# Are two finite groups of the same order always isomorphic?

Are two finite groups of the same order always isomorphic? Some simple example would be great!

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Did you check any examples yourself? –  Tobias Kildetoft Sep 29 '13 at 11:43
In fact, this is ONLY true if $(n,\varphi(n))=1$. –  Alex Youcis Sep 29 '13 at 12:13

The smallest counterexample are the groups of order $4$.

The Klein four group $\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$ and the cyclic group $\mathbb Z/4\mathbb Z$ both have order $4$. However, they are not isomorphic, since only the latter one contains an element of order $4$.

In the representation $\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$, the Klein four group is the set $$\{(0,0),\quad (1,0),\quad (0,1),\quad (1,1)\}$$ together with the entry-wise addition mod $2$.

Another quite common representation of the Klein four group is the subgroup of the symmetric group $S_4$ consisting of the identity and the $3$ double transpositions: $$\{\operatorname{id},\quad (12)(34),\quad (13)(24),\quad (14)(23)\}$$

A quite concrete representation of the Klein four group is given by the symmetry group of a rectangle.

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can you please define $Z/2ZXZ/2Z$ ? –  Aman Mittal Sep 29 '13 at 11:37
It's just the usual direct product of the cyclic group $\mathbb Z/2\mathbb Z$ with itself. –  azimut Sep 29 '13 at 11:38
okie, Thanks, thats all i needed :) –  Aman Mittal Sep 29 '13 at 11:39
@AmanMittal: I've added this information to the answer, as well as a second representation of the Klein four group. –  azimut Sep 29 '13 at 11:42
Not to be connfused with this Klein Four Group. –  David Zhang Sep 29 '13 at 16:28
• 56 092 pairwise non-isomorphic groups of order 256
• 10 494 213 pairwise non-isomorphic groups of order 512
• 49 487 365 422 pairwise non-isomorphic groups of order 1024
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woh !! Thank you ! –  Aman Mittal Sep 29 '13 at 11:39
Addition: The exact number of isomorphism types of groups of order 2048 is still not known. But it's huge, for sure. –  azimut Sep 29 '13 at 11:45
@AmanMittal, you're welcome! –  Andreas Caranti Sep 29 '13 at 12:01
+1 for the categorization. May I know your point of view about my post? Thanks Andreas. ;-) –  Babak S. Oct 16 '13 at 6:56
@BabakS., it's a nice complement, +1 to you. –  Andreas Caranti Oct 16 '13 at 11:18

All finite groups of prime order $p$ are isomorphic, and they are all isomorphic to $\langle \mathbb Z_p, +_p\rangle$, where $+_p$ is addition modulo p.

This makes a nice exercise to prove it. I'm sure you can search math.stackexchange to check your proof and compare with others.

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Why $p \geq 5$? –  azimut Sep 29 '13 at 12:28
@azimut You're right, I was simply assuming common knowledge that there is only one group up to isomorphism of order 2, and ditto for order 3. But the statement surely applies to those orders as well. –  amWhy Sep 29 '13 at 12:38
@amWhy and thenfor a given $n$ all groups of order $n$ are isomorphic (cyclic) if and only if $\gcd(n, \varphi(n)) = 1$, see yiminge.wordpress.com/2009/01/22/… –  Andreas Caranti Sep 29 '13 at 13:13
@amWhy: $98\to 108$ ;-) –  Babak S. Oct 16 '13 at 6:40

Not at all...$\mathbb{Z}_4$ and the Klein's 4-group $K_4$ both are finite groups of order 4 but not isomorphic.

Many other examples are there. There are two non-isomorphic groups of order 6, viz. $\mathbb{Z}_6$ and $S_3$.

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I don't know if you know the group action but, you can review this counterexample for the futhure. Consider the following groups:

$$G_1=\{\text{id},(1,2),(3,4),(1,2)(3,4)\}\\ G_2=\{\text{id},(1,3)(2,4),(1,4)(2,3),(1,2)(3,4)\}$$

Although, both are isomorphic to $\mathbf{V}$ of order $4$; they are not isomorphic of permutation groups point of view. In fact, it easy to check that $(G_2\mid\Omega)$ transitively while $(G_1\mid\Omega)$ is not wherein $\Omega=\{1,2,3,4\}$.

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