Properties of Isomorphism

Theorem 1:
If isomorphism exists between two groups, then the identities correspond, i.e. if $f:G \to G'$ is an isomorphism and $e,e'$ are respectively the identities in $G,\,G'$, then $f\left( e \right) = e'$.

Theorem 2:
If isomorphism exists between two groups, then the identities correspond, i.e. if $f:G \to G'$ is an isomorphism and $f\left( a \right) = a'$, where $a \in G,\,\,\,a' \in G'$ then $f\left( {{a^{ - 1}}} \right) = {a'^{ - 1}} = {\left[ {f\left( a \right)} \right]^{ - 1}}$.

Theorem 3:
In an isomorphism the order of an element is preserved, i.e. if $f:G \to G'$ is an isomorphism, and the order of $a$ is $n$, then the order of $f\left( a \right)$ is also $n$.

Proof:
As $f\left( a \right) = a'$, then we have $f\left( {a \cdot a} \right) = f\left( a \right) \cdot f\left( a \right) = a' \cdot a' = {a'^2}$ and in general we can write it as $f\left( {{a^n}} \right) = {a'^n}$.

But $f\left( {{a^n}} \right) = f\left( e \right) = e'$, by using the statement of Theorem 1,

therefore ${a'^n} = e'$. Also ${a'^m} \ne e'$ for $m < n$, i.e. $o\left( {a'} \right) = n$.

It follows that the order of an element of $G$, if finite, is equal to the order of its image in $G'$. If the order of $a$ is infinite, we can similarly show that the order of $a'$ cannot be finite.

Theorem 4:
The relation of isomorphism in the set of groups is an equivalence relation.