Subgroups and cosets

Section 2.3 Subgroups and cosets

Subsection 2.3.1

Definition 2.3.1. Subgroups and cosets.

A subset \(H\) of a group \(G\) is called a subgroup of \(G\) if \(H\) itself is a group under the group operation of \(G\) restricted to \(H\text{.}\) We write \(H\leq G\) to indicate that \(H\) is a subgroup of \(G\text{.}\) Given a subgroup \(H\) of \(G\text{,}\) and given an element \(g\) in \(G\text{,}\) the set

\begin{equation*} gH := \{gh \colon h\in H\} \end{equation*}

is called a (left) coset of \(H\text{.}\) The set of all cosets of \(H\) is denoted \(G/H\text{.}\)

\begin{equation*} G/H := \{gH\colon g\in G\} \end{equation*}

Checkpoint 2.3.2.

Consider \(D_4\) as described in Checkpoint 2.1.6.

\begin{equation*} D_4=\{R_0,R_{1/4},R_{1/2},R_{3/4},F_H,F_V,F_D,F_{D'}\} \end{equation*}

Is the subset \(\{R_0,R_{1/4},R_{1/2},R_{3/4}\}\) of rotations a subgroup of \(D_4\text{?}\) Why or why not?
Is the subset \(\{F_H,F_V,F_D,F_{D'}\}\) of reflections a subgroup of \(D_4\text{?}\) Why or why not?

Answer.

Yes. The composition of any two rotations is a rotation, and every rotation has an inverse that is also a rotation.
No. Just observe that \(F_H^2=R_0\) is not a reflection. The group operation on \(D_4\) does not restrict properly to the subset of reflections.

Checkpoint 2.3.3.

Find \(G/H\) for \(G=S_3\text{,}\) \(H=\{e,(12)\}\text{.}\)

Answer.

\begin{align*} G/H \amp =\{eH, (12)H, (13)H, (23)H, (123)H, (132)H\}\\ \amp = \{\{e,(12)\},\{(12),e\},\{(13),(123)\},\{(23),(132)\},\{(13),(123)\},\{(132),(23)\}\}\\ \amp = \{H,\{(13),(123)\},\{(23),(132)\}\} \end{align*}

Proposition 2.3.4. Subgroup tests.

Let \(H\) be a subset of a group \(G\text{.}\) The following are equivalent.

\(H\) is a subgroup of \(G\)
(2-step subgroup test) \(H\) is nonempty, \(ab\) is in \(H\) for every \(a,b\) in \(H\) (\(H\) is closed under the group operation), and \(a^{-1}\) is in \(H\) for every \(a\) in \(H\) (\(H\) is closed under group inversion)
(1-step subgroup test) \(H\) is nonempty and \(ab^{-1}\) is in \(H\) for every \(a,b\) in \(H\)

Proposition 2.3.5. Subgroup generated by a set of elements.

Let \(S\) be a nonempty subset of a group \(G\text{,}\) and let \(S^{-1}\) denote the set \(S^{-1}=\{s^{-1}\colon s\in S\}\) of inverses of elements in \(S\text{.}\) We write \(\langle S\rangle\) to denote the set of all elements of \(G\) of the form

\begin{equation*} s_1s_2\cdots s_k \end{equation*}

where \(k\) ranges over all positive integers and each \(s_i\) is in \(S\cup S^{-1}\) for \(1\leq i\leq k\text{.}\) The set \(\langle S\rangle\) is a subgroup of \(G\text{,}\) called the subgroup generated by the set \(S\) , and the elements of \(S\) are called the generators of \(\langle S\rangle\text{.}\)

Comment on notational convention: If \(S=\{s_1,s_2,\ldots,s_k\}\) is finite, we write \(\langle s_1,s_2,\ldots,s_k\rangle\) for \(\langle S\rangle\text{,}\) instead of the more cumbersome \(\langle \{s_1,s_2,\ldots,s_k\}\rangle\text{.}\)

Observation 2.3.6.

If \(G\) is a cyclic group with generator \(g\text{,}\) then \(G=\langle g\rangle\text{.}\)

Checkpoint 2.3.7.

Show that \(\langle S\rangle\) is indeed a subgroup of \(G\text{.}\) How would this fail if \(S\) were empty?

Checkpoint 2.3.8.

Find \(\langle F_H,F_V\rangle\subseteq D_4\text{.}\)
Find \(\langle 6,8\rangle \subseteq \Z\text{.}\)

Answer.

\(\displaystyle \langle F_H,F_V\rangle=\{R_0,R_{1/2},F_H,F_V\}\)
\(\displaystyle \langle 6,8\rangle =\langle 2\rangle = 2\Z\)

Proposition 2.3.9. Cosets as equivalence classes.

