Simple question: does it suffice to compare the element orders of 2 groups to conclude they're not isomorphic I'm currently studying some basics of group theory and earlier I had to make a proof saying that if $G$ and $H$ are both groups and $F:G\to H$ is an injective function then there applies for all  $g\in$ G that $|g| = |F(g)|$. 
So now there's an exercise that's asking me to compare groups and prove that they're not isomorphic. So I thought I'd use the fact that some of them are abelian and others aren't and the fact that isomorphism implies monomorphism (injective), so the statement of the proof I mentioned above counts. But now I'm not sure that this counts since the exercise is terribly easy now... Am I doing something wrong? If so what's that? 
Image containing what I did and the groups in question:
Here you can see how I wrote down all the orders for every element of the groups
 A: Any property you can find in one group but not in the other (at least which isn't tied to the specific names of the elements) makes the groups not isomorphic.
Different number of elements? Different number of elements of order $24$? One of them abelian and not the other? Different number of order $9$ subgroups? Different number of index $3$ subgroups? One has a subgroup isomorphic to some third group and the other one doesn't? One of them has a subgroup of order $18$ generated by two elements but the other group doesn't? Yes any one of these means that the groups are not isomorphic.
How many such questions you have to ask to prove that two groups are isomorphic, on the other hand, is as far as I know an open problem.
A: This has been solved all at MSE here, you will find it easily. Here is the summary:
1.) The groups $D_4$ and $Q_8$ both have $8$ elements, but they are not isomorphic, because every proper subgroup of $Q_8$ is cyclic, which is not true for $D_4$.
2.) The abelian groups are pairwise non-isomorphic, because of the order of elements. This is what you wanted. I find this a good way of proving that they are not isomorphic. You also could have said, that one group is cyclic, the others not.
3.) An abelian group cannot be isomorphic to a non-abelian one; both $Q_8$ and $D_4$ are non-abelian.
Edit: To the title question there is a counterexample. It does not suffice in general, for two finite groups, that all the element orders coincide, see here:
If I know the order of every element in a group, do I know the group?
