About the Order of Groups

Solution 1:

No, this is unimaginably not true. In fact, there is a theorem which says that almost the opposite of what you have just said:

Theorem: Let $n\in\mathbb{N}$. Then, there exists only one group (up to isomorphism) of order $n$ if and only if $n=p_1\cdots p_m$ for distinct primes $p_i$, such that $p_i\not\equiv 1\text{ mod }p_j$ for any $i,j$.

This is a standard set of exercises in most algebra textbooks--ask if you'd like an explicit reference.

(I prove this on my blog, here)

Solution 2:

Not all groups of order n are the same.

$\mathbb{Z}_6$ and $S_3$ are both of order 6. However $S_3$ has 3 subgroups of order 2, where $\mathbb{Z}_6$ has only one subgroup of order 2.

Therefore they cannot be isomorphic.

Solution 3:

Consider the group $C_2\times C_2$ and $C_4$. Both of order $4$ but the latter has an element of order $4$ while the former does not

Solution 4:

You might want to read the paper of Besche, Eick and O'Brien http://www.math.auckland.ac.nz/~obrien/research/2000.pdf which contains a table of the number of groups of order $n<2001$.