I have been going through Gallian's Abstract Algebra Book and ran across a problem that showed how U(8), the group of integers less than 8 that are relatively prime to 8, under multiplication mod 8 is not isomorphic to U(10), the group of integers less than 10 that are relatively prime to 10, under multiplication mod 10. So although the two groups have an equal number of elements (U(8)=1,3,5,7 and U(10)=1,3,7,9), any bijective mapping doesn't preserve structure as required by an isomorphism.
The proof is by counterexample, and centers on the fact that the square of any element in U(8) is the unity. So when mapping from U(10) to U(8), we see two distinct inputs leading to the same output, violating the condition of injectivity.
For example, if the mapping is f:U(10)->U(8), then f(ab mod 10)=f(a)f(b) mod 8 by the definition of an isomorphism. But take a,b=3 and a,b=1 and you'll see that both 9 and 1 map to 1.
My question is if a group isomorphism somehow depends on the order of the elements in the group. So in the above case, we don't see an isomorphism because every element in U(8) is of order 2, but not every element in U(10) is of order 2.