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Section 1.6 Cosets

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Subsection 1.6.1 Definition

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Now that we understand subgroups, we can talk about cosets. In fact we have already seen the idea of a coset when we stated the rearrangement theorem, but now instead of taking the whole group G, we will consider subgroups H⊂G.

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Definition 1.6.1. Cosets.

Let H={e,h1,…,hr} be a subgroup of G, and pick an element g∈G. Then the set:

Hg={eg,h1g,…,hrg}

is a right coset of H. Similarly, the set

gH={ge,gh1,…,ghr}

is a left coset of H. Cosets are generally not subgroups of G; they will be only if g∈H⊂G.

Consider the group \(\mathbb{Z}_4\) represented as \(G =\{1,-i,-1,i\}\) under multiplication. Consider the subgroup \(H=\{1,-1\}\text{.}\) Then the left cosets are:

\begin{align*} 1 \cdot H \amp = \{1, -1\}, \qquad \amp i \cdot H \amp = \{i, -i\},\\ -1 \cdot H \amp = \{-1, 1\}, \qquad \amp -i \cdot H \amp = \{-i, i\}. \end{align*}

So there are really two distinct cosets, given by \(H\) itself and the set \(\{i, -i\}\text{.}\) Since the group is abelian, the left cosets are the same as the right cosets. In general however left cosets and right cosets are not the same. We also notice that all elements of \(G\) appear exactly once in the distinct cosets; as we will see, this is a general result.

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Cosets will turn out to be very useful in physics. Often we need to deal with the fact that many different mathematical quantities give the same physical observable. For instance, the electromagnetic potential is not uniquely fixed; add to it the gradient of any function, and you get the same electric and magnetic fields. Similarly, in quantum mechanics, any two states in a Hilbert space that only differ by a phase give the same physical system. The way to deal with this kind of redundancy mathematically is to define equivalence classes of objects that give the same physics. It turns out that cosets are perfectly suited for that; under suitable conditions, we can see the cosets as being equivalence classes, hence from a physics point of view what we are often really interested in are the cosets of a given group, rather than the group itself.

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Thus what we would like to see is whether we can upgrade the cosets to groups. It is clear from the definition that cosets of a subgroup H⊂G are generally not subgroups of G themselves. However, what we can do is look at the “set of cosets” all together; these in fact form a group, which will be called the quotient group. This will be the group of equivalence classes formed by the cosets, which is often the relevant group of objects that we are interested in physically. But we are jumping ahead of ourselves; we will come back to quotient groups shortly.

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Subsection 1.6.2 Cosets and Lagrange's theorem

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Going back to cosets, the cool thing about them is that, given a subgroup, there is some sort of decomposition of a group into cosets of the subgroup, as you can see in the example above. This is because of the following simple Lemma:

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This lemma will be important to make sense of cosets as equivalence classes, since two elements cannot be in two distinct but not identical equivalence classes.

Let \(H = \{e, h_1, \ldots, h_r\}\) be a subgroup of \(G\text{.}\) Clearly, two cosets either have no common elements or have at least one common element. Let us show that if they have a common element, then they are identical.

Suppose that the left cosets generated by \(g_i\) and \(g_j\text{,}\) \(i \neq j\text{,}\) have a common element, say \(g_i h_m = g_j h_n\) for some \(m\) and \(n\text{.}\) Then \(g_j^{-1} g_i = h_n h_m^{-1}\text{,}\) therefore \(g_j^{-1} g_i \in H\text{.}\) It thus follows that the left coset \(g_j^{-1} g_i H\) is in fact equal to \(H\) itself as a set, since multiplying by an element of \(H\) on the left is only rearranging terms (this is the statement of the Rearrangement Theorem 1.4.2 for the subgroup \(H\)). Therefore \(g_j^{-1} g_i H = H\) as sets, and multiplying by \(g_j\) on the left, we get \(g_i H = g_j H\text{,}\) that is, the two cosets are identical as sets.

We conclude that left cosets are either identical as sets or contain not common elements. The same proof goes through for right cosets.

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This lemma allows us to prove an important theorem about subgroups of finite groups, known as Lagrange's theorem:

First, since the identity element \(e\) is in the subgroup \(H\text{,}\) and to construct the cosets of \(H\) we multiply by all elements of \(G\text{,}\) we see that all elements of \(G\) must appear in at least one coset of \(H\text{.}\) But since two cosets are either identical as sets or have no common elements, it follows that all elements of \(G\) must appear in exactly one distinct coset.

Furthermore, one can prove that all cosets have the same number of elements, which is equal to the order of the subgroup \(|H|\text{.}\) Consider a coset \(g H \) for some \(g \in G\text{.}\) Suppose that two elements of the coset are identical, \(g h_i = g h_j\) for some \(h_i \neq h_j\text{.}\) Multiplying by \(g^{-1}\) on the left, we get \(h_i = h_j\text{,}\) which is a contradiction. Therefore all elements of \(g H\) must be distinct, and hence the number of elements in \(g H\) is the same as the number of elements in \(H\text{.}\)

Putting this together, we conclude that the distinct cosets are partitioning the group \(G\) into non-intersecting bins of size equal to \(|H|\text{.}\) It follows that the number of distinct cosets times \(|H|\) must be equal to \(|G|\text{,}\) that is, \(|H|\) divides \(|G|\text{.}\)

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Definition 1.6.5. The index of a subgroup.

The number of distinct cosets of a subgroup H⊂G is called the index of the subgroup.

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Lagrange's theorem is in fact quite powerful to rule out when a subset cannot be a subgroup of a finite group. For instance, Lagrange's theorem implies the following corollaries:

We know that the centre of a group is a subgroup. Thus, by Lagrange's theorem, we know that if the order of the centre of the group is larger than half of the order of the group, then the centre must be the whole group, i.e. the group is abelian.

By Lagrange's theorem, if a group \(G\) has order given by a prime number, then its only subgroups must have either order one (the trivial subgroup) or the same order as the group itself (the group itself). Thus it cannot have proper subgroups. Moreover, if you pick any element \(g \in G\text{,}\) then it generates a cyclic subgroup \(\langle g \rangle \subseteq G\text{.}\) But since \(G\) has no proper subgroups, then \(\langle g \rangle\) is either the trivial subgroup if \(g\) is the identity element, or it is the whole group \(G\) for any other element \(g \in G\text{.}\) That is, \(G\) is a cyclic group, as it is equal to its non-trivial cyclic subgroups.

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