Subgroup

A subgroup of a group $G$ is a subset of $G$ that forms a group with the same law of composition. For example, the even numbers form a subgroup of the group of integers with group law of addition.

Any group $G$ has at least two subgroups: the trivial subgroup $\{1\}$ and $G$ itself. It need not necessarily have any other subgroups however; for example, $\mathbb{Z}_5$ has no nontrivial proper subgroup.

• The group of integers equipped with addition is a subgroup of the real numbers equipped with addition; i.e. $(\mathbb{Z}, +) \subset (\mathbb{R}, +)$.
• The group of real matrices with determinant 1 is a subgroup of the group of invertible real matrices, both equipped with matrix multiplication. To prove this, it is necessary to prove closure, meaning that it must be shown that the product of two matrices with determinant 1 is another matrix with determinant 1, but this is immediate from the multiplicative property of the determinant. This group is denoted $SL(n, \mathbb{R})$, and is often seen.
• The set of complex numbers with magnitude 1 is a subgroup of the nonzero complex numbers equipped with multiplication. It is known as the circle group as its elements form the unit circle.

Another useful example is the subgroup $\mathbb{Z}a$, the set of multiples of $a$ equipped with addition:

$$\{\ldots,-2a, -a, 0, a, 2a, \ldots\},$$

where $a$ is any positive integer. These are the only nontrivial subgroups of the set of integers, and they help establish some classical number theory including the concept of greatest common divisor and least common multiple.

The subgroups of the additive group of integers $($denoted $\mathbb{Z}^+)$ are of the form $\mathbb{Z}a$ for some positive integer $a$. $_\square$

Consider a subgroup $S$ of $\mathbb{Z}^{+}$. Then 0 must be in $S$, as 0 is the identity of $\mathbb{Z}^+$. If $S$ does not contain any element other than 0, then $S$ is the trivial group. Otherwise, suppose $S$ contains another element $a$. Either $a$ or $-a$ will be positive, so $S$ contains a positive element. Let $a$ be the smallest positive integer in $S$.

It is necessary to show that

• every integer multiple of $a$ is in $S$;
• no other numbers are in $S$.

The first is easy to show: since $S$ is a group, the axiom of closure applies, so $a + a + \cdots + a = ka$ is in $S$ for any positive integer $k$. The inverse of $ka$ is $-ka$, so $-ka \in S$ as well. Hence any multiple of $a$ is in $S$.

Now suppose some other number $b$ were in $S$. Then $b$ can be written in the form $qa + r$ (for example, by the division algorithm), and since $b$ is not a multiple of $a$, $r \neq 0$. But $-qa \in S$, so $b - qa = r \in S$, which violates the assumption that $a$ is the smallest positive integer in $S$.

So $S$ consists precisely of the multiples of $a$. $_\square$

The above proof shows that any subgroup is equal to $\mathbb{Z}a$, where $a$ is the smallest integer in the subgroup. This gives the following corollary:

Let $\mathbb{Z}a + \mathbb{Z}b$ be defined as

$$\mathbb{Z}a + \mathbb{Z}b = \{m + n | m \in \mathbb{Z}a , n \in \mathbb{Z}b\},$$

i.e. the elements are the sums of an element in $\mathbb{Z}a$ and an element in $\mathbb{Z}b$. This is a subgroup of $\mathbb{Z}^{+}$. Thus

$$\mathbb{Z}a + \mathbb{Z}b = \mathbb{Z}d$$

for some integer $d$, where $d$ is the greatest common divisor of $a$ and $b$.

This also shows Bezout's identity:

for any $a,b$, there exist integers $r,s$ such that

$$ar + bs = \text{gcd}(a,b),$$

since the elements of $\mathbb{Z}a$ are $ar$ for some integer $r$, and the elements of $\mathbb{Z}b$ are $bs$ for some integer $s$.

$d$ is named the greatest common divisor because

• $d$ divides $a$ and $b$ $($as $a,b$ are elements of $\mathbb{Z}d$, since $a + 0, 0 + b \in \mathbb{Z}d$, and the elements of $\mathbb{Z}d$ are multiples of $d)$, i.e. $d$ is a common divisor of $a,b$;
• If $e$ divides $a$ and $b$, it also divides $d$ $($since if $e \mid a,b,$ then $e \mid ar + bs = d)$, i.e. there is no greater common divisor than $d$.

