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A computer program outputs the following Cayley tables for groups of order 4.

Wikipedia tells me that there are only two groups of order 4, the cylic group ($Z_4$) and the Klein four-group ($Dih_2 = Z_2 \times Z_2$), so some of these groups must be isomorphic.

 * [e][a][b][c]
[e] e  a  b  c  
[a] a  e  c  b  
[b] b  c  e  a  
[c] c  b  a  e  

 * [e][a][b][c]
[e] e  a  b  c  
[a] a  e  c  b  
[b] b  c  a  e  
[c] c  b  e  a  

 * [e][a][b][c]
[e] e  a  b  c  
[a] a  c  e  b  
[b] b  e  c  a  
[c] c  b  a  e  

 * [e][a][b][c]
[e] e  a  b  c  
[a] a  b  c  e  
[b] b  c  e  a  
[c] c  e  a  b

Which of these groups are isomorphic and why?

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  • $\begingroup$ read which groups have elements of order $4$ and $2$ $\endgroup$ – Jeb May 30 '14 at 11:35
  • $\begingroup$ Related: Any group of order four is either cyclic or isomorphic to $\mathbb Z_2\times \mathbb Z_2$ $\endgroup$ – amWhy May 30 '14 at 11:37
  • $\begingroup$ This reminds me that if you set an exam question like this, "which of the following groups are isomorphic", then there will often be some students who give answers like "the first, second and fourth groups are isomorphic, and the third and fifth are not". $\endgroup$ – Derek Holt May 30 '14 at 13:10
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An isomorphism can be interpreted as a relabelling of the elements (except for the identity $e$ which is fixed by an isomorphism). So, if you can find two tables which are the same after interchanging the roles of $a, b, c$ then you will have the multiplication tables of isomorphic groups. For example, the third table is the same as the second if you replace $c$ by $a$ and $a$ by $c$. Therefore, these two tables correspond to isomorphic groups.

Suppose now that you're willing to take as given that there are only two groups of order four: $\mathbb{Z}_4$ and $\mathbb{Z}_2\times\mathbb{Z}_2$. Note that $\mathbb{Z}_4$ has one element of order two, but $\mathbb{Z}_2\times\mathbb{Z}_2$ has three. As $e^2 = e$, we can tell which group corresponds to each table by counting how many times $e$ appears on the diagonal, two for $\mathbb{Z}_4$ and four for $\mathbb{Z}_2\times\mathbb{Z}_2$.

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  • $\begingroup$ Thanks, this answer was the most useful. $\endgroup$ – user154230 May 30 '14 at 17:47
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You can see that in the first one, every non-identity element $x$ is of order 2, i.e., $x^2 = e$.

This is not true in the other three, each of which contains one element of order 2. Those three have to be isomorphic to $\mathbb{Z}_4$. Why? Well, because it is a fact that there are only two groups of order 4, and the Klein four-group is exactly that group of order four with all non-identity elements having order 2. So the last three groups have to be isomorphic to $\mathbb{Z}_4$ (if they are groups). Of course, you can meticulously verify this.

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  • $\begingroup$ $\mathbb Z_4$ has one element of order $2$, as do the last three groups. They do not have two elements of order $2$. $\endgroup$ – amWhy May 30 '14 at 11:42
  • $\begingroup$ @amWhy Whoops, that's right. I was thinking "two elements $x$ with $x^2 = e$". Of course, one of them is the identity. $\endgroup$ – M. Vinay May 30 '14 at 11:46
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HINT: Look at the diagonal. A diagonal entry is $e$ if and only if the corresponding element has order two.

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  • $\begingroup$ Thanks, but then what does that tell you about the group? $\endgroup$ – user154230 May 30 '14 at 13:02
  • $\begingroup$ It tells you how many elements of the group have order 2. $\endgroup$ – Lee Mosher May 30 '14 at 14:16
  • $\begingroup$ ...and as the isomorphism class is determined precisely by the number of elements of order two... $\endgroup$ – user1729 May 30 '14 at 15:29

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