Say the group $D_n$, the dihedral group, has a subgroup $\langle s\rangle$. How would you describe the left and right cosets of $\langle s\rangle$? What are these cosets used for?

edit: $\langle s\rangle$ is the flip symmetry

  • $\begingroup$ Proper notation is $\langle s\rangle$, not $<s>$. I changed it. $\qquad$ $\endgroup$ Commented Oct 5, 2016 at 23:28

2 Answers 2


Cosets partition a group $G$ in an interesting way. Here, "partition" means to organize elements into distinct, nonoverlapping subsets, where every element of the group is in one of these subsets.

Finding a coset of the subgroup $S$ in $G$ is done by multiplying every member of $S$ on the left or on the right by a fixed group element $a \in G$, denoting, respectively, the left coset $aS$ and the right coset $Sa$ (in a commutative group, this directionality won't matter). For this post, let's multiply on the right for sake of example.

If you multiply every element of the subgroup $S$ by all members of the group $G$ and record each set of elements of $G$, you'll find that some sets have exactly the same elements as other sets, but do not share a single element with the rest of the sets; to see this, consider if two distinct such sets $Sa$ and $Sb$ share a common element - you can use the laws for inverses to show that $Sa$ and $Sb$ must be the same in this case. Now, we see that we've partitioned the group by recording each distinct subset $Sa$ for all $a \in G$ and noticing that each member of the group appears in exactly one of these subsets (again, use the notion of group inverse to convince yourself of this ...).

In particular (since, remember, $S$ itself was assumed to be a subgroup), all distinct cosets $Sa$ have exactly the same number of elements as all the other distinct cosets! And, since every element of $G$ appears in exactly one of these cosets, we have Lagrange's theorem - that the order of $S$ must divide the order of $G$. Be careful with this theorem: it doesn't work in reverse, i.e. if some number divides the order of $G$ there doesn't necessarily have to be a subgroup of $G$ of that size.

Now for your question about dihedral groups. Let's take an easy example, $D_8.$ Informally, this is the set of rotations and reflections of a square. If you label each corner of a square as $x_1, x_2, x_3,$ and $x_4,$ say, starting with the upper left hand corner and working around clockwise in order, $D_8$ acts on these vertices by either rotating clockwise (bringing $x_1$ to $x_2,$ $x_2$ to $x_3,$ $x_3$ to $x_4,$ and $x_4$ over to where $x_1$ was), or by reflecting across the horizontal line cutting "halfway across" the middle of the square (swapping $x_1$ with $x_4$ and $x_2$ with $x_3$). These are denoted in the group as, respectively, $r$ and $s.$

Since two reflections leaves you with the same indices you started with in order, $s^2=e$ where $e$ is the group element that does nothing with the square. If you rotate the square clockwise four times, it also does nothing, therefore $r^4=e$ (and $4$ is the least natural number with this property). So our distinct elements of $G$ in this case are $\{e, r, r^2, r^3, s, sr, sr^2, sr^3\}.$ Note the last three elements involve a certain number of clockwise rotations followed by a "flip."

The subgroup generated by $s$ is $<s> = \{e, s\}.$ We'll call that $S.$ What are the right cosets? Multiply $\{e, s\}$ by every element of $G$ to find out! You will get $\{r, sr\}, \{r^2, sr^2\}, \{r^3, sr^3\},$ and $\{e, s\}$. Since every element is in one and only one coset, we won't have any confusion shortening the notation here: call these cosets $Sr, Sr^2, Sr^3,$ and $Se$ (or just $S$). We could have correctly called these cosets $S(sr), S(sr^2),$ etc., but that would be a notational nightmare.

See how there are four cosets? The order of the group is $8$ elements. Since $8=4\cdot 2,$ the order of the subgroup divides the order of the group. Also, we say that $S$ is of index $4$ in $G=D_8$ as there are four cosets. Can you write the cosets for $<r>?$ What is that subgroup's index?

Of special importance are subgroups where all the left cosets are the same sets as all the right cosets; these subgroups are called normal subgroups. Elements in normal subgroups can "kind of" commute with all elements in a group: that is, if $s$ is in a normal subgroup, $a\cdot s = s \cdot b$ for $b$ a group element in the same coset as $a.$ The crowning achievement with this fact is that, if $S$ is a subgroup that is "normal," the set of cosets of S themselves forms a group! This leads to a lot of rich theory, especially surrounding the notion of homomorphism, which are similar to linear transformations (and normal subgroups correspond to the special role null spaces play with linear transformations).

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    $\begingroup$ Your example $S$ is of index $4$, not $2$. $\endgroup$ Commented Oct 6, 2016 at 2:58
  • $\begingroup$ Whoops! Good catch!!! $\endgroup$ Commented Oct 6, 2016 at 3:05

Take the group of all integers under usual addition. The integers that are multiples of 10 form a subgroup ere. A coset corresponding to 7 would be simple the subset of positive integers with terminal digit 7 and negative integers with terminal digit 3.

For some purposes only the last digit of the number might be important, and the whole integer. For example to be a prime number the last digit has to be necessarily from among 1,3,7 or 9.

In non-zero complex numbers which form a group under multiplication numbers of modulus 1 form a subgroup. The coset for a number like $3-7i$ here would be numbers having the same modulus as $3-7i$, that is those in a circle centred at origin and passing through $3-7i$. SOme property of numbers may depend only on how far away the complex number is from the origin and not in what direction. Then this coset is the right thing.

The appropriate concept is quotient groups. When you study them this will become clearer.


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