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I have the following set $G=\lbrace a,b,e \rbrace$ and I successfully computed the following Cayley-Table \begin{align} \begin{array}{|c|c|c|c|} \hline \circ& a & b & e \\ \hline a& b&e &a\\ \hline b& e &a &b\\ \hline e& a& b&e\\ \hline \end{array} \end{align}

Now $(G, \circ)$ forms a group. When talking about homomorphisms I learned that a mapping is considered a homomorphism if $\varphi: G \to G'$ is a mapping such that $\forall a,b \in G, \varphi(a \circ b) = \varphi(a) \circ' \varphi(b)$.

So I was asked to show that the group $(G, \circ)$ of order 3 is isomorph to the group $(G', \circ ')$ where $G' = \lbrace c,d,e' \rbrace$. So what I did was define the mapping \begin{align} \varphi: \begin{cases} &G \longrightarrow G' \\ &a \longmapsto c \\&b \longmapsto d \\ &e \longmapsto e' \end{cases} \end{align}

So it is trivial to notice that $\varphi$ is bijective. It also follows easily (I've done the calculations) that $\varphi$ is homomorphism, using the same Cayley table as above and replacing all letters by $c,d,e'$ respectively. Therefore $\varphi$ is an isomorphism.


My question now is, what have I done? If this approach was even remotely correct, then I wonder what my statement now is saying. I have found an isomorphism between $G$ and $G'$ and therefore they have the same structure. But I defined them from the beginning as groups of the same order and even used the same Cayley-table.

The only relevance I can see in this is that I made no statement about $c,d,e'$ at all, so I never said that they must be equal or even related to $a,b,e$.

To really hit the point home about my confusion and the triviality I see in what I have done let me get a bit more verbal. I consider the Cayley table above as regular grid, or as some sort of game in which for whatever reasons the field are labeled just as above. Now I meet my friend and replace the abstract letters $a,b,e$ by $c,d,e'$ and tell him, "look, game 1 and game 2 have the same structure!"... well, yes?

Also my tutor said that this doesn't work for groups of order 4, can someone maybe tell me why or link me something where I can read into that?

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    $\begingroup$ One substantial catch in what you've said: simply saying $G'=\{c,d,e'\}$ does not define a group in and of itself; you need to define the group operation. If the question is 'show that all groups of order 3 are isomorphic' then what you've described can go towards showing that statement because you can show what such a group operation must satisfy, but the way you've written it is a bit misleading. $\endgroup$ – Steven Stadnicki Oct 14 '14 at 19:11
  • $\begingroup$ Thanks Steven for pointing that out to me, you're absolutely right! $\endgroup$ – Spaced Oct 14 '14 at 20:13
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The concept of isomorphic groups are indeed that the two are the same up to the labeling of the elements.

If $$ \varphi:\,(G,*_{1})\to(G',*_{2}) $$

is an isomorphism then instead of talking about $a,b\in G$ we will talk about $a'b'\in G'$ where $$ a'=\varphi(a),b'=\varphi(b) $$

and everything should be OK since the structure of the group is defined only by the multiplication so $$ \varphi(a*_{1}b)=a'*_{2}b' $$

You saw this illustration in your example - you literally changed the label and got an isomorphic group.

Regarding groups of order $4$ - the point is that there are two different groups of order $4$.

That is - there are groups $G_{1},G_{2}$ both of order $4$ but no relabeling of the elements of $G_{1}$ will give you the same structure of $G_{2}$ (it is actually a nice exercise to find both Cayley tables of those groups and I encourage you to do so).

On a final note: although this concept is easy, it is incredibly useful - it allows us to describe a group using groups we already know - i.e take some $G$ and say it is isomorphic to $G'$ we already know.

I assume that very soon you will learn about the group $Z_{n}$ which is the group of order $n$, I can then make statements such as:

  1. Every group of order $k\leq3$ is isomorphic to $Z_{k}$

  2. If $p$ is prime and $G$ is of order $p$ then $G$ is isomorphic to $Z_{p}$

(The proof is usually done in any course about groups)

Now - A key point is that since isomorphism is structure preserving everything I know about the group $Z_{p}$ will translate into knowledge about that group $G$ that is isomorphic to it.

So I can study the single group $Z_{p}$ and make statements about all groups of order $p$ .

The concept of a structure preserving map is a theme throughout mathematics (mainly in all sort of algebraic structure) and group isomorphism is one (great) example of it.

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Every group of three elements is isomorphic to one another, and is cyclic. Put differently, there is only one group of order $3$ up to isomorphism.

This is not true of groups of order four, of which there are two, up to isomorphism: The group $(\mathbb Z_4, +)$ and the Klein 4-group $\{e, a, b, c\} = \mathbb Z_2\times \mathbb Z_2$, in wich every element is of order $2$.

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  • $\begingroup$ and that I have shown successfully with the abstraction above? Or did I just write down some inclusive thoughts? $\endgroup$ – Spaced Oct 14 '14 at 18:41
  • $\begingroup$ If $G$ is a group of order $3$ described by your table, then every group of order three is given by the same table, with the only difference being the renaming of $a, b, e$. $\endgroup$ – amWhy Oct 14 '14 at 18:45
  • $\begingroup$ And yes, the abstraction above is successful. $\endgroup$ – amWhy Oct 14 '14 at 18:51
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Just an addition to the answer by @amWhy: If you want to construct the Cayley tables of the two non-isomorphic groups of order 4, proceed as follows:

Place the elements $e,a,b,c$ in that order. The first row and the first column of the table must also be $e,a,b,c$. Now you are left with a $3 \times 3$ square that you have to fill. There are just two ways of filling that square, so that there are no repeated elements on any row or any column.

For the first option, fill the main diagonal with the element $e$ (the identity). Once you do that, there is only one way to complete the table. You will thus get the group $\mathbb{Z}_2 \times \mathbb{Z}_2$.

For the second option (the cyclic group $\mathbb{Z}_4$), fill the other diagonal of the $3 \times 3$ square with $e$. Again you will have only one way to complete the square. You will then see that both tables are essentially different.

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