\phi(b)$ for general $a, b$ and compare. 3. As a quick disqualifier, check $\phi(e) = e'$; a map that moves the identity fails. 4. For an isomorphism, check injectivity and surjectivity as well. 5. To show two groups are not isomorphic, compare element orders.
A handful of maps do most of the work in this subject, and each turns one operation into a different one.
| Map | From | To | What it turns into what |
|---|---|---|---|
| $k \mapsto k \bmod n$ | $\mathbb{Z}$ | $\mathbb{Z}_n$ | addition into addition |
| $\det$ | $GL_n(\mathbb{R})$ | $\mathbb{R}^{\times}$ | matrix product into product |
| $\operatorname{sgn}$ | $S_n$ | $\{1, -1\}$ | composition into product |
| $t \mapsto e^{t}$ | $(\mathbb{R}, +)$ | $(\mathbb{R}^{>0}, \times)$ | addition into multiplication |
| $g \mapsto \lambda_g$ | any $G$ | permutations of $G$ | the operation into composition |
The last is Cayley's theorem, which was a homomorphism all along — an injective one, which is what embedding means.
The exponential row deserves a second look, because it is the historical origin of the word. Logarithms were invented to turn multiplication into addition, and the statement $\log(xy) = \log x + \log y$ is exactly the homomorphism condition. A slide rule is a physical realisation of an isomorphism between two groups.
Two maps that are not homomorphisms are worth as much: $x \mapsto x^{2}$ on a non-abelian group, since $(ab)^{2} \ne a^{2}b^{2}$ in general; and $x \mapsto x + 1$ on $\mathbb{Z}$, which moves the identity. The first is a homomorphism exactly when the group is abelian, which is a small theorem in itself.
Using one operation on both sides. $\phi(a + b) = \phi(a)\phi(b)$ is the right shape when the source is additive and the target multiplicative. Writing the same symbol twice hides which group each product lives in.
Assuming a homomorphism is onto. The image is a subgroup of the target and usually a proper one. Surjectivity is an extra hypothesis whenever it is needed.
Assuming an injective map is an isomorphism. It is an isomorphism onto its image, which is smaller than the target unless it is also surjective.
Thinking two groups of the same size are isomorphic. There are two of order $4$ and five of order $8$. Size is necessary and nowhere near sufficient.
Forgetting that the order of an image can drop. $\phi(g)$ has order dividing the order of $g$, not equal to it. Everything collapses to the identity under the trivial homomorphism, which is a homomorphism.
$\det : GL_2(\mathbb{R}) \to \mathbb{R}^{\times}$. The condition to check is $\det(AB) = \det A \det B$, which is a standard fact of linear algebra.
The homomorphism condition, already known.
It is onto: the matrix with diagonal $r, 1$ has determinant $r$ for any non-zero $r$.
Surjective.
It is far from injective: everything of determinant $1$ goes to $1$. What it collapses is exactly the special linear group, and the next lesson calls that the kernel.
Not injective, and the failure is the interesting part.
Suppose $\phi : \mathbb{Z}_4 \to U(8)$ were an isomorphism. In $\mathbb{Z}_4$ the element $1$ has order $4$.
Pick an element and note its order.
Then $\phi(1)$ would have order $4$ in $U(8)$. But every element of $U(8)$ squares to $1$, so no element there has order $4$.
Order is preserved by an isomorphism.
Contradiction, so the two are not isomorphic — although both are abelian of order $4$. Comparing element orders settled it in three lines.
The standard method.
Check the condition: $\phi(x + y) = 2(x + y) = 2x + 2y = \phi(x) + \phi(y)$, all modulo $6$.
The defining equation.
So yes. It is not injective, though: $\phi(0) = \phi(3) = 0$.
A homomorphism, not an isomorphism.
Its image is $\{0, 2, 4\}$, a subgroup of order $3$, and the two elements sent to $0$ form a subgroup of order $2$. Three times two is six, which is not a coincidence — it is the next two lessons.
Let $\phi$ send $x$ to $5x$ in the integers modulo $10$. Write down $\phi(1)$, $\phi(2)$ and $\phi(3)$.
| Its image | |
|---|---|
| 1 | |
| 2 | |
| 3 |
For a homomorphism $\phi$ from $G$ to $H$, match each expression to what it must equal or be.
| the inverse of $\phi(a)$ | the identity of $H$ | a divisor of the order of $g$ | a subgroup of $H$ | the whole of $H$ | |
|---|---|---|---|---|---|
| $\phi(e)$ | |||||
| $\phi(a^{-1})$ | |||||
| The order of $\phi(g)$ | |||||
| The image of $\phi$ |
Two groups are isomorphic. Select every statement that must then be true.
This task has no paper form; do it on a device.
Is squaring, on the integers under addition a homomorphism?
How many homomorphisms are there from the integers modulo $17$ to the integers modulo $10$, both under addition?
Answer:
Build the proof that a homomorphism sends the identity to the identity and inverses to inverses.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Let $\phi$ send $x$ to $5x$ in the integers modulo $15$. Write down $\phi(1)$, $\phi(2)$ and $\phi(3)$.
| Its image | |
|---|---|
| 1 | |
| 2 | |
| 3 |
You can test a map for being a homomorphism and say what follows automatically once it is one. Say in your own words why comparing element orders shows two groups are not isomorphic. Next: the kernel, which measures exactly how much a homomorphism collapses.
9. Your turn: is $\phi(x) = 2x$ a homomorphism from the integers modulo $6$ to itself?, step 3