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The subgroup shifted: why every coset has the same size, why two cosets never partly overlap, and how a subgroup tiles the group it sits in.
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 list the cosets of a subgroup, say why each has as many elements as the subgroup, test whether two elements give the same coset with a single computation, explain why cosets partition the group and why only one of them is a subgroup, distinguish left from right cosets with an example where they differ, and recognise congruence modulo $n$ as a partition into cosets.
Unit 1 found subgroups; the alternating group came with the observation that it is exactly half of the symmetric group, with the odd permutations making up the other half. That was a coset without the name. This lesson gives the name and shows that the same split happens for every subgroup of every group.
For a subgroup $H \le G$ and $g \in G$, the left coset is $gH = \{gh : h \in H\}$ and the right coset is $Hg = \{hg : h \in H\}$. Any element of a coset is a representative of it. The index $[G : H]$ is the number of distinct left cosets. In additive notation a coset is written $g + H$.
Take $H \le G$ and shift it: $gH = \{gh : h \in H\}$. Three facts, and the third follows from the first two.
Every coset has $|H|$ elements. The map $h \mapsto gh$ is injective by cancellation and onto $gH$ by definition, so it is a bijection $H \to gH$.
Two cosets are equal or disjoint. If they share an element, each can be rewritten with the other's representative, so they coincide. And every $g$ lies in $gH$, since $g = ge$.
So the cosets partition $G$ into pieces of equal size. The group is tiled by copies of $H$, one of which is $H$ itself.
The test for equality is a single computation:
$$aH = bH \iff a^{-1}b \in H,$$
which additively reads $b - a \in H$. Two cosets are never compared by listing them.
Only one coset is a subgroup: $H$ itself. Any other misses the identity.
Left and right. $Hg$ need not equal $gH$. In $S_3$ with $H = \{e, (1\,2)\}$, the left coset $(1\,3)H$ and the right coset $H(1\,3)$ are different sets. The subgroups for which they always agree are the normal ones, and they are the subject of the last lesson of this unit. Until then, coset means left coset.
Most of the theorems in this course are counting arguments wearing algebraic clothes. Cosets all have the same size, so a subgroup's order divides the group's; the remainders below a degree are a finite list, so a quotient by an irreducible polynomial is a finite field.
Another way: picture
Think of $H$ as a tile and $G$ as a floor. Sliding the tile to the position of $g$ gives $gH$, still the same size and shape. Slide it to every position in turn: some slides give a tile you have already laid, and the ones that do not never overlap what is there. When you stop, the floor is exactly covered. Nothing is left over, because no position is uncovered, and nothing is doubled, because tiles never partly overlap.
Another way: steps
To list the cosets of $H$ in $G$: 1. Write $H$ itself; it is the coset of the identity. 2. Pick any element not yet used and form its coset. 3. Repeat until every element of $G$ has appeared. 4. Check: every coset has $|H|$ elements, and the number of cosets times $|H|$ is $|G|$. 5. To test whether $a$ and $b$ give the same coset, compute $a^{-1}b$ and ask whether it is in $H$.
The partition is not a coincidence; it is an equivalence relation in disguise. Define
$$a \sim b \iff a^{-1}b \in H.$$
It is reflexive because $a^{-1}a = e \in H$; symmetric because $H$ is closed under inverses and $(a^{-1}b)^{-1} = b^{-1}a$; and transitive because $H$ is closed under the operation and $(a^{-1}b)(b^{-1}c) = a^{-1}c$. Each of the three subgroup properties does exactly one job.
The equivalence classes of $\sim$ are precisely the left cosets, and every equivalence relation partitions its set — so the partition needed no separate proof at all. Anyone who has met equivalence relations has met this argument already; what is new is which relation to write down.
The familiar case is $G = \mathbb{Z}$ and $H = n\mathbb{Z}$. Then $a \sim b$ says $b - a$ is a multiple of $n$: congruence modulo $n$. Its classes are the residue classes, and $\mathbb{Z}_n$ is the set of cosets. So modular arithmetic is coset arithmetic, and it has been all along.
