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Section 1.2 Subgroups

Subsection 1.2.1 Definition

The notion of a subgroup is rather obvious. But these are important in physics, for instance in the context of symmetry breaking. Suppose that a physical system is symmetric under a certain group of transformation. Then one could think of adding a perturbation to the system, for instance an extra term in the Lagrangian or Hamiltonian, that lowers the symmetry group to a subgroup of symmetries. This is captured by the concept of a subgroup.

The idea of subgroups is just like subspaces for vector spaces, which consist of subspaces that are vector spaces in their own right.

Definition 1.2.1. Subgroups.

A subset \(H\) of a group \(G\) is a subgroup if it is a group in its own right. We write \(H \subset G\text{.}\)

Since \(H\) is a group in its own right, it follows that any subgroup of \(G\) contains the identity element \(e \in G\text{.}\) According to the definition, the group \(G\) is a subgroup of itself. Also, for any \(G\text{,}\) the subset \(\{e\}\) containing only the identity element is a subgroup, known as the trivial subgroup. Those two subgroups are rather boring; we call them improper. We are interested in studying proper subgroups.

Subsection 1.2.2 Examples

Some examples of subgroups:

Consider only rotations in three dimensions around a fixed axis. This gives a subgroup \(SO(2) \subset SO(3)\text{.}\)

Consider only permutations of \(n\) objects that leave \(n-m\) objects fixed. This gives a subgroup \(S_m \subset S_n\text{,}\) with \(m \lt n\text{.}\)

We have not defined the parity of a permutation yet, but if we consider the subgroup of permutations that are even, we get the so-called alternating subgroup \(A_n \subset S_n\text{.}\)

Consider the cyclic group \(\mathbb{Z}_n\text{,}\) represented as the \(n\)'th roots of unity under multiplication. Let \(m\) be an integer that divides \(n\text{.}\) Then there are a subset of \(n\)'th roots of unity that are also \(m\)'th roots of unity; those form the subgroup \(\mathbb{Z}_m \subset \mathbb{Z}_n\text{.}\) For instance, if we write \(\mathbb{Z}_4 = \{1, -i, -1, i\}\text{,}\) then we have a subgroup \(\mathbb{Z}_2 =\{1,-1 \} \subset \mathbb{Z}_4\text{.}\)

The real numbers \(\mathbb{R}\) form a group under addition. If we restrict to the integers \(\mathbb{Z}\text{,}\) then we get a subgroup \(\mathbb{Z} \subset \mathbb{R}\) under addition.

The set of all multiples of a positive integer \(m\text{,}\) denoted by \(m \mathbb{Z}\text{,}\) is a subgroup of \(\mathbb{Z}\) with respect to addition. In fact, all subgroups of \(\mathbb{Z}\) are of this form.

We can in fact construct many subgroups of \(GL(n,\mathbb{R})\) and \(GL(n,\mathbb{C})\text{.}\) Recall that we can think of \(GL(n,\mathbb{R})\) and \(GL(n,\mathbb{C})\) either as groups of real and complex invertible matrices (respectively), or as groups of invertible linear transformations on \(\mathbb{R}^n\) and \(\mathbb{C}^n\text{.}\) We can construct subgroups by restricting to matrices of special types, or by looking at linear transformations that preserve special structures on \(\mathbb{R}^n\) or \(\mathbb{C}^n\text{.}\)

Since \(SL(n, \mathbb{R})\) and \(SL(n,\mathbb{C})\) consist of \(n \times n\) matrices with determinant 1, they form subgroups of \(GL(,n\mathbb{R})\) and \(GL(n,\mathbb{C})\) respectively.

We can start with \(GL(n,\mathbb{R})\text{,}\) and consider the subset of orthogonal matrices, which we denote by \(O(n)\text{.}\) Recall that a matrix \(M \in GL(n,\mathbb{R})\) is orthogonal if \(M^T M = I\text{.}\)

The subset \(O(n)\) is a subgroup of \(GL(n,\mathbb{R})\text{,}\) which we now check. First, the identity \(I \in O(n)\text{.}\) Second, if \(A,B \in O(n)\text{,}\) then \(A B \in O(n)\text{,}\) since \((A B)^T A B = B^T A^T A B = B^T B = I\text{.}\) Therefore the subset \(O(n)\) is closed under matrix multiplication. Finally, if \(A \in O(n)\text{,}\) then its inverse \(A^{-1}\) is also orthogonal, and hence in \(O(n)\text{.}\) Indeed, since \(A^{-1} = A^T\text{,}\) we have \((A^{-1})^T A^{-1} = (A^T)^T A^T = A A^T = I\text{.}\)