Let \(G\) be a group and let \(H\) be a subgroup of \(G\text{.}\) Let \(\sim_H\) be the relation on \(G\) defined by \(x\sim_H y\) if and only if \(x^{-1}y \in H\text{.}\) The relation \(\sim_H\) is an equivalence relation on \(G\text{,}\) and the equivalence classes are the cosets of \(H\text{,}\) that is, we have \(G/\!\!\sim_H= G/H\text{.}\)

Corollary 2.3.10. Cosets as a partition.

Let \(G\) be a group and let \(H\) be a subgroup of \(G\text{.}\) The set \(G/H\) of cosets of \(H\) form a partition of \(G\text{.}\)

Exercises 2.3.2 Exercises

1.

Prove Proposition 2.3.4.

2.

Find all the subgroups of \(S_3\text{.}\)

Answer.

In the "list of values" permutation notation of Checkpoint 2.1.2, the subgroups of \(S_3\) are \(\{[1,2,3]\}\text{,}\) \(\{[1,2,3],[2,1,3]\}\text{,}\) \(\{[1,2,3],[1,3,2]\}\text{,}\) \(\{[1,2,3],[3,2,1]\}\text{,}\) \(\{[1,2,3],[2,3,1],[3,1,2]\}\text{,}\) and \(S_3\text{.}\) In cycle notation, the subgroups of \(S_3\) (in the same order) are \(\{e\}\text{,}\) \(\{e,(12)\}\text{,}\) \(\{e,(23)\}\text{,}\) \(\{e,(13)\}\text{,}\) \(\{e,(123),(132)\}\text{,}\) \(S_3\text{.}\)

3.

Find all the cosets of the subgroup \(\{R_0,R_{1/2}\}\) of \(D_4\text{.}\)

4. Subgroups of \(\Z\) and \(\Z_n\).

Let \(H\) be a subgroup of \(\Z\text{.}\) Show that either \(H=\{0\}\) or \(H=\langle d\rangle\text{,}\) where \(d\) is the smallest positive element in \(H\text{.}\)
Let \(H\) be a subgroup of \(\Z_n\text{.}\) Show that either \(H=\{0\}\) or \(H=\langle d\rangle\text{,}\) where \(d\) is the smallest positive element in \(H\text{.}\)
Let \(n_1,n_2,\ldots,n_r\) be positive integers. Show that
\begin{equation*} \langle n_1,n_2,\ldots,n_r \rangle = \langle \gcd(n_1,n_2,\ldots,n_r)\rangle. \end{equation*}

Hint.
Suggestion: Do the case \(r=2\) first.

Consequence of this exercise: The greatest common divisor \(\gcd(a,b)\) of integers \(a,b\) is the smallest positive integer of the form \(sa+tb\) over all integers \(s,t\text{.}\) Two integers \(a,b\) are relatively prime if and only if there exist integers \(s,t\) such that \(sa+tb=1\text{.}\)

5. Centralizers, Center of a group.

The centralizer of an element \(a\) in a group \(G\text{,}\) denoted \(C(a)\text{,}\) is the set

\begin{equation*} C(a) = \{g\in G\colon ag=ga\}. \end{equation*}

The center of a group \(G\text{,}\) denoted \(Z(G)\text{,}\) is the set

\begin{equation*} Z(G) = \{g\in G\colon ag=ga \;\; \forall a\in G\}. \end{equation*}

Show that the centralizer \(C(a)\) of any element \(a\) in a group \(G\) is a subgroup of \(G\text{.}\)
Show that the center \(Z(G)\) of a group \(G\) is a subgroup of \(G\text{.}\)

6. The order of a group element.

Let \(g\) be an element of a group \(G\text{.}\) The order of \(g\text{,}\) denoted \(|g|\text{,}\) is the smallest positive integer \(n\) such that \(g^n=e\text{,}\) if such an integer exists. If there is no positive integer \(n\) such that \(g^n=e\text{,}\) then \(g\) is said to have infinite order. Show that, if the order of \(g\) is finite, say \(|g|=n\text{,}\) then

\begin{equation*} \langle g \rangle = \{g^0,g^1,g^2,\ldots,g^{n-1}\}\text{.} \end{equation*}

Consequence of this exercise: If \(G\) is cyclic with generator \(g\text{,}\) then \(|G|=|g|\text{.}\)

7. Cosets of a subgroup partition the group, Lagrange’s Theorem.

Prove Proposition 2.3.9.
Now suppose that a group \(G\) is finite. Show that all of the cosets of a subgroup \(H\) have the same size.
Prove the following.
Lagrange’s Theorem.

If \(G\) is a finite group and \(H\) is a subgroup, then the order of \(H\) divides the order of \(G\text{.}\)

Hint.

For part (b), let \(aH,bH\) be cosets. Show that the function \(aH\to bH\) given by \(x\to ba^{-1}x\) is a bijection.

8. Consequences of Lagrange’s Theorem.

Show that the order of any element of a finite group divides the order of the group.
Let \(G\) be a finite group, and let \(g\in G\text{.}\) Show that \(g^{|G|}=e\text{.}\)
Show that a group of prime order is cyclic.

Prev Top Next