##### Definition15.2.2Coset

If \([G;*]\) is a group, \(H \leq G\) and \(a \in G\), the left coset of \(H\) generated by \(a\) is \[a*H = \{ a*h | h \in H\}\] and the right coset of \(H\) generated by \(a\) is \[H*a = \{h*a | h\in H\}\].

Consider the group \(\left[\mathbb{Z}_{12},+_{12}\right]\). As we saw in the previous section, we can picture its cyclic properties with the string art of 15.1.3. Here we will be interested in the non-generators, like 3. The solid lines in 15.2.1 show that only one-third of the tacks have been reached by starting at zero and jumping to every third tack. The numbers of these tacks correspond to \(\langle 3 \rangle = \{0, 3, 6, 9\}\).

What happens if you start at one of the unused tacks and again jump to every third tack? The two broken paths on 15.2.1 show that identical squares are produced. The tacks are thus partitioned into very similar subsets. The subsets of \(\mathbb{Z}_{12}\) that they correspond to are \(\{0, 3, 6, 9\}\), \(\{1, 4, 7, 10\}\), and \(\{2, 5, 8, 11\}\). These subsets are called cosets. In particular, they are called cosets of the subgroup \(\{0, 3, 6, 9\}\). We will see that under certain conditions, cosets of a subgroup can form a group of their own. Before pursuing this example any further we will examine the general situation.

If \([G;*]\) is a group, \(H \leq G\) and \(a \in G\), the left coset of \(H\) generated by \(a\) is \[a*H = \{ a*h | h \in H\}\] and the right coset of \(H\) generated by \(a\) is \[H*a = \{h*a | h\in H\}\].

\(H\) itself is both a left and right coset since \(e*H = H*e = H\).

If \(G\) is abelian, \(a*H = H*a\) and the left-right distinction for cosets can be dropped. We will normally use left coset notation in that situation.

Any element of a coset is called a representative of that coset.

One might wonder whether \(a\) is in any way a special representative of \(a*H\) since it seems to define the coset. It is not, as we shall see.

A duality principle can be formulated concerning cosets because left and right cosets are defined in such similar ways. Any theorem about left and right cosets will yield a second theorem when “left” and “right” are exchanged for “right” and “left.”

If \(b\in a*H\), then \(a*H = b*H\), and if \(b \in H*a\), then \(H*a = H*b\).

In 15.2.1, you can start at either 1 or 7 and obtain the same path by taking jumps of three tacks in each step. Thus, \[1+_{12} \{0, 3, 6, 9\} = 7 +_{12} \{0, 3, 6, 9\} = \{1, 4, 7, 10\}\].

The set of left (or right) cosets of a subgroup partition a group in a special way:

If \([G; *]\) is a group and \(H\leq G\), the set of left cosets of \(H\) is a partition of \(G\text{.}\) In addition, all of the left cosets of \(H\) have the same cardinality. The same is true for right cosets.

The function \(\rho\) has a nice interpretation in terms of our opening example. If \(a \in \mathbb{Z}_{12}\), the graph of \(\{0, 3, 6, 9\}\) is rotated \((30a)^{\circ}\) to coincide with one of the three cosets of \(\{0, 3, 6, 9\}\).

If \(\lvert G\rvert < \infty\) and \(H\leq G\), the number of distinct left cosets of \(H\) equals \(\frac{\lvert G\rvert }{\lvert H\rvert }\). For this reason we use \(G/H\) to denote the set of left cosets of \(H\) in \(G\)

The set of integer multiples of four, \(4\mathbb{Z}\), is a subgroup of \([\mathbb{Z}, +]\). Four distinct cosets of \(4\mathbb{Z}\) partition the integers. They are \(4\mathbb{Z}\), \(1+4\mathbb{Z}\), \(2+4\mathbb{Z}\), and \(3+4\mathbb{Z}\), where, for example, \(1+4\mathbb{Z} = \{1+4k | k \in \mathbb{Z}\}\). \(4\mathbb{Z}\) can also be written \(0+4\mathbb{Z}\).

Although we have seen that any representative can describe a coset, it is often convenient to select a distinguished representative from each coset. The advantage to doing this is that there is a unique name for each coset in terms of its distinguished representative. In numeric examples such as the one above, the distinguished representative is usually the smallest nonnegative representative. Remember, this is purely a convenience and there is absolutely nothing wrong in writing \(-203+4\mathbb{Z}\), \(5+4\mathbb{Z}\), or \(621+4\mathbb{Z}\) in place of \(1+4\mathbb{Z}\) because \(-203, 5, 621 \in 1+4\mathbb{Z}\).

