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Subgroups

Sometimes we wish to investigate smaller groups sitting inside a larger group. The set of even integers 2{\mathbb Z} = \{\ldots, -2, 0, 2, 4, \ldots \} is a group under the operation of addition.

Definitions and Examples

Sometimes we wish to investigate smaller groups sitting inside a larger group. The set of even integers \(2{\mathbb Z} = \{\ldots, -2, 0, 2, 4, \ldots \}\) is a group under the operation of addition. This smaller group sits naturally inside of the group of integers under addition. We define a subgroup \(H\) of a group \(G\) to be a subset \(H\) of \(G\) such that when the group operation of \(G\) is restricted to \(H\), \(H\) is a group in its own right. Observe that every group \(G\) with at least two elements will always have at least two subgroups, the subgroup consisting of the identity element alone and the entire group itself. The subgroup \(H = \{ e \}\) of a group \(G\) is called the trivial subgroup. A subgroup that is a proper subset of \(G\) is called a proper subgroup. In many of the examples that we have investigated up to this point, there exist other subgroups besides the trivial and improper subgroups.

Example

Consider the set of nonzero real numbers, \({\mathbb R}^*\), with the group operation of multiplication. \(\mathbb R^*\) the multiplicative group of real numbers The identity of this group is \(1\) and the inverse of any element \(a \in {\mathbb R}^*\) is just \(1/a\). We will show that \[\begin{aligned}\end{aligned}\] is a subgroup of \({\mathbb R}^*\). \(\mathbb Q^*\) the multiplicative group of rational numbers The identity of \({\mathbb R}^*\) is \(1\); however, \(1 = 1/1\) is the quotient of two nonzero integers. Hence, the identity of \({\mathbb R}^*\) is in \({\mathbb Q}^*\). Given two elements in \({\mathbb Q}^*\), say \(p/q\) and \(r/s\), their product \(pr/qs\) is also in \({\mathbb Q}^*\). The inverse of any element \(p/q \in {\mathbb Q}^*\) is again in \({\mathbb Q}^*\) since \((p/q)^{-1} = q/p\). Since multiplication in \({\mathbb R}^*\) is associative, multiplication in \({\mathbb Q}^*\) is associative.

Example

Recall that \({\mathbb C}^{\ast}\) is the multiplicative group of nonzero complex numbers. Let \(H = \{ 1, -1, i, -i \}\). Then \(H\) is a subgroup of \({\mathbb C}^{\ast}\). It is quite easy to verify that \(H\) is a group under multiplication and that \(H \subset {\mathbb C}^{\ast}\).

Condensed — the full section is in Judson, Abstract Algebra: Theory and Applications.

Some Subgroup Theorems

Let us examine some criteria for determining exactly when a subset of a group is a subgroup.

Proposition

A subset \(H\) of \(G\) is a subgroup if and only if it satisfies the following conditions.

The identity \(e\) of \(G\) is in \(H\).
If \(h_1, h_2 \in H\), then \(h_1h_2 \in H\).
If \(h \in H\), then \(h^{-1} \in H\).

Proof

First suppose that \(H\) is a subgroup of \(G\). We must show that the three conditions hold. Since \(H\) is a group, it must have an identity \(e_H\). We must show that \(e_H = e\), where \(e\) is the identity of \(G\). We know that \(e_H e_H = e_H\) and that \(ee_H = e_H e = e_H\); hence, \(ee_H = e_H e_H\). By right-hand cancellation, \(e =e_H\). The second condition holds since a subgroup \(H\) is a group. To prove the third condition, let \(h \in H\). Since \(H\) is a group, there is an element \(h' \in H\) such that \(hh' = h'h = e\). By the uniqueness of the inverse in \(G\), \(h' = h^{-1}\).

Conversely, if the three conditions hold, we must show that \(H\) is a group under the same operation as \(G\); however, these conditions plus the associativity of the binary operation are exactly the axioms stated in the definition of a group.

Proposition

Let \(H\) be a subset of a group \(G\). Then \(H\) is a subgroup of \(G\) if and only if \(H \neq \emptyset\), and whenever \(g, h \in H\) then \(gh^{-1}\) is in \(H\).

Proof

First assume that \(H\) is a subgroup of \(G\). We wish to show that \(gh^{-1} \in H\) whenever \(g\) and \(h\) are in \(H\). Since \(h\) is in \(H\), its inverse \(h^{-1}\) must also be in \(H\). Because of the closure of the group operation, \(gh^{-1} \in H\).

Conversely, suppose that \(H \subset G\) such that \(H \neq \emptyset\) and \(g h^{-1} \in H\) whenever \(g, h \in H\). If \(g \in H\), then \(gg^{-1} = e\) is in \(H\). If \(g \in H\), then \(eg^{-1} = g^{-1}\) is also in \(H\). Now let \(h_1, h_2 \in H\). We must show that their product is also in \(H\). However, \(h_1(h_2^{-1})^{-1} = h_1 h_2 \in H\). Hence, \(H\) is a subgroup of \(G\).

Condensed — the full section is in Judson, Abstract Algebra: Theory and Applications.

Symbols used here

x \in A,\ A \subseteq B

element of, subset
x belongs to A; every element of A is in B.

imaginary unit
i² = −1.

\neq

not equal
The two sides are different.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

\blacksquare\ \text{or}\ \square

end of proof (halmos)
Marks the point where the statement has been established.

a \equiv b \pmod n

congruent modulo n
n divides a − b; a and b have the same remainder.

a \mid b,\ \gcd(a,b)

divides, greatest common divisor
b is a multiple of a; the largest number dividing both.

(G, \cdot),\ e,\ g^{-1}

group, identity, inverse
A set with an operation; the do-nothing element; the element that undoes g.

G \cong H,\ G / N

isomorphic, quotient group
Same structure; the group of cosets of a normal subgroup N.

\mathbb{Z}/n\mathbb{Z},\ \mathbb{Z}_n

integers modulo n
The remainders 0…n−1 with clock arithmetic.

\operatorname{Hom}(A, B),\ f \circ g

arrows from A to B, composition
The set of morphisms; do g then f.

Questions people ask

What is a group, in plain words?

A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group — that is where the idea came from.

What is the difference between a ring and a field?

A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.

Обиди се со себе.

Parts of this page are adapted from Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). Condensed and re-explained here; errors are ours.

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