Orthogonal matrices also have a special interpretation in terms of linear transformations of \(\mathbb{R}^n\text{.}\) We want to look at linear transformations that preserve special structures on \(\mathbb{R}^n\text{.}\) In this case, we are interested in preserving the Euclidean scalar product on \(\mathbb{R}^n\text{.}\) Recall that given two column vectors \(u, v \in \mathbb{R}^n\text{,}\) the Euclidean scalar product is given by

\begin{equation*} u^T v = u_1 v_1 + u_2 v_2 + \ldots + u_n v_n . \end{equation*}

If we apply a linear transformation \(M: \mathbb{R}^n \to \mathbb{R}^n\text{,}\) to get new vectors \(u' = M u\) and \(v' = M v\text{,}\) then

\begin{equation*} (u')^T v' = (M u)^T (M v) = u^T M^T M v. \end{equation*}

Thus \((u')^T v' = u^T v \) for all \(u,v \in \mathbb{R}^n\) if and only if \(M^T M = I\text{.}\) In other words, \(M\) is orthogonal, i.e. in \(O(n)\text{.}\) Therefore, \(O(n)\) can be thought of as the subgroup of invertible linear transformations of \(\mathbb{R}^n\) that preserve the Euclidean scalar product.

These transformations preserve length and angle. Indeed, the length square of a vector \(u\) is \(|u|^2 = u^T u\text{,}\) thus if \(u' = M u\) with \(M\) orthogonal, then \(|u'|^2 = |u|^2\text{,}\) and the length is preserved. But then, since \(u^T v = |u| |v| \cos \theta\) with \(\theta\) the angle between the vectors \(u\) and \(v\text{,}\) it follows from \((u')^T v' = u^T v\) and the fact that \(|u'| = |u|\) and \(|v'| = |v|\) that \(\theta'= \theta\text{,}\) i.e. the angle between the vectors is also preserved.

But what transformations preserve length and angle? Rotations and reflections! So \(O(n)\) is the group generated by rotations and reflections in \(n\) dimensions.

Consider \(n \times n\) orthogonal matrices \(M\text{.}\) Then \(M^T M = I\text{,}\) and \(\det(M^T M) = \det(M)^2 = \det(I) = 1\text{,}\) and hence \(\det(M) = \pm 1\text{.}\) It is easy to convince yourself that rotations correspond to the subgroup of transformations with \(\det(M) = 1\) (which is the component of \(O(n)\) that is connected to the identity). We call this subgroup the special orthogonal group \(SO(n) \subset GL(n,\mathbb{R})\text{.}\) This is the group of rotations in \(n\) dimensions.

Check that \(SO(n)\) is a subgroup of \(O(n)\text{.}\)

We can do a similar construction for \(GL(n,\mathbb{C})\text{.}\) Thinking of it as the group of \(n \times n\) invertible matrices with complex entries, we can first restrict to the subset of unitary matrices, namely matrices \(M \in GL(n,\mathbb{C})\) such that \(M^\dagger M = I\text{.}\) Here we defined the adjoint (or Hermitian conjugate) of a matrix \(M\) as \(M^\dagger = (M^*)^T\text{,}\) where \(M^*\) denotes the matrix \(M\) with the entries complex conjugated. Following the exact same steps as in the orthgonal case for real matrices, one can show that the subset of unitary matrices is a subgroup, which we call the unitary group \(U(n) \subset GL(n,\mathbb{C})\text{.}\) As for orthogonal matrices, \((\det M)^2 = 1\text{,}\) and if we restrict to matrices with \(\det M = 1\text{,}\) we get the special unitary group \(SU(n) \subset GL(n,\mathbb{C})\text{.}\)

We can also think of unitary matrices as linear transformations of \(\mathbb{C}^n\text{.}\) What kind of transformations are those? They are the transformations that preserve the standard inner product on \(\mathbb{C}^n\text{.}\) Recall that given two vectors \(u,v \in \mathbb{C}^n\text{,}\) we define the inner product as

\begin{equation*} (u^*)^T v = u_1^* v_1 + u_2^* v_2 + \ldots + u_n^* v_n. \end{equation*}

Thus, just as for orthogonal transformations, it is easy to see that unitary transformations preserve the inner product on \(\mathbb{C}^n\text{.}\) So the unitary subgroup \(U(n) \subset GL(n,\mathbb{C})\) can be understood as the subgroup of invertible linear transformations on \(\mathbb{C}^n\) that preserve the standard inner product.

Recall that the orthogonal group \(O(n)\) can be thought of as the subgroup of invertible linear transformations of \(\mathbb{R}^n\) that preserve the Euclidean scalar product. But instead of looking at the Euclidean notion of distance on \(\mathbb{R}^n\text{,}\) we could have started with a different scalar product. For instance, we could take \(\mathbb{R}^4\) to be Minkowski space, that is, equipped with the Minkowski scalar product:

\begin{equation*} u \cdot v = - u_0 v_0 + u_1 v_1 + u_2 v_2 + u_3 v_3. \end{equation*}

Note the different sign for the first term, which corresponds to the time direction in Minkowski space. We can look at the subset of linear transformations of \(\mathbb{R}^4\) that leave this notion of scalar product invariant. It forms a subgroup, which is denoted by \(O(1,3) \subset GL(4,\mathbb{R})\text{.}\) This is nothing but the Lorentz group of special relativity! Indeed, one can check that elements of \(O(1,3)\) are Lorentz transformations. So one can think of Lorentz transformations as "rotations in Minkowski space".

