2.6 Powers of elements
Suppose \(M\) is a monoid with identity element \(e\), \(m\in M\) and \(k\in \mathbb {N}\). Suppose further that the binary operation on \(M\) is written multiplicatively as \((m_1,m_2)\mapsto m_1m_2\).
We define an element \(m^k\in M\) inductively as follows.
We set \(m^0 = e\).
Assuming \(m^j\) is already defined for \(j\leq k\), we set \(m^k = m m^{k-1}\).
So \(m^0 = e\), \(m^1 = me = m\), \(m^2 = m m^1 = m m\), \(m^3 = m m^2 = m m m\), etc. Intuitively \(m^k\) is just the product of \(m\) with itself \(k\) times.
Suppose \(M\) is a monoid and \(m\in M\) as in Definition 2.40. Let \(i,j\in \mathbb {N}\). Then
\(m^{i+j} = m^im^j\).
\((m^i)^j = m^{ij}\).
Exercise.
If the binary operation on \(M\) is written additively, then, we write \(km\) for \(m^k\).
Now suppose that \(G\) is a group with identity element \(e\) and binary operation \((g_1,g_2)\mapsto g_1g_2\).
Suppose \(g\in G\) and \(n\) is a positive integer. We set \(g^{-n} := (g^{-1})^n\).
So, now for \(g\in G\), we have \(g^n\) defined for all integers \(n\).
Suppose \(G\) is a group, \(g\in G\) and \(i,j\in \mathbb {Z}\). Then
\(g^{i+j} = g^ig^j\).
\((g^i)^j = g^{ij}\).
Suppose \(G\) is a group and \(g\in G\). Then \(\langle g\rangle = \{ g^n:n\in \mathbb {Z}\} \).
Set \(H = \langle g\rangle \). By Theorem 2.39, the elements of \(H\) are exactly the element of the form \(h = g^{i_1}g^{i_2} \cdots g^{i_k}\) where \(i_1,\ldots , i_k\in \{ -1,0,1\} \). So any such element can be written as \(h = g^{i_1+\cdots i_k}\). This shows that \(H \subseteq \{ g^n: n\in \mathbb {Z}\} \).
On the other hand, for any \(n\in \mathbb {Z}\), we can find a positive integer \(k\) and \(i_1,\ldots , i_k\in \{ -1,0,1\} \) such that \(n = i_1 + \cdots + i_k\). This shows that \(\{ g^n: n\in \mathbb {Z}\} \subseteq H\).