Before completing the main thrust of this section, we will make note of a significant implication of Theorem 15.2.8. Since a finite group is divided into cosets of a common size by any subgroup, we can conclude:

The order of a subgroup of a finite group must divide the order of the group.

One immediate implication of Lagrange's Theorem is that if \(p\) is prime, \(\mathbb{Z}_p\) has no proper subgroups.

We will now describe the operation on cosets which will, under certain circumstances, result in a group. For most of this section, we will assume that \(G\) is an abelian group. This is one sufficient (but not necessary) condition that guarantees that the set of left cosets will form a group.

Let \(C\) and \(D\) be left cosets of \(H\text{,}\) a subgroup of \(G\) with representatives \(c\) and \(d\text{,}\) respectively. Then \[C\otimes D = (c*H) \otimes (d*H) = (c*d)*H \] The operation \(\otimes\) is called the operation induced on left cosets by \(*\).

In Theorem 15.2.18, later in this section, we will prove that if \(G\) is an abelian group, \(\otimes\) is indeed an operation. In practice, if the group \(G\) is an additive group, the symbol \(\otimes\) is replaced by \(+\), as in the following example.

Consider the cosets described in Example 15.2.10. For brevity, we rename \(0+4\mathbb{Z}\), \(1+4\mathbb{Z}\), \(2+4\mathbb{Z}\), and \(3+4\mathbb{Z}\) with the symbols \(\bar{0}\), \(\bar{1}\), \(\bar{2}\), and \(\bar{3}\). Let's do a typical calculation, \(\bar{1}+\bar{3}\). We will see that the result is always going to be \(\bar{0}\) , no matter what representatives we select. For example, \(9 \in \bar{1}\), \(7\in \bar{3}\), and \(9+7=16 \in \bar{0}\). Our choice of the representatives \(\bar{1}\) and \(\bar{3}\) were completely arbitrary.

In general, \(C \otimes D\) can be computed in many ways, and so it is necessary to show that the choice of representatives does not affect the result. When the result we get for \(C \otimes D\) is always independent of our choice of representatives, we say that “\(\otimes\) is well defined.” Addition of cosets is a well-defined operation on the left cosets of 4\(\mathbb{Z}\) and is summarized in the following table. Do you notice anything familiar? \begin{equation*} \begin{array}{c|cccc} \otimes & \bar{0} & \bar{1} & \bar{2} & \bar{3}\\ \hline \bar{0} & \bar{0} & \bar{1} & \bar{2} & \bar{3}\\ \bar{1} & \bar{1} & \bar{2} & \bar{3} & \bar{0}\\ \bar{2} & \bar{2} &\bar{3} & \bar{0} & \bar{1}\\ \bar{3} & \bar{3}& \bar{0} & \bar{1} & \bar{2} \\ \end{array} \end{equation*}

Consider the real numbers. \([\mathbb{R}; +]\), and its subgroup of integers, \(\mathbb{Z}\text{.}\) Every element of \(\mathbb{R}/\mathbb{Z}\) has the same cardinality as \(\mathbb{Z}\text{.}\) Let \(s, t\in \mathbb{R}\). \(s\in t+\mathbb{Z}\) if \(s\) can be written \(t+n\) for some \(n \in \mathbb{Z}\). Hence \(s\) and \(t\) belong to the same coset if they differ by an integer. (See Exercise 15.2.1.6 for a generalization of this fact.)

Now consider the coset \(0.25+\mathbb{Z}\). Real numbers that differ by an integer from 0.25 are \(1.25, 2.25, 3.25, \ldots\) and \(-0.75, -1.75, -2.75, \ldots\). If any real number is selected, there exists a representative of its coset that is greater than or equal to 0 and less than 1. We will call that representative the distinguished representative of the coset. For example, 43.125 belongs to the coset represented by 0.125; \(-6.382+\mathbb{Z}\) has 0.618 as its distinguished representative. The operation on \(\mathbb{R}/\mathbb{Z}\) is commonly called addition modulo 1. A few typical calculations in \(\mathbb{R}/\mathbb{Z}\) are \begin{equation*} \begin{array}{c} (0.1+\mathbb{Z})+(0.48+\mathbb{Z}) = 0.58+\mathbb{Z}\\ (0.7+\mathbb{Z})+(0.31+\mathbb{Z}) = 0.01+\mathbb{Z}\\ -(0.41+\mathbb{Z}) = -0.41+\mathbb{Z} = 0.59+\mathbb{Z}\\ \textrm{and in general, } -(a+\mathbb{Z}) = (1 - a)+\mathbb{Z} \end{array} \text{.} \end{equation*}

Consider \(F = (\mathbb{Z}_4\times \mathbb{Z}_2 )/H\), where \(H=\{(0,0),(0,1)\}\). Since \(\mathbb{Z}_4 \times \mathbb{Z}_2\) is of order 8, each element of \(F\) is a coset containing two ordered pairs. We will leave it to the reader to verify that the four distinct cosets are \((0, 0)+H\), \((1,0) +H\), \((2, 0)+H\) and \((3, 0)+H\). The reader can also verify that \(F\) is isomorphic to \(\mathbb{Z}_4\) , since \(F\) is cyclic. An educated guess should give you a generator.