(More precisely, \(O(1,3)\) is obtained by composing Lorentz transformations with parity and time reversal transformations, just as \(O(n)\) is obtained by composing reflections with rotations. Lorentz transformations correspond to the component of \(O(1,3)\) connected to the identity.)

In general, if we start with \(\mathbb{R}^n\) with a scalar product of the form

\begin{equation*} u \cdot v = - u_1 v_1 - \ldots - u_p v_p + u_{p+1} v_{p+1} + \ldots + u_n v_n, \end{equation*}

the subset of linear transformations that preserve this inner product is denoted by \(O(p,n-p)\text{,}\) and is a subgroup of \(GL(n,\mathbb{R})\text{.}\)

Another interesting subgroup that plays an important role in physics is the symplectic group. Let \(x, y \in \mathbb{R}^{2n}\text{,}\) and \(J\) be the \(2n \times 2n\) symplectic matrix \(J = \begin{pmatrix} 0 \amp I \\ - I \amp 0 \end{pmatrix}\text{,}\) where \(I\) is the \(n \times n\) identity matrix. The subset of \(GL(2n, \mathbb{R})\) that leave the “antisymmetric bilinear form” \(x^T J y\) invariant is a subgroup of \(GL(2n, \mathbb{R})\text{,}\) which is called the sympletric group and is denoted by \(Sp(2n, \mathbb{R})\text{.}\) It is for instance fundamental in the more geometrical treatment of Hamiltonian mechanics.

Many of those groups will play a fundamental in physics. In particular, we will see that \(SO(3)\) and \(SU(2)\) are intimately related, and so are \(SO(1,3)\) and \(SL(2,\mathbb{C})\text{.}\)

Subsection 1.2.3 A few universal subgroups

There are a few particular subgroups that are rather universal. In some way, what we are doing now parallels what we you did in linear algebra for vector spaces. We want to define the analog of the span of a subset, and the commutator of two elements, but in the context of abstract groups.

Definition 1.2.15. Subgroup generated by a subset of a group.

Let \(S\) be a subset of finite group \(G\text{.}\) The subgroup generated by \(S\), denoted by \(\langle S \rangle\text{,}\) is the union of \(S\) and all inverses and products of the elements in \(S\)

In particular, if \(S\) consists of a single element \(S = \{a\}\) with \(a \in G\text{,}\) then we write \(\langle a \rangle\) for the subgroup generated by \(a\text{,}\) which consists in the collection of integer powers of \(a\text{.}\) Note that because \(G\) is finite, we know that at some point \(a^k = e\) for some integer \(k\text{.}\) \(\langle a \rangle\) is called the cyclic subgroup generated by \(a\).

In fact, based on this, we can define the notion of cyclic groups.

Definition 1.2.16. Cyclic group.

A cyclic group is a group generated by a single element. In other words, it is equal to one of its cyclic subgroups \(G = \langle g \rangle \) for some element \(g \in G\text{,}\) called a generator of \(G\text{.}\)

Given a particular element of a group, we can also look at all elements that commute with it.

Definition 1.2.17. Centralizer and the centre of a group.

Let \(a \in G\text{.}\) The set of elements of \(G\) that commuate with \(a\text{,}\) denoted by \(C_G(a)\text{,}\) is called the centralizer of \(a\) in \(G\). The set of elements of \(G\) that commute with all elements of \(G\) is called the centre of \(G\), and denoted by \(Z(G)\text{.}\)

It is easy to to show that both \(C_G(a)\) and \(Z(G)\) are subgroups of \(G\text{.}\) Moreover, \(Z(G)\) is abelian by definition, and, clearly, \(G\) is abelian if and only if \(Z(G) = G\text{,}\) since all its elements commute with each other.

Remark 1.2.18.

The definitions that we are making look very similar to the analogous definitions for vector places, but with \(0\) replaced by the identity element \(e\text{.}\) Indeed, the definitions that we are proposing reduce to the previous propositions for vector spaces when the group operation is considered to be vector space addition. In this case, the identity element for addition in a vector space is the zero vector, so \(e\) would be replaced by \(0\text{,}\) as needed.

We can also define the “commutator” of two elements of a group \(G\text{.}\) But we need to keep in mind that we only have one operation to work with, namely group multiplication. The commutator should somehow measure how non-abelian a group \(G\) is.

Definition 1.2.19. The commutator subgroup of a group.

Let \(a,b \in G\text{.}\) The commutator \([a,b]\) of \(a\) and \(b\) is defined by

\begin{equation*} [a,b] = a^{-1} \cdot b^{-1} \cdot a \cdot b. \end{equation*}

If \(a\) and \(b\) commute, then \([a,b]=e\text{,}\) thus we can think of the commutator as measuring how far from being abelian a group is. The subgroup generated by all commutators of elements \(G\) is called the commutator subgroup of \(G\text{.}\)

If \(G\) is abelian, then \([a,b]=e\) for all \(a,b \in G\text{.}\) The implication goes in the other direction as well. Thus a group \(G\) is abelian if and only if its commutator subgroup is the trivial subgroup \(\{e\} \subset G\text{.}\)

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