I was asked to construct a Cayley table for this Cayley diagram and it occurred to me that it will be impossible.
My reasoning is that at first glance this appears to be the group $D_4$. It has to be, since it has only two generators. But now, the group $D_4$ has 5 self inverses. This will be impossible to achieve with the Cayley diagram above since we have $$ea=I$$ and $$cg=I$$ In the group $D_4$ there is only one unequal pair that are inverses of each other, here there are two. So you can't generate 5 self inverses. Perhaps this is easier to see if you construct the identity skeleton for $D_4$ and this graph.
So I was wondering where have I gone wrong? Is this graph in fact erroneous in that it doesn't represent an actual group?
Thanks.
The Cayley diagram is indeed not contradictory. It represents the Quaternion group (order 8) with the following table:
For more information you can see a neater table here:
http://escarbille.free.fr/group/?g=8_4c&y=8&z=0|4|1|5|2|6|3|7