TITLE: Groups of Order $n$
QUESTION [5 upvotes]: Is there a formula for finding the number of groups of order n. For example, if a group $G$ has an order n, is there a formula in which someone can find the number of groups with that order. I suppose this would act similar to prime factorization since prime groups have only 1 group.

REPLY [2 votes]: In general there is no formula $f(n)$ for the number of groups of order $n$ up to isomorphism. However, if $n$ is squarefree (no prime to the power $2$ divides $n$), then Otto Hölder in 1893 proved the following amazing formula. 
$$f(n)=\sum_{m \mid n}\prod_p \frac{p^{c(p)}-1}{p-1}$$ where $p$ is a prime divisor of $n/m$ and $c(p)$ is the number of prime divisors $q$ of $m$ that satisfy $q \equiv 1$
(mod $p$). From this formula it follows for example that $f(n)=1$ if and only if gcd$(n,\varphi(n))=1$, where $\varphi$ is the Euler totient function.
Recommended further reading is the paper of J.H. Conway, H. Dietrich and E.A. O'Brien - Counting groups:gnu's, moas and other exotica, (google this title and find the .pdf!), where you will encounter more formula's and also a table of the values of $f(n)$ up to $n=2048$. You will note that for $n$ is a prime power, $f(n)$ becomes very very large.