TITLE: Is it always possible to create a intuition for abstract algebra theorems?
QUESTION [16 upvotes]: I am a Ph.D student in computer science, and I work on graph isomorphism. My research work requires some level of mathematics (mostly group theory ). I have done basic level abstract algebra course. I try to write down the theorems on peace of paper and try to understand them; I usually repeat this process four five times. Some time by doing this I understand the theorem and its proof, but there are times when I find it difficult. The biggest problem I have faced is that theorems related to abstract algebra are really abstract I mean there is no way to create an intuition (Is it true ?). 

My question : How to create an intuition for abstract algebra theorems?

For Example

REPLY [6 votes]: One way to build intuition is to reframe results in terms of objects and relationships that you know of and are familiar with. Since it seems as though you're looking at group action in the context of graph theory, this might be a good route to go.
For example, consider the first theorem you state about the primitivity of a transitive group action. For a graph $X$, think of $\Omega = V(X)$ and $G=\operatorname{Aut}(X)$. Then a system of imprimitivity can be seen as a partition of $X$ into subgraphs that 'look the same.' Now, the theorem states

Assume that $\operatorname{Aut}(X)$ acts transitivity on $V(X)$, and consider $v\in V(X)$. Then $\operatorname{Aut}(X)$ has a system of imprimitivity if and only if there is some nontrivial subgroup $H$ containing $\operatorname{stab}(v)$. 

This might start to seem more concrete. Since $\operatorname{Aut}(X)$ acts transitively on $X$, the subgroup that sends these subgraphs to themselves not only contains $\operatorname{stab}(v)$, but strictly contains it, since $v$ can also be mapped to any other vertex in its partition.