Power of Element/Semigroup
Theorem
Let $\struct {S, \oplus}$ be a magma.
Let $a \in S$.
Let $n \in \N_{>0}$.
Let $\tuple {a_1, a_2, \ldots, a_n}$ be the ordered $n$-tuple defined by $a_k = a$ for each $k \in \N_n$.
Then:
- $\ds \bigoplus_{k \mathop = 1}^n a_k = \oplus^n a$
where:
- $\ds \bigoplus_{k \mathop = 1}^n a_k$ is the composite of $\tuple {a_1, a_2, \ldots, a_n}$ for $\oplus$
- $\oplus^n a$ is the $n$th power of $a$ under $\oplus$.
Proof
The proof will proceed by the Principle of Mathematical Induction on $\N$.
Let $T$ be the set defined as:
- $\ds T := \set {n \in \N: \bigoplus_{k \mathop = 1}^n a_k = \oplus^n a}$
First, recall the definition of the composite of $\tuple {a_1, a_2, \ldots, a_n}$ for $\oplus$:
- $\ds \bigoplus_{k \mathop = 1}^n a_k = \begin{cases} a: & n = 1 \\ \map {\oplus_m} {a_1, \ldots, a_m} \oplus a_{m + 1}: & n = m + 1 \end{cases}$
Secondly, recall the definition of the $n$th power of $a$ under $\oplus$:
- $\forall n \in \N_{>0}: \oplus^n a = \begin{cases} a & : n = 1 \\ \paren {\oplus^m a} \oplus a & : n = m + 1 \end{cases}$
Basis for the Induction
We have that:
\(\ds \bigoplus_{k \mathop = 1}^1 a_k\) | \(=\) | \(\ds a_1\) | Definition of Composite of $\paren {a_1}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds a\) | Definition of $\tuple {a_1, a_2, \ldots, a_n}$: $\forall k \in \N_n: a_k = a$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \oplus^1 a\) | Definition of Power of Element of Semigroup |
So $1 \in T$.
This is our basis for the induction.
Induction Hypothesis
It is to be shown that, if $j \in T$ where $j \ge 1$, then it follows that $j + 1 \in T$.
This is the induction hypothesis:
- $\ds \bigoplus_{k \mathop = 1}^j a_k = \oplus^j a$
It is to be demonstrated that it follows that:
- $\ds \bigoplus_{k \mathop = 1}^{j + 1} a_k = \oplus^{j + 1} a$
Induction Step
This is our induction step:
\(\ds \bigoplus_{k \mathop = 1}^{j + 1} a_k\) | \(=\) | \(\ds \map {\oplus_j} {a_1, \ldots, a_j} \oplus a_{j + 1}\) | Definition of Composite of $\tuple {a_1, \ldots, a_j, a_{j + 1} }$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\bigoplus_{k \mathop = 1}^j a_k} \oplus a_{j + 1}\) | Definition of Composite of $\tuple {a_1, \ldots, a_j}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\oplus^j a} \oplus a_{j + 1}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\oplus^j a} \oplus a\) | Definition of $\tuple {a_1, a_2, \ldots, a_{j + 1} }$: $\forall k \in \N_{j + 1}: a_k = a$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \oplus^{j + 1} a\) | Definition of Power of Element of Semigroup |
So $k \in T \implies k + 1 \in T$ and the result follows by the Principle of Mathematical Induction:
- $\ds \forall n \in \N: \bigoplus_{k \mathop = 1}^n a_k = \oplus^n a$
Notation
Let $\paren {S, \circ}$ be a semigroup.
Let $a \in S$.
Let $\circ^n a$ be the $n$th power of $a$ under $\circ$.
The usual notation for $\circ^n a$ in a general algebraic structure is $a^n$, where the operation is implicit and its symbol omitted.
In an algebraic structure in which $\circ$ is addition, or derived from addition, this can be written $n a$ or $n \cdot a$, that is, $n$ times $a$.
Thus:
- $a^1 = \circ^1 a = a$
and in general:
- $\forall n \in \N_{>0}: a^{n + 1} = \circ^{n + 1} a = \paren {\circ^n a} \circ a = \paren {a^n} \circ a$
When the operation is addition of numbers or another commutative operation derived from addition, the following symbology is often used:
- $n a = \begin{cases} a & : n = 1 \\ \paren {n - 1} a + a & : n > 1 \end{cases}$
Sometimes, for clarity, $n \cdot a$ is preferred to $n a$.
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.2$