TITLE: Why there are $11$ non-isomorphic graphs of order $4$?
QUESTION [25 upvotes]: I'm new to graph theory and I don't plan to become a serious student of graph theory either. My book suggests that there are $11$ non-isomorphic graphs of order $4$, but I can't see why.
I know that the most naive approach is that I can use brute force and draw four vertices and then try to connect two vertices by an edge in such a way that I don't get isomorphic graphs in each step. But then the question is, how should I know I have counted all possibilities at the end? And when the size of the graph becomes larger, how should I know I'm not counting two isomorphic structures more than once?
To be more precise, I have 5 questions about determining the number of non-isomorphic graphs with $n$ vertices:

Is there an algorithmic approach for creating all non-isomorphic graphs of a certain order $n$? 
Does it have anything to do with graphic sequences and Havel-Hakimi algorithm or something?
Is it possible to count the number of non-isomorphic graphs with $n$ vertices without actually knowing them or it's still an open problem?
Is it possible to partially solve this by using group theory?
How can I see that there are $11$ non-isomorphic graphs with $4$ vertices without brute force?

REPLY [11 votes]: We can compute this using the Polya Enumeration Theorem. What you want here is the cycle index of the edge permutation group $G$ of the complete graph $K_4$, which is a tetrahedron (this observation helps visualize the cycle structure of the permutations). The permutations of $G$ can be enumerated as follows.
There is the identity, which contributes $$a_1^6.$$ We can fix one vertex and rotate the other three, which gives, when applied to the edges, $$4\times 2\times a_3^2.$$ We can rotate by 180 degrees about the midpoints of two opposite edges, fixing those edges, for $$3\times a_1^2 a_2^2.$$ We can reflect, mapping the midpoints of two opposite edges onto each other, followed by a 90 or 270 degree rotation, giving $$3\times 2\times a_2 a_4.$$ Finally we can do a single squish, exchanging the vertices of one edge, which gives $$6\times a_1^2 a_2^2.$$
This yields the cycle index
$$Z(G) = \frac{1}{24}
\left(a_1^6 + 8 a_3^2 + 3 a_1^2 a_2^2 + 6  a_2 a_4 + 6  a_1^2 a_2^2\right)$$
which is
$$Z(G) = \frac{1}{24}
\left(a_1^6 + 8 a_3^2 + 9 a_1^2 a_2^2 + 6  a_2 a_4\right).$$
It follows that the ordinary generating function of unlabelled graphs on four vertices by the number of edges is given by
$$Z(G)(1+z) = 1/24\, \left( 1+z \right) ^{6}+3/8\, \left( 1+{z}^{2} \right) ^{2} \left( 1+z
 \right) ^{2}+1/3\, \left( 1+{z}^{3} \right) ^{2}\\+1/4\, \left( 1+{z}^{2} \right)
 \left( 1+{z}^{4} \right)$$
which expands to
$$Z(G)(1+z) = 1+z+2\,{z}^{2}+3\,{z}^{3}+2\,{z}^{4}+{z}^{5}+{z}^{6}.$$
It follows that there are
$$\left.Z(G)(1+z)\right|_{z=1} = 11$$
nonisomorphic graphs on four vertices. Here we have classified by the number of edges. If you are just after a count you can use Burnside on the cycle index, setting all variables to two, again for a value of $11.$
The following link does this Polya Enumeration computation for graphs of an arbitrary number of vertices. Here is another interesting Polya Enumeration II.