In general, what techniques can be used to show that 2 groups are not isomorphic? Say I have 2 groups $G$ and $H$, what techniques can be used to show that they are not isomorphic?
A simple one I can think of is proving that their order is different, thus showing there cannot be a bijection in between the 2. However I am interested in other approaches as well.

The Group Isomorphism problem for finitely presentable groups is undecidable. Restricting to finite groups, the problem is certainly decidable, with the trivial algorithm of enumerating all $n!$ permutations (assuming of course, the two groups have the same order). 
If the input groups $G, H$ are given by their Cayley tables (which is a generous assumption), we can improve the bound to $n^{O(\log(n))}$ in the following way. First, observe that for a group $G$ of order $n$, there exists a generating set of size $\log_{2}(G)$. To see this, we consider the following algorithm: 
-Set $K := \emptyset$. 
-While $\langle K \rangle \neq G$, select $g \in G \setminus \langle K \rangle$, and add $g$ to $K$.
Each new element we add to $K$ at least doubles the size of $\langle K \rangle$. So $|K| = \log_{2}(n)$. 
As $\langle K \rangle = G$, we seek to map the elements of $K$ into a generating set of $H$. There are $|H|^{|K|} = n^{\log(n)}$ such functions. Accounting for some overhead in testing such functions, we achieve the $n^{O(\log(n))}$ bound. This bound has been relatively unscathed, though several classes of groups have polynomial time algorithms for the isomorphism problem. Some of these classes include abelian groups (https://www.sciencedirect.com/science/article/pii/S0022000007000293), groups without abelian normal subgroups (https://link.springer.com/chapter/10.1007%2F978-3-642-31594-7_5), and groups with abelian Sylow towers (http://drops.dagstuhl.de/opus/volltexte/2012/3400/). 
