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Is there any theorem that states the all the finite groups of order n are the same? or some sort of theorem that refers to the order of two finite groups? If anyone can post a reference to this topic will be great... 10x in advance

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This is boldly not true for $n=4$. – Asaf Karagila May 7 '12 at 8:22
Take any noncommutative group, compare it with the cyclic group of the same order. – anon May 7 '12 at 8:23
"some sort of theorem that refers to the order of two finite groups"? I'm pretty sure there's more than one such theorem... – Zev Chonoles May 7 '12 at 8:24
Side remark: On Wikipedia for "finite field" I found: "Any two finite fields with the same number of elements are isomorphic.". But that's about fields. – Gerenuk May 7 '12 at 8:39
@Gerenuk: your comment about finite fields is correct. However, there are very few finite fields: there is a field of order $n$ if and only if $n=p^m$ for some prime $p$, $m\in\mathbb{N}$. – user1729 May 7 '12 at 9:29
up vote 14 down vote accepted

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)

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which is equivalent to $gcd(n,\varphi(n))=1$, where $\varphi$ is the Euler totient. – Nicky Hekster May 7 '12 at 8:32
See also cyclic number. – lhf May 7 '12 at 11:56

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.

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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

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I use $C_n$ to denote the cyclic group of $n$ elements. – Asaf Karagila May 7 '12 at 8:26

You might want to read the paper of Besche, Eick and O'Brien which contains a table of the number of groups of order $n<2001$.

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