In practice, it is much easier to determine that two groups are not isomorphic than to show that they are. To show that they're not isomorphic, it suffices to find a single different property.
For example, there is no element of the Klein $4$ group that has order $4$. The largest order of an element is $2$. The cyclic group $\mathbb{Z}/4\mathbb{Z}$ has an element of order $4$. Thus these two groups are not isomorphic.
Other useful aspects that can frequently be used to show that two groups are not isomorphic include cardinality, conjugacy classes, sizes and numbers of subgroups, among others.
But it is still possible to find pairs of groups that share practically all easily computable aspects noted above, but still aren't isomorphic. This is a pretty annoying problem and very slow to compute for general finite groups. For general infinite groups, the isomorphism problem is undecidable! (M.O. Rabin, Recursive unsolvability of group theoretic problems. Ann. Math 67:172--194, 1958).
The abelianess without any appeal to orders of elements, Cauchy's Theorem or Lagrange's or type of group: assume that $G$ has $4$ elements and is not abelian. Then we can find two non-identity elements $a,b$ that do not commute, so $ab \neq ba$. Note that this implies $ab \notin \{e,a,b\}$. Because $G$ is closed under the group operation we must have $G=\{e,a,b,ab\}$. Now $ba$ belongs to this set. Since $a$ and $b$ do not commute $ba \notin \{e,a,b\}$. Hence we must have $ab=ba$, a contradiction. So $G$ is abelian after all.
OK, so we have $\{e,a,b,ab\}$ is an abelian group of $4$ elements, so each of them does not equal one of the others. So were is $a^2$? Assume for the moment that $a^2 \neq e$. Can $a^2=a$? No, since then it would follow $a=e$. Can it be $a^2=b$? Yes, and it that case $ab=a^3$ and the group looks like $\{e,a,a^2,a^3\}$. Observe that $a^4$ must be the identity since if $a^4=a$, $a^4=a^2$ or $a=a^3$, the set reduces to less than $4$ elements. So the group is cyclic of order $4$ generated by $a$, in this case.
Similarly, if $b^2=e$, we would arrive at the group $\{e,b,b^2,b^3\}$, which is of course again cyclic of order $4$ and isomorphic to the one we already found.
We are left with the case where $a^2=e=b^2$. But then (using abelianess) $(ab)^2=abab=aabb=a^2b^2=e.e=e$ and now all elements of the group have order $2$. This one is isomorphic to $C_2 \times C_2$, a direct product of two groups of order $2$, also called the Klein $4$-group $V_4$.
So you see in this small case everything can still be figured out without using more sophisticated theorems.
Best Answer
Note that if $a$ is the generator of a cyclic group,i.e. $G=\langle a\rangle$, then $\phi(G)=\langle \phi(a)\rangle$, for any homomorphism $\phi$.