Is it true that an isomorphism maps elements of the same order to each other? I know that for two groups to be isomorphic each group must contain elements of the same order.  So then are elements of the same order mapped to each other?  If so, why?
 A: Yes,
because for isomorphism it must hold that $f( u \times v) = u \cdot v$.
If order of the element $a$ is $m$ then $a^m = e$
$f(a^m)= f(a\times a \times \dots \times a) = f(a) \cdot f(a) \dots f(a)=f(a)^m=f(e) = e$
Since $f(a)^m = e$ the order of $f(a)$ is less than or equal to $m$. Let us say that the order of $f(a)$ is $k$, $k \le m$
$f$ is an isomorphism, therefore the same argument holds for $f^{-1}$
$f^{-1}(f(a)^k)= f^{-1}(f(a) \cdot f(a) \dots f(a))=f^{-1}(f(a)) \cdot f^{-1}(f(a)) \dots f^{-1}(f(a)) = a \cdot a \dots a = a^k = f^{-1}(e) = e = f(e)$
$a^k=e$ therefore $k \ge m$. But also $k \le m$, so $k=m$
Note
$f(e) = e$ because $e\times e= e$ therefore $f(e)\cdot f(e)= f(e)$. Therefore $f(e)$ must be the neutral element.
A: "Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed." – PiccolMan
The order of any member of a finite group must have finite order.  Let G be a group of order n and a member of G.  Look at all a, $a^2$, "$a^3$", $a^4$, ..., $a^{n+1}$.  Since there are only n different members of G, at two of those n+1 values must be the same.  Suppose $a^i= a^j$ with i>j.  Multiplying on both sides by $a^{-j}$, $a^ja^i= a^{i-j}= e$.
A: If a group $G$ contains an element of infinite order then $G$ has infinite order.
Proof: Let $a \in G$ with $|a|=\infty$. Consider the cyclic group generated by $a$ denoted $\langle a \rangle$. The order of a cyclic group is the order of a generator of that group, so $|\langle a \rangle|=|a|=\infty$. Thus $\langle a \rangle$ has infinite order. And $\langle a \rangle$ is a subgroup of $G$, so we must have $|\langle a \rangle|<|G|$. $\blacksquare$
