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I'm trying to understand what is required to show that two groups are isomorphic. My understanding is that two groups $G_1$ and $G_2$ are isomorphic if:

  1. $ |G_1| = |G_2| $ (equal cardinality)

  2. $G_1$ has element of order $n \Leftrightarrow G_2$ has element of order n

  3. $G_1$ and $G_2$ have the same number of order n elements

Is this the most appropriate way to determine $G_1$ and $G_2$ are isomorphic? Am I missing any other properties?

My confusion is that I'm unable to deduce how I should go about applying these rules.

For example, say that my groups are $S_3$ (the abstract permutation group of order 3) and $D_3$ (the Dihedral group of order 3), I know from my textbook that these two groups are isomorphic. It's also easy enough to see that $|S_3|=3!=6$ and that $|D_3|=2n=6$ and thus each group has the same cardinality.

But what is the appropriate way to show that conditions (2) and (3) are satisfied? How would I determine the order of each element in $S_3$ and $D_3$ such that I can show that each element maps to another element of equal order (conditions (2) and (3))?

user1160465
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    Those conditions are not enough to conclude that $G_1$ and $G_2$ are isomorphic. See this: https://math.stackexchange.com/questions/693163/groups-with-same-number-of-elements-of-each-order You will have a hard time finding such list of properties, without ultimately ending up with constructing an isomorphism. – freakish Mar 16 '23 at 09:54
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    Entries in such lists are consequences of the two groups being isomorphic. Logic then tells that they can only be used in the contrapositive way of proving that two groups are NOT isomorphic, if one such litmus test fails. Usually it is simpler to describe an isomorphism. There are some exceptions, e.g. those relying on the consequences of the structure theory of finite abelian groups, but at this point in your studies such tools are not available. – Jyrki Lahtonen Mar 16 '23 at 10:10
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    The elements of $D_3$ by definition can be written as $(1),(123),(132),(12),(13),(23)$, which is exactly $S_3$. So $D_3\cong S_3$. Here $(123)$ denotes the rotation by $120$ degrees, etc. – Dietrich Burde Mar 16 '23 at 10:24
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    Property 2 is, of course, implied by property 3, so there are only two properties here. – David A. Craven Mar 16 '23 at 13:43

2 Answers2

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Two finite groups even need not be isomorphic if they have the same number of elements of order $d$ for each order $d$, see here:

If I know the order of every element in a group, do I know the group?

So 1.,2.,3., are far from being enough to conclude that $G$ and $H$ are isomorphic. On the other hand, there are many algebraic properties which can be used to conclude that $G$ and $H$ are not isomorphic. Here are a few further simple examples:

  1. $G$ is abelian, $H$ is not.

  2. $G$ is solvable, $H$ is not.

  3. The center $Z(G)$ has a different cardinality from the center $Z(H)$.

  4. $G$ is simple, $H$ is not.

  5. $\mu(G)$ is different from $\mu(H)$, where $\mu$ denotes the degree of a minimal permutation representation.

  6. The cohomology groups $H^n(G,\Bbb Z)$ and $H^n(H,\Bbb Z)$ are not always isomorphic.

  7. The homology groups $H_n(G,\Bbb Z)$ and $H_n(H,\Bbb Z)$ are not always isomorphic.

  8. In case, both $G$ and $H$ are solvable, the derived length is different.

  9. In case, both $G$ and $H$ are nilpotent, the nilpotency class is different.

  10. The character groups of $G$ and $H$ are different.

Dietrich Burde
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The properties 1,2,3 you mentioned are not sufficient to conclude, weather $G_1$ and $G_2$ are isomorphic or not .

Take this counter example a) $Q_8$ Group of quaternions b) $Z_4 \times Z_4$

these groups satisfies conditions 1,2,3 mentioned by you, but they are not isomorphic since $Q_8$ is non abelian but $Z_4 \times Z_4$ is abelian.

A Narode
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