# 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.

jinsong li

May 2@ 1:49 pmTheorem2: the “identities” should be “inverses”