Back to the on-screen lesson ·
The cosets of a normal subgroup as a group in their own right: multiplying through representatives, why normality is exactly what makes that well defined, and what a quotient forgets.
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 build the operation table of a quotient group, compute the size of a quotient as the index, multiply cosets through representatives, prove that the operation is well defined exactly when the subgroup is normal, identify small quotients with familiar groups, and say what information a quotient deliberately discards.
The cosets of a subgroup have been sets so far — tiles covering the group. This lesson promotes them: the cosets of a normal subgroup are themselves the elements of a new group. The construction has been in use since school, because the integers modulo $n$ are exactly this, with $N$ the multiples of $n$.
For $N \trianglelefteq G$, the quotient group $G/N$ has the cosets of $N$ as its elements and $(aN)(bN) = abN$ as its operation. A definition made through a choice is well defined when the result does not depend on the choice. The canonical projection $\pi : G \to G/N$ sends $g$ to $gN$; it is a surjective homomorphism with kernel $N$.
Let $N \trianglelefteq G$. Define a product on the set of cosets by
$$(aN)(bN) = abN.$$
The rule names the answer through representatives, so the first question is whether it is a definition at all. If $a' = an_1$ and $b' = bn_2$ with $n_1, n_2 \in N$, then
$$a'b' = an_1bn_2 = ab(b^{-1}n_1b)n_2,$$
and $b^{-1}n_1b \in N$ because $N$ is normal. So $a'b' \in abN$ and the answer is unchanged. Without normality this step fails and the product really does depend on the choice.
With that settled, $G/N$ is a group: associativity comes from $G$, the identity is $N$ itself, and the inverse of $aN$ is $a^{-1}N$. Its size is the index,
$$|G/N| = \frac{|G|}{|N|}.$$
The projection. $\pi : G \to G/N$, $\pi(g) = gN$, is a surjective homomorphism with kernel $N$. So every normal subgroup is a kernel, which completes the correspondence begun last lesson.
What a quotient is for. It is a controlled loss of information: $a$ and $b$ become equal in $G/N$ exactly when they differ by an element of $N$. Questions that do not care about $N$ can then be asked in a smaller group. The integers modulo $12$ are the integers with the multiples of $12$ forgotten, and that is why clock arithmetic works.
A quotient forgets on purpose. Everything in one coset of $N$ becomes a single element of $G/N$, so a question about $G$ that does not depend on $N$ can be asked in a smaller group — and the price is that questions which do depend on $N$ can no longer be asked at all.
Another way: picture
Think of the cosets as boxes and the elements as objects sorted into them. To multiply two boxes, reach into each, take one object from each, multiply them, and see which box the result lands in. The construction is only sensible if the box you land in does not depend on which objects you happened to grab — and normality is exactly the promise that it does not. Quotienting is then relabelling every object by its box and throwing the objects away.
Another way: steps
To work in $G/N$: 1. Check $N$ is normal; without it there is no quotient. 2. The elements are the cosets; there are $|G|/|N|$ of them. 3. To multiply, multiply representatives and name the coset the answer lies in. 4. The identity is $N$, and the inverse of $aN$ is $a^{-1}N$. 5. To identify the quotient, count its elements and compare with the groups of that size.
The quotient is not a subgroup of $G$ and is not a subset of it: its elements are sets. What it is, is $G$ seen at a coarser resolution.
A useful pair of examples sits inside the integers. $\mathbb{Z}/n\mathbb{Z}$ forgets everything about an integer except its remainder, which is why it can answer is this number odd and cannot answer how large is it. The forgetting is the feature: proving that $x^{2} + y^{2} = 3z^{2}$ has no non-zero integer solution is hard directly and easy modulo $4$, because passing to the quotient throws away exactly the information that was in the way.
A second pattern worth naming: a quotient can be simpler than the group. $S_4$ is non-abelian of order $24$; quotienting by the normal subgroup $\{e, (1\,2)(3\,4), (1\,3)(2\,4), (1\,4)(2\,3)\}$ of order $4$ gives a group of order $6$ isomorphic to $S_3$, and quotienting $S_n$ by $A_n$ gives a group of order $2$. Each step trades detail for tractability.
And it can be less simple than either part: knowing $N$ and knowing $G/N$ does not determine $G$. Both $\mathbb{Z}_4$ and the Klein four-group have a normal subgroup of order $2$ with quotient of order $2$, and they are not isomorphic. The question of how many groups have a given $N$ and $G/N$ is the extension problem, and it is genuinely hard.
