Isomorphism $f$ preserves the order of an element?

Let's say $f$ is an isomorphism $f:G \rightarrow G'$, where $G$ and $G'$ are multiplicative groups.

Then for $x\in G$, if $f(x) = x' \in G'$. Do we have always have $\text{ord}(x) = \text{ord}(x')$? Why?

Solution 1:

Put $y = f(x)$. If $o(x) = n$, then we have $y^n = f(x)^n = f(x^n) = 1$, so that $o(y)\le n$. If $m < n$ and $1 = y^m = f(x)^m = f(x^m)$, then applying $f^{-1}$ to both sides of this equation and noting that isomorphisms send $1\mapsto 1$, we have $1 = x^m$, absurd.

Solution 2:

"Isomorphism" means that the elements of $G$ obtain new names taken from the set $G'$, but everything operationwise stays the same. It is therefore innate in the very essence of "isomorphism" that the claim you are told to prove is true. Any "formal proof" would make the claim less believable.