Inverse of Algebraic Structure Isomorphism is Isomorphism
Theorem
Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be algebraic structures.
Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be a mapping.
Then $\phi$ is an isomorphism iff $\phi^{-1}: \left({T, *}\right) \to \left({S, \circ}\right)$ is also an isomorphism.
Proof
Let $\phi$ be an isomorphism.
Then by definition $\phi$ is a bijection.
Thus $\exists \phi^{-1}$ such that $\phi^{-1}$ is also a bijection from Bijection iff Inverse is Bijection.
That is:
- $\exists \phi^{-1}: \left({T, *}\right) \to \left({S, \circ}\right)$
It follows that:
| \(\displaystyle \) | \(\displaystyle \forall s \in S, t \in T:\) | \(\displaystyle \) | \(\displaystyle \) | \(\) | \(\displaystyle \phi \left({s}\right) = t \iff \phi^{-1} \left({t}\right) = s\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Inverse Element of Bijection | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\implies\) | \(\displaystyle \phi \left({s_1 \circ s_2}\right) = t_1 * t_2\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Morphism Property | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\implies\) | \(\displaystyle \phi^{-1} \left({t_1 * t_2}\right) = s_1 \circ s_2 = \phi^{-1} \left({t_1}\right) \circ \phi^{-1} \left({t_2}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Inverse Element of Bijection |
So $\phi^{-1}: \left({T, *}\right) \to \left({S, \circ}\right)$ is a homomorphism, and (from above) bijective, and thus an isomorphism.
Applying the same result in reverse, we have that if $\phi^{-1}: \left({T, *}\right) \to \left({S, \circ}\right)$ is an isomorphism, then $\left({\phi^{-1}}\right)^{-1}: \left({S, \circ}\right) \to \left({T, *}\right)$ is an isomorphism.
But by Inverse of Inverse of Bijection, $\left({\phi^{-1}}\right)^{-1} = \phi$ and hence the result.
$\blacksquare$
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Sources
- J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 7.2$: Lemma $\text{(v)}$
- Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 6$: Theorem $6.1: \ 2^\circ$
- B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra (1970): $\S 2.2$
- Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 62$
- Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): Exercise $8.3 \ \text{(ii)}$
