There is a problem in Pinter's "A Book of Abstract Algebra" (Chpt 9 Exercise D4) that has the following picture (some arbitrary group's Cayley Diagram representation):
In fact, this picture is literally on the cover of Pinter's Second Edition book. Based on my experience with Cayley Diagrams...I am pretty confused. I want to assume this is an error, but, once again, it is literally on the cover of the book (so I'm fairly skeptical about this being a wrong drawing).
This diagram is effectively saying that the element $a \in G$ is its own inverse...but also that $a \circ a = b \neq e$.
Should there be unidirectional arrows (either clockwise or counter-clockwise) on this drawing? ... Or is this just something that is implicitly understood within the math community? Thanks for the sanity check.
Edit 1: Just as a side note, I am operating off of the convention that Pinter states earlier in his book which is, "It is also convention that if $a^2 = e$, then no arrow head is used"...found on pages 51-52.
Edit 2:
I have included this picture to address one of the answers presented by Verret...who tried to offer an alternative interpretation to the above drawing. i.e. I do not think Pinter is trying to indicate $S_3$ with the initial picture...because he correctly depicts $S_3$ in the second picture.
Edit 3: I will include the full problem, by request. The objective is to determine which of the 4 cayley diagrams (if any) are isomorphic to one another.
If you are using the convention that no arrow means the corresponding generator is an involution, than this must be a diagram for $S_3$, which is generated by two involutions (and is the only group of order $6$ with this property).
So indeed, $a^2=e$, but I don't see why you think that $a^2=b\neq e$. There are TWO generators, $a$ and $b$ (corresponding to, say, $(12)$ and $(23)$ in $S_3$), and, in cyclic order, the element are $1-a-ab-aba=bab-ba-b$.