Index Laws for Monoids

Theorem
These results are an extension of the results in Index Laws for Semigroups in which the domain of the indices is extended to include all integers.

Let $$\left ({S, \odot}\right)$$ be a monoid whose identity is $$e_S$$.

Let $$a \in S$$ be invertible for $$\odot$$.

Let $$n \in \mathbb{N}$$.

Let $$a^n = \odot^n \left({a}\right)$$ be defined as in Power of an Element:

$$ a^n = \begin{cases} e_S : & n = 0 \\ a^x \odot a : & n = x + 1 \end{cases} $$

... that is, $$a^n = a \odot a \odot \cdots \left({n}\right) \cdots \odot a = \odot^n \left({a}\right)$$.

Also, for each $$n \in \mathbb{N}$$ we can define:

$$a^{-n} = \left({a^{-1}}\right)^n$$

(Note that the notation $$a^n$$ and $$\odot^n \left({a}\right)$$ mean the same thing. The former is more compact and readable, but the latter is more explicit and can be more useful when more than one structure is under consideration.)

Then we have the following results:

Negative Index
Let $$a \in S$$ be invertible for $$\odot$$. Then:

$$\forall n \in \mathbb{Z}: \left({a^n}\right)^{-1} = a^{-n} = \left({a^{-1}}\right)^n$$

Sum of Indices
Let $$a \in S$$ be invertible for $$\odot$$. Then:

$$\forall m, n \in \mathbb{Z}: a^{n+m} = a^n \odot a^m$$

Product of Indices
Let $$a \in S$$ be invertible for $$\odot$$. Then:

$$\forall m, n \in \mathbb{Z}: a^{n m} = \left({a^m}\right)^n = \left({a^n}\right)^m$$

Product of Commutative Elements
Let $$a, b \in S$$ be invertible elements for $$\odot$$ that also commute. Then:

$$\forall n \in \mathbb{Z}: \left({a \odot b}\right)^n = a^n \odot b^n$$

Commutation of Powers
Let $$a, b \in S$$ be invertible elements for $$\odot$$ that also commute. Then:

$$\forall m, n \in \mathbb{Z}: a^m \odot b^n = b^n \odot a^m$$

Negative Index
We have $$a^0 = e$$ so it follows trivially that $$a^{-0} = \left({a^{-1}}\right)^0$$.


 * From the generalized inverse of product, we have:

$$\left({a_1 \odot a_2 \odot \cdots \odot a_n}\right)^{-1} = a_n^{-1} \odot \cdots \odot a_2^{-1} \odot a_1^{-1}$$

where $$a_1, a_2, \ldots, a_n \in S$$ are all invertible for $$\circ$$.

So we can put $$a_1, a_2, \ldots, a_n = a$$ and we have that


 * $$a^n$$ is invertible for all $$n \in \mathbb{N}$$;
 * $$\forall n \in \mathbb{N}: \left({a^n}\right)^{-1} = \left({a^{-1}}\right)^n$$.


 * From the above, we have $$a^{-n} = \left({a^{-1}}\right)^n$$. Thus:

Similarly, if $$a$$ is invertible then $$a^{-1}$$ is also invertible, so we also have $$\odot^{-n} \left({a^{-1}}\right) = \odot^n \left({\left({a^{-1}}\right)^{-1}}\right)$$. Thus:

Thus the result holds for all $$n \in \mathbb{Z}$$.

Sum of Indices
For each $$c \in S$$ which is invertible for $$\odot$$, we define the mapping $$g_c: \mathbb{Z} \to S$$ as:

$$\forall n \in \mathbb{Z}: g_c \left({n}\right) = \odot^n \left({c}\right)$$

By the Index Law for Negative Index above, $$g_a \left({n}\right)$$ is invertible for all $$n \in \mathbb{Z}$$.

By Recursive Mapping to Semigroup, the restriction of $$g_a$$ to $$\mathbb{N}$$ is a homomorphism from $$\left({\mathbb{N}, +}\right)$$ to $$\left({S, \odot}\right)$$.

From Recursive Mapping to Semigroup, $$g_a \left({0}\right)$$ is the identity for $$\odot$$ by definition.

Hence, by the Extension Theorem for Homomorphisms, there is a unique homomorphism $$h_a: \left({\mathbb{N}, +}\right) \to \left({S, \odot}\right)$$ which coincides in $$\mathbb{N}$$ with $$g_c$$.

But by Homomorphism with Identity Preserves Inverses:

Hence $$h_a = g_a$$ and so $$g_a$$ is a homomorphism and so the result follows.

Product of Indices

 * Let $$m \in \mathbb{N}, c = a^m, d = \left({a^{-1}}\right)^m$$.

Let $$g_c$$ be as defined in the proof of Index Law for Sum of Indices.

Let $$h: \mathbb{Z} \to \mathbb{Z}$$ be the mapping defined as:

$$\forall z \in \mathbb{Z}: h \left({z}\right) = z m$$

Then:

By Index Law for Sum of Indices and Naturally Ordered Semigroup Power Law, $$g_a \circ h$$ and $$g_c$$ are homomorphisms from $$\mathbb{Z}$$ to $$S$$ which coincide on $$\mathbb{N}$$.

So by the Extension Theorem for Homomorphisms, $$g_a \circ h = g_c$$.

Therefore:

$$\forall n \in \mathbb{Z}, m \in \mathbb{N}: a^{n m} = \left({a^m}\right)^n$$

Also:

So, by the same reasoning as before, $$g_{a^{-1}} \circ h = g_d$$.

Therefore:

$$\forall n \in \mathbb{Z}, m \in \mathbb{N}: a^{n \left({-m}\right)} = \left({a^{-m}}\right)^n$$

Thus:

$$\forall n, m \in \mathbb{Z}: a^{n m} = \left({a^m}\right)^n$$

As $$n m = m n$$, the result follows.

Product of Commutative Elements
From Semigroup Cancellable Commutative, this holds if $$n \ge 0$$.

Since $$a$$ and $$b$$ commute, then so do $$a^{-1}$$ and $$b^{-1}$$ by Commutation with Inverses: Theorem 2.

Hence, if $$n > 0$$:

The result follows.

Commutation of Powers
By Powers of Commutative Elements, if $$m > 0$$ and $$n > 0$$ then $$a^m$$ commutes with $$b^n$$.

By Commutation with Inverses: Theorem 1, again if $$m > 0$$ and $$n > 0$$ then $$a^m$$ commutes with $$\left({b^n}\right)^{-1} = b^{-n}$$.

Similarly $$b^n$$ commutes with $$a^{-m}$$.

But as $$a^{-m}$$ commutes with $$b^n$$, it also commutes with $$\left({b^n}\right)^{-1} = b^{-n}$$, again by Commutation with Inverses: Theorem 1.

Hence the result.