Skip to main content

Section 7.2 Subgroups

Subsection 7.2.1 Subgroups

A subgroup \(H\) of a group \(G\) is a subset of \(G\) that is itself a group under the operation of \(G\text{.}\) We write \(H \leq G\text{.}\) Every group \(G\) has the trivial subgroup \(\{e\}\) and the improper subgroup \(G\) itself; any other subgroup is proper.

Proof.

If \(H \leq G\) and \(g, h \in H\text{,}\) then \(h^{-1} \in H\) and so \(gh^{-1} \in H\) by closure.
Conversely, suppose \(H\) is nonempty and closed under \(gh^{-1}\text{.}\) Taking \(g = h\) gives \(e \in H\text{.}\) Taking \(g = e\) gives \(h^{-1} \in H\) for every \(h \in H\text{.}\) Finally, for \(h_1, h_2 \in H\) we have \(h_2^{-1} \in H\text{,}\) so \(h_1 (h_2^{-1})^{-1} = h_1 h_2 \in H\text{.}\) Thus \(H\) contains \(e\text{,}\) is closed under inverses, and is closed under the operation; associativity is inherited from \(G\text{.}\)

Example 7.2.2.

The positive real numbers \({\mathbb R}^{+}\) form a subgroup of \({\mathbb R}^{\ast}\text{:}\) \(1 \in {\mathbb R}^{+}\text{,}\) and if \(a, b \gt 0\) then \(a/b \gt 0\text{.}\)

Example 7.2.3.

The unit circle \({\mathbb T} = \{ z \in {\mathbb C} : |z| = 1 \}\) is a subgroup of \({\mathbb C}^{\ast}\text{:}\) \(|1| = 1\text{,}\) and if \(|z| = |w| = 1\) then \(|z w^{-1}| = |z|/|w| = 1\text{.}\)

Example 7.2.4.

For each \(n\text{,}\) the complex numbers satisfying \(z^n = 1\) are the nth roots of unity,
\begin{equation*} z = \cos\left( \frac{2 k \pi}{n} \right) + i \sin\left( \frac{2 k \pi}{n} \right), \qquad k = 0, 1, \ldots, n-1\text{.} \end{equation*}
They form a finite subgroup \(\mu_n = \{ z \in {\mathbb C} : z^n = 1 \}\) of \({\mathbb T}\) (hence of \({\mathbb C}^{\ast}\)) of order \(n\text{:}\) if \(z^n = w^n = 1\text{,}\) then \((zw^{-1})^n = z^n (w^n)^{-1} = 1\text{.}\) Writing \(\omega = \cos(2\pi/n) + i \sin(2\pi/n)\text{,}\) every element of \(\mu_n\) is a power of \(\omega\text{,}\) so \(\mu_n\) is in fact cyclic of order \(n\text{.}\) A generator of \(\mu_n\) is called a primitive nth root of unity.

Subsection 7.2.2 Cosets and Lagrange’s Theorem

Let \(H \leq G\text{.}\) For \(g \in G\text{,}\) the left coset of \(H\) with representative \(g\) is \(gH = \{gh : h \in H\}\text{;}\) right cosets \(Hg\) are defined similarly.

Proof.

If \(g_1 H = g_2 H\text{,}\) then \(g_2 = g_2 e \in g_2 H = g_1 H\text{,}\) so \(g_2 = g_1 h\) for some \(h \in H\text{,}\) and \(g_1^{-1} g_2 = h \in H\text{.}\) Conversely, if \(g_1^{-1} g_2 = h \in H\text{,}\) then \(g_2 = g_1 h\text{,}\) and for any \(h' \in H\text{,}\) \(g_2 h' = g_1 (h h') \in g_1 H\) and \(g_1 h' = g_2 (h^{-1} h') \in g_2 H\text{;}\) hence \(g_1 H = g_2 H\text{.}\)

Proof.

Every \(g \in G\) lies in the coset \(gH\text{,}\) so the cosets cover \(G\text{.}\) If \(g_1 H \cap g_2 H \neq \emptyset\text{,}\) say \(a = g_1 h_1 = g_2 h_2\text{,}\) then \(g_1^{-1} g_2 = h_1 h_2^{-1} \in H\text{,}\) so \(g_1 H = g_2 H\) by Lemma 7.2.5. Hence distinct cosets are disjoint.
The number of left cosets of \(H\) in \(G\) is the index \([G:H]\text{.}\)

Proof.

It is onto by definition of \(gH\text{.}\) If \(gh_1 = gh_2\text{,}\) then \(h_1 = h_2\) by left cancellation, so it is one-to-one.
For \(a \in G\text{,}\) the set
\begin{equation*} \langle a \rangle = \{ a^k : k \in {\mathbb Z} \} \end{equation*}
is a subgroup of \(G\text{:}\) for \(a^m, a^n \in \langle a \rangle\) we have \(a^m (a^n)^{-1} = a^{m-n} \in \langle a \rangle\text{,}\) so \(\langle a \rangle \leq G\) by Proposition 7.2.1. We call \(\langle a \rangle\) the cyclic subgroup generated by \(a\text{,}\) and \(a\) a generator of \(\langle a \rangle\text{.}\) If \(G = \langle a \rangle\) for some \(a \in G\text{,}\) we call \(G\) a cyclic group.
The order of \(a \in G\) is the smallest positive integer \(n\) such that \(a^n = e\text{,}\) written \(|a| = n\text{;}\) if no such \(n\) exists, the order of \(a\) is infinite.
The two results below make use of the Division Algorithm for integers, Theorem B.2.1 of Section B.2.

Proof.

The elements \(e, a, a^2, \ldots, a^{n-1}\) are distinct: if \(a^i = a^j\) with \(0 \leq i \lt j \lt n\text{,}\) then \(a^{j-i} = e\) with \(0 \lt j - i \lt n\text{,}\) contradicting that \(n\) is the smallest positive power sending \(a\) to \(e\text{.}\) So \(\langle a \rangle\) contains at least \(n\) elements.
Conversely, every element of \(\langle a \rangle\) is of this form: given \(a^k\text{,}\) write \(k = nq + r\) with \(0 \leq r \lt n\) by the Division Algorithm. Then \(a^k = (a^n)^q a^r = a^r\text{,}\) so \(a^k \in \{ e, a, \ldots, a^{n-1} \}\text{.}\) Hence \(\langle a \rangle = \{ e, a, \ldots, a^{n-1} \}\) has exactly \(n\) elements.

Proof.

By the Division Algorithm, \(k = nq + r\) for some integers \(q, r\) with \(0 \leq r \lt n\text{.}\) Then
\begin{equation*} e = a^k = a^{nq+r} = (a^n)^q a^r = a^r. \end{equation*}
Since \(n\) is the smallest positive integer with \(a^n = e\) and \(0 \leq r \lt n\text{,}\) we must have \(r = 0\text{.}\) Hence \(k = nq\text{,}\) so \(n\) divides \(k\text{.}\)

Proof.

By Theorem 7.2.9, \(|\langle g \rangle|\) is the order of \(g\text{.}\) Applying Theorem 7.2.8 to \(H = \langle g \rangle\) shows this divides \(|G|\text{.}\) Writing \(|G| = k \cdot |g|\text{,}\) we get \(g^{|G|} = (g^{|g|})^k = e^k = e\text{.}\)
PrevTopNext