Graph Combinatorics - Upper and lower bound on the number of group isomorphism classes I have this problem which seems to be hard, and of course I am looking for a simple solution, which is quite absurd, but that's mathematics anyways. The problem states that if $f(n)$ is the number of group isomorphism classes for graphs with n vertices, then there are $a>1, b>0,c>0$ such that 
$$a^nc\leq f(n)\leq b^{n^2}  \tag{1} $$ 
Finding $f(n)$ using the Bernstein theorem gives a starting point, but this in no way shows that the bounds (1) are possible. 
I thought about checking on the edges that a graph with n vertices will have. The extreme cases are the cases where are no edges (which could lead to the lower bound) and the case where every vertex communicates with every other (which could lead to the upper bound). But I cannot see how the exponents come to be...
Is this approach feasible?
Is there a simple way to prove the bounds? The bounds are possibly not sharp. 
Thanks in advance for any replies...
 A: The number of graphs on $n$ distinct vertices is $2^{(n^2 - n)/2}$, giving the upper bound.
To show the lower bound, consider a set $\mathcal{S}$ of non-isomorphic graphs on $n$ vertices. Construct two sets of graphs on $n+1$ vertices: the first set $\mathcal{A}$ by adding a disconnected vertex to each graph in $\mathcal{S}$, the second set $\mathcal{B}$ by adding a vertex and connecting it to each of the old vertices.
We claim that $\mathcal{A} \cup \mathcal{B}$ contains no pair of isomorphic graphs:


*

*No graph in $\mathcal{A}$ can be isomorphic to a graph in $\mathcal{B}$, because every graph in $\mathcal{B}$ and no graph in $\mathcal{A}$ has a vertex of order $n$.

*Any isomorphism between two graphs in $\mathcal{A}$ must align an isolated vertex of each; removing that vertex gives an isomorphism between two graphs in $\mathcal{S}$, a contradiction.

*Any isomorphism between two graphs in $\mathcal{B}$ must align an order-$n$ vertex of each; removing those vertices also gives an isomorphism between two graphs in $\mathcal{S}$. (It doesn't matter which order-$n$ vertices are aligned, as any graph is left unchanged if two vertices of maximum possible order—i.e., two isolated vertices of the complementary graph—are interchanged.)


The recurrence relation $f(n+1) \geq 2 f(n)$ follows, so from $f(1) = 1$ we have $f(n) \geq 2^{n-1}$. As a side note, a previous answer by Chris Godsil points out a lower bound of $2^{(n^2 - n)/2}/n!$, following from the fact that the size of any isomorphism class of a graph on $n$ vertices is at most $n!$.