Consider the group \(\mathbb{Z}_2{}^4 = \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2\) . Let \(H\) be \(\langle (1,0,1, 0)\rangle\), the cyclic subgroup of \(\mathbb{Z}_2{}^4\) generate by (1,0,1,0). Since \[(1,0,1, 0)+(1,0,1, 0)=(1+_21,0+_20,1+_21,0+_20) = (0,0,0,0)\] the order of \(H\) is 2 and , \(\mathbb{Z}_2{}^4/H\) has \(\lvert \mathbb{Z}_2^4 /H\rvert =\frac{\lvert \mathbb{Z}_2^4\rvert }{\lvert H\rvert }=\frac{16}{2}= 8\) elements. A typical coset is \[C = (0, 1, 1, 1)+H = \{(0, 1, 1, 1), (1, 1, 0, 1)\}\] Note that \(2(0, 1, 1, 1) = (0, 0, 0, 0)\), \(2C = C\otimes C = H\), the identity for the operation on \(\mathbb{Z}_2{}^4/H\). The orders of all non-identity elements of of this factor group are all 2, and it can be shown that the factor group is isomorphic to \(\mathbb{Z}_2{}^3\).

If \(G\) is an abelian group, and \(H \leq G\), the operation induced on cosets of \(H\) by the operation of \(G\) is well defined.

Let \(G\) be a group and \(H \leq G\). If the operation induced on left cosets of \(H\) by the operation of \(G\) is well defined, then the set of left cosets forms a group under that operation.

Let \(G\) be a group and \(H \leq G\). If the set of left cosets of \(H\) forms a group, then that group is called the factor group of “\(G\) modulo \(H\text{.}\)” It is denoted \(G/H\).

If \(G\) is abelian, then every subgroup of \(G\) yields a factor group. We will delay further consideration of the non-abelian case to Section 15.4.

It is customary to use the same symbol for the operation of \(G/H\) as for the operation on \(G\text{.}\) The reason we used distinct symbols in this section was to make the distinction clear between the two operations.

Consider \(\mathbb{Z}_{10}\) and the subsets of \(\mathbb{Z}_{10}\), \(\{0, 1, 2, 3, 4\}\) and \(\{5, 6, 7, 8, 9\}\). Why is the operation induced on these subsets by modulo 10 addition not well defined?

AnswerCan you think of a group \(G\text{,}\) with a subgroup \(H\) such that \(\lvert H\rvert = 6\) and \(\lvert G/H\rvert = 6\)? Is your answer unique?

For each group and subgroup, what is \(G/H\) isomorphic to?

\(G = \mathbb{Z}_4 \times \mathbb{Z}_2\) and \(H = \langle (2, 0)\rangle\). Compare to Example 15.2.16.

\(G = [\mathbb{C}, +]\) and \(H = \mathbb{R}\).

\(G\) = \(\mathbb{Z}_{20}\) and \(H = \langle 8\rangle\).

For each group and subgroup, what is \(G/H\) isomorphic to?

\(G = \mathbb{Z}\times \mathbb{Z}\) and \(H = \{\{a, a) | a \in \mathbb{Z}\}\).

\(G = \left[\mathbb{R}^*, \cdot \right]\) and \(H = \{1, -1\}\).

\(G =\mathbb{Z}_2{}^5\) and \(H = \langle (1, 1, 1, 1, 1)\rangle\).

Prove that if \(G\) is a group, \(H \leq G\) and \(a, b \in G\), \(a*H= b*H\) if and only if \(b^{-1}*a \in H\).

AnswerReal addition modulo r, . \(r > 0\), can be described as the operation induced on cosets of \(\langle r\rangle\) by ordinary addition. Describe a system of distinguished representatives for the elements of \(\mathbb{R}/\langle r\rangle\).

Consider the trigonometric function sine. Given that \(\sin (x+2\pi k) = \sin x\) for all \(x\in \mathbb{R}\) and \(k\in \mathbb{Z}\), show how the distinguished representatives of \(\mathbb{R}/\langle 2\pi \rangle\) can be useful in developing an algorithm for calculating the sine of a number.

Complete the proof of Theorem 15.2.8 by proving that if \(a \in G\), \(\rho:H \to a*H\) defined by \(\rho(h)= a*h\) is a bijection.