Inverse of Product
Contents |
Theorem
Monoid
Let $\left({S, \circ}\right)$ be a monoid whose identity is $e$.
Let $x, y \in S$ be invertible for $\circ$, with inverses $x^{-1}, y^{-1}$.
Then $x \circ y$ is invertible for $\circ$, and $\left({x \circ y}\right)^{-1} = y^{-1} \circ x^{-1}$.
Generalized Result
Let $\left({S, \circ}\right)$ be a monoid whose identity is $e$.
Let $a_1, a_2, \ldots, a_n \in S$ be invertible for $\circ$, with inverses $a_1^{-1}, a_2^{-1}, \ldots, a_n^{-1}$.
Then $\left({a_1 \circ a_2 \circ \cdots \circ a_n}\right)^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$.
Group
For any $a$ and $b$ in a group $G$, $(ab)^{-1}=b^{-1}a^{-1}$.
In general:
- $\left({a_1 a_2 \cdots a_{n-1} a_n}\right)^{-1} = a_n^{-1} a_{n-1}^{-1} \cdots a_2^{-1} a_1^{-1}$
Proof
Proof for Monoid
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \left({x \circ y}\right) \circ \left({y^{-1} \circ x^{-1} }\right)\) | \(=\) | \(\displaystyle \left({\left({x \circ y}\right) \circ y^{-1} }\right) \circ x^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Associativity | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({x \circ \left({y \circ y^{-1} }\right)}\right) \circ x^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Associativity | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({x \circ e}\right) \circ x^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Behaviour of Inverse | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle x \circ x^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Behaviour of Identity | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle e\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Behaviour of Inverse |
Similarly for $\left({y^{-1} \circ x^{-1}}\right) \circ \left({x \circ y}\right)$.
$\blacksquare$
Proof of Generalized Result
Proof by induction:
We have, from above, $\left({a_1 \circ a_2}\right)^{-1} = a_2^{-1} \circ a_1^{-1}$, and (trivially) $\left({a_1}\right)^{-1} = a_1^{-1}$.
Assume that $\left({a_1 \circ a_2 \circ \cdots \circ a_k}\right)^{-1} = a_k^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$.
Then:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \left({a_1 \circ a_2 \circ \cdots \circ a_k \circ a_{k+1} }\right)^{-1}\) | \(=\) | \(\displaystyle \left({\left({a_1 \circ a_2 \circ \cdots \circ a_k}\right) \circ \left({a_{k+1} }\right)}\right)^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle a_{k+1}^{-1} \circ \left({a_1 \circ a_2 \circ \cdots \circ a_k}\right)^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle a_{k+1}^{-1} \circ \left({a_k^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle a_{k+1}^{-1} \circ a_k^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
So the assumption being true for $n = k$ implies that it is also true for $n = k + 1$.
It is also true for $n = 1$ (trivially) and $n = 2$ (proved above).
So, by the Principle of Mathematical Induction, it is true for all $n \in \N^*$.
$\blacksquare$
Proof for Group
As a group is also a monoid, then the main result applies directly.
The general result follows from the same source.
$\blacksquare$
Comment
This theorem is also known as the Socks-Shoes Property.
If one thinks of $a$ as putting on socks, $b$ as putting on shoes, $a^{-1}$ as taking off socks, and $b^{-1}$ as taking off shoes, the theorem demonstrates the order in which one must perform these actions. $ab$ would represent putting on socks followed by shoes. In order to take them off, they must be removed in reverse order, that is, $b^{-1}a^{-1}$.
Less informally it is known as the reversal rule (for group inverses).
Sources
- J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 4.6$: Theorem $\text{(iv)}$
- Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 4$: Theorem $4.4$
- Richard A. Dean: Elements of Abstract Algebra (1966): $\S 1.4$: Theorem $2$
- George McCarty: Topology: An Introduction with Application to Topological Groups (1967): Chapter $\text{II}$
- Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 28 \ (3), \ \S 28 \alpha$
- Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 35.6, \ \S 35.7$
- John F. Humphreys: A Course in Group Theory (1996): $\S 3$: Corollary $3.4$