That example also shows what is coming. In $\mathbb{Z}$ the cosets can be added — $(a + n\mathbb{Z}) + (b + n\mathbb{Z}) = (a + b) + n\mathbb{Z}$ — and the result does not depend on which representatives were used. Whether that works in general is exactly the question of unit 3's last lesson, and the answer is: only when the subgroup is normal.
Expecting a coset to contain the identity. Only $H$ does. A coset is a shifted copy, and the shift moves the identity out of it.
Thinking a coset has one name. $aH = bH$ whenever $b \in aH$, so a coset of $k$ elements has $k$ representatives. A statement about a coset must not depend on which one is chosen — that requirement is what well defined will mean in unit 4.
Comparing cosets by listing them. Use $a^{-1}b \in H$. Listing is only available when the group is small.
Assuming $gH = Hg$. Not in general. Writing one and meaning the other is the commonest error in the next two lessons.
Thinking cosets can partly overlap. They cannot: sharing one element forces equality. That is the whole reason the counting works.
$H = \langle 4 \rangle = \{0, 4, 8\}$ inside $\mathbb{Z}_{12}$: three elements.
The tile.
$1 + H = \{1, 5, 9\}$, $2 + H = \{2, 6, 10\}$, $3 + H = \{3, 7, 11\}$. Every element of $\mathbb{Z}_{12}$ has now appeared.
Four cosets of three.
$4 + H$ is $\{4, 8, 0\} = H$ again, as it must be: $4 \in H$, so it adds no new coset. Four cosets, twelve elements.
A representative already used gives nothing new.
In $S_3$ take $H = \{e, (1\,2)\}$ and $g = (1\,3)$.
A subgroup of order two.
$gH = \{(1\,3), (1\,3)(1\,2)\} = \{(1\,3), (1\,3\,2)\}$, composing right to left.
The left coset.
$Hg = \{(1\,3), (1\,2)(1\,3)\} = \{(1\,3), (1\,2\,3)\}$ — a different set. So $gH \ne Hg$, and $H$ is not normal.
The two sides disagree.
The subgroup has $12$ elements and the whole group has $24$.
Sizes first.
Every coset has $12$ elements, and together they cover all $24$ without overlapping.
Equal tiles, exact cover.
So there are $24/12 = 2$ cosets: the even permutations and the odd ones. A subgroup of index two, and the odd permutations are the coset that is not a subgroup.
Inside the integers modulo $9$ under addition, let $H$ be the multiples of $3$. Complete the table of its three cosets; the first column is given.
| Smallest element | Next element | How many elements | |
|---|---|---|---|
| the subgroup itself | 0 | ||
| the coset of 1 | 1 | ||
| the coset of 2 | 2 |
A subgroup $H$ with $4$ elements sits inside a group with $16$. How many elements does the coset $gH$ have, for a $g$ outside $H$?
Answer:
Select every statement about the cosets of a subgroup that is true.
This task has no paper form; do it on a device.
Inside the integers modulo $12$, let $H$ be the subgroup $\{0, 4, 8\}$. Match each element to the coset it lies in.
| the subgroup itself | the coset of $1$ | the coset of $2$ | the coset of $3$ | the coset of $9$ | |
|---|---|---|---|---|---|
| $5$ | |||||
| $6$ | |||||
| $7$ | |||||
| $8$ |
Inside the integers modulo $12$, let $H = \{0, 4, 8\}$. Is the coset of $2$ the same set as the coset of $11$?
Build the proof that two cosets of the same subgroup are either equal or disjoint.
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 $9$ under addition, let $H$ be the multiples of $3$. Complete the table of its three cosets; the first column is given.
| Smallest element | Next element | How many elements | |
|---|---|---|---|
| the subgroup itself | 0 | ||
| the coset of 1 | 1 | ||
| the coset of 2 | 2 |
You can list the cosets of a subgroup and test two elements for lying in the same one. Say in your own words why cosets cannot partly overlap, and why only one of them is a subgroup. Next: counting the tiles, which is Lagrange's theorem.
9. Your turn: how many cosets does the subgroup of even permutations have inside all the permutations of four letters?, step 3