# Is it always possible to create a intuition for abstract algebra theorems?

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  • You may see this. (ps: However I haven't read this.) – Eric Jul 31 '17 at 12:37
• stating a theorem carefully with proper terminology etc. is actually easier than developing/ having the intuitive idea of the theorem. Intuition is a goal not always easily attained (at least for me) But, being an optimist, I do think it's possible, with work, to attain intuition. – James S. Cook Jul 31 '17 at 15:23
• @new_bee it's different for different folks. For one of my students, intuition flows from his ability to restate the problem in terms of category theory. For me, it's more about comparing and contrasting to things I already know. Often understanding the limitations or generalizations of a given statement brings a sense of intuition. But, for others, there is a secret formula or picture which, while a lie, gives insight into the source of the idea. For others, its meaning in physics has merit. What is intuition? What is dear to you? Answer that and answer the question. – James S. Cook Jul 31 '17 at 17:39
• @Eric - that's an excellent book for building intuition. I definitely recommend it. – Nagase Jul 31 '17 at 22:33
• @JamesS.Cook for Ramanujan, it's from his connection to the Hindu Goddess Namagiri while dreaming. The Secrets of Ramanujan’s Garden – Ooker Nov 12 '17 at 3:31

I think the best way to understand abstractions intuitively is to study lots of examples. To understand group actions, write some down. Then think about whether each is transitive, whether it's primitive, look at the stabilizers of elements. Verify that the theorem is true; try to understand why in each particular case. Look for examples where some hypothesis fails and see whether (and why) the conclusion fails.

That learning strategy reproduces (in part) the explorations that led to the useful abstractions of group theory. Mathematicians studying the symmetries of various geometrical objects realized that they could reason about the symmetry of just about anything by inventing an abstract language whose definitions captured the essence of the properties of symmetries.

Unfortunately, often the abstractions - the definitions and theorems - take center stage in teaching and learning. Students find it hard to understand what's going on without the examples. I would tell my students that for me the category of groups consisted of those groups with which I was personally familiar. I assigned homework that called for working out examples at least as much as proving theorems.

In your own discipline (computer science) there are analogous historical trends leading from examples to abstractions: the concepts of object oriented or functional programming languages, the development of abstract tools to reason about databases.

• For me too, a big part of intuition is definitely a reportoire of working examples and counterexamples, and the understanding necessary to know why they are (counter)examples. – Arthur Jul 31 '17 at 12:42

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.