Two properties do pass down. A quotient of an abelian group is abelian, and a quotient of a cyclic group is cyclic, because the image of a generator generates the image. Neither converse holds: a non-abelian group can have abelian quotients, and the sign homomorphism shows it.
Treating an element of $G/N$ as an element of $G$. It is a coset — a set. Writing $aN$ rather than $a$ is a discipline worth keeping until the habit is fixed.
Thinking $G/N$ sits inside $G$. It does not. It is a different group built from $G$, and its elements are not elements of $G$ at all.
Quotienting by a subgroup that is not normal. There is no group there. The set of cosets exists and can be counted, but no operation on it is well defined.
Believing $|G/N| = |G| - |N|$. It is a division: $|G|/|N|$, the index.
Expecting $G$ to be recoverable from $N$ and $G/N$. It is not. Two different groups can have the same normal subgroup and the same quotient.
$N = \{0, 4, 8\}$ inside $\mathbb{Z}_{12}$, which is abelian so $N$ is normal. The index is $12/3 = 4$.
Four cosets.
Cosets: $N$, $1 + N$, $2 + N$, $3 + N$. Adding: $(1 + N) + (2 + N) = 3 + N$, and $(1 + N) + (3 + N) = 4 + N = N$.
Add representatives and reduce.
So $1 + N$ has order $4$ and the quotient is cyclic: $\mathbb{Z}_{12}/N \cong \mathbb{Z}_4$.
Identified by its size and by being cyclic.
Inside $S_4$, the four permutations $e$, $(1\,2)(3\,4)$, $(1\,3)(2\,4)$, $(1\,4)(2\,3)$ form a normal subgroup $V$ of order $4$ — normal because it is a union of conjugacy classes.
A normal subgroup of order four.
The quotient has $24/4 = 6$ elements. There are two groups of order $6$, and this one is not abelian: the images of $(1\,2)$ and $(1\,2\,3)$ do not commute.
Six cosets.
So $S_4/V \cong S_3$. A group of order $24$ has been replaced by one of order $6$ with most of its structure intact — which is the step that makes the quartic solvable by radicals.
A quotient with real content.
The group is abelian so the subgroup is normal, and the index is $10/2 = 5$.
Count the cosets.
So the quotient has five elements, and a quotient of a cyclic group is cyclic.
Cyclic, of order five.
There is only one cyclic group of order $5$, so the quotient is the integers modulo $5$. Note the cosets are $\{0,5\}, \{1,6\}, \{2,7\}, \{3,8\}, \{4,9\}$ — the residues modulo $5$, as expected.
Inside the integers modulo $12$, let $N$ be the subgroup $\{0, 4, 8\}$. Its four cosets are named by $0, 1, 2, 3$. Complete the addition table of the quotient; the row and column of $0$ are given.
| 0 | 1 | 2 | 3 | |
|---|---|---|---|---|
| 0 | 0 | 1 | 2 | 3 |
| 1 | 1 | |||
| 2 | 2 | |||
| 3 | 3 |
A normal subgroup $N$ with $4$ elements sits inside a group with $20$. How many elements does the quotient have?
Answer:
Put the steps of computing the product of two cosets in a quotient group into order.
Number the steps in order (write the number in the box):
Match each quotient to the familiar group it turns out to be.
| the integers modulo $n$ | the integers modulo $4$ | a group with two elements | the non-zero reals under multiplication | the trivial group | |
|---|---|---|---|---|---|
| The integers, quotiented by the multiples of $n$ | |||||
| The integers modulo $12$, quotiented by the multiples of $4$ | |||||
| All the permutations, quotiented by the even ones | |||||
| The invertible matrices, quotiented by those of determinant one |
Select every statement about a quotient group that is true.
This task has no paper form; do it on a device.
Build the proof that coset multiplication is well defined when the subgroup is normal.
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.
Inside the integers modulo $12$, let $N$ be the subgroup $\{0, 4, 8\}$. Its four cosets are named by $0, 1, 2, 3$. Complete the addition table of the quotient; the row and column of $0$ are given.
| 0 | 1 | 2 | 3 | |
|---|---|---|---|---|
| 0 | 0 | 1 | 2 | 3 |
| 1 | 1 | |||
| 2 | 2 | |||
| 3 | 3 |
You can compute in a quotient group and say why its operation needs the subgroup to be normal. Say in your own words what an element of a quotient group is. Next: the first isomorphism theorem, which names a quotient without any coset arithmetic at all.
9. Your turn: what is the quotient of the integers modulo $10$ by the subgroup $\{0, 5\}$?, step 3