Similarly, $\mathbb{Z}a \cap \mathbb{Z}b = \mathbb{Z}\ell$ for some integer $\ell$. It is known as the least common multiple of $a$ and $b$. The reasons are analogous to the above analysis.

First, it is important to have a criterion for checking whether some subset is in fact a subgroup. In particular, while the axiom of associativity is inherited from $G$, both closure and the existence of identity/inverses need to be manually checked; since a given subset might not be closed, it might not contain the identity, and the inverse of an element may not be in the subset. Hence, a subgroup $H$ must satisfy the following:

• For all $a,b \in H$, $ab \in H$ as well.
• For all $a \in H$, $a^{-1} \in H$ as well.

These can be combined into a single form:

• For all $a,b \in H$, $ab^{-1} \in H$ as well.

For this reason, the identity of $G$ is necessarily the identity of $H$ as well.

Given a way to recognize subgroups, the next natural question is how to go about finding them. Fortunately, this is easy to accomplish:

Any subset $S$ of $G$ defines a subgroup formed by taking the intersection of all subgroups containing $S$. It is called the subgroup generated by $S$.

The case where $S$ is a single element is of particular interest. In this case, the subgroup generated by $S$ is a cyclic group, and its size is called the order of the element (which may be familiar from modular arithmetic).

Given a subgroup, the next natural question is to examine precisely how the subgroup "fits" into the original. Cosets provide a framework for analyzing this.

A left coset of a subgroup $H$ is the set $\{gh | h \in H\}$, where $g$ is an element of $G$. The set of all left cosets of $H$ w.r.t. $G$ is denoted by $G/H$.

A right coset of a subgroup $H$ is the set $\{hg | h \in H\}$, where $g$ is an element of $G$. The set of all right cosets of $H$ w.r.t. $G$ is denoted by $H\backslash G$. In general, right cosets are not independently very interesting. $_\square$

For example, consider the subgroup $H = \{0, 2, 4, 6\}$ of $G=\mathbb{Z}_8$ (which is an additive group). The left cosets are

$$\begin{aligned} \{0 + h | h \in H\} &= \{2 + h | h \in H\} = \{4 + h | h \in H\} = \{6 + h | h \in H\} = \{0, 2, 4, 6\}\\ \{1 + h | h \in H\} &= \{3 + h | h \in H\} = \{5 + h | h \in H\} = \{7 + h | h \in H\} = \{1, 3, 5, 7\}, \end{aligned}$$

so $G/H = \{\{0, 2, 4, 6\}, \{1, 3, 5, 7\}\}$. The right cosets are

$$\begin{aligned} \{h + 0 | h \in H\} &= \{h + 2 | h \in H\} = \{h + 4 | h \in H\} = \{h + 6 | h \in H\} = \{0, 2, 4, 6\}\\ \{h + 1 | h \in H\} &= \{h + 3 | h \in H\} = \{h + 5 | h \in H\} = \{h + 7 | h \in H\} = \{1, 3, 5, 7\}, \end{aligned}$$

so $H\backslash G = \{\{0, 2, 4, 6\}, \{1, 3, 5, 7\}\}$.

This example provides motivation for some properties of cosets:

• Every left coset has the same number of elements.
• The left cosets partition $G$; i.e. each element of $G$ appears in exactly one left coset.

Taken together, these properties combine to form Lagrange's theorem:

Lagrange's theorem:

$|G| = |H|\Big|G/H\Big|,$ where $|G|$ is the order of $G.$

Equivalently, $\Big|G/H\Big| = \frac{|G|}{|H|}.\ _\square$

Note that this also demonstrates that the order of a subgroup necessarily divides the order of the original group.

In the above example, it was also true that the left cosets and right cosets were identical; i.e. for any $g$, $gH = Hg$. This is not generally the case, but when it occurs, the subgroup $H$ is called normal. Normal subgroups are used to generate quotient groups, as $\{0, 2, 4, 6\}$ was in the above example. The subgroups of abelian groups are always normal.

• Group theory
• Abelian group
• Normal subgroup