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I am doing a problem from Fraleigh's abstract algebra text. I am on question 20, section 4. It asks to draw the Cayley tables for a group of order 4. I fill in the required sections (top row and far left column) and have drawn the three possible tables that guarantee a group structure. My problem is trying to find a way to determine if the two are isomorphic to eachother.

From trying some mapping functions, I have sort of determined that it cannot be a transposition of naming elements. Eg, let (a,b) denote mapping a to b. Then given {e,a,b,c} any function like {(e,a),(a,e),(b,c),(c,b)} will only produce a group that can have the table rearranged to give the same group. Thus the mapping function must be at least cyclic... Here I run out of ideas. Try all cyclic permutations of the elements?

What can I do to determine not only IF two groups are isomorphic via cayley table but also a mapping function defining the isomorphism.

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It should be clear that one (and only one) of the elements is the identity element of your respective tables, and that an isomorphism must map these elements to each other. So the question remains, what can we deduce from the remaining three elements?

You are correct that IF the two groups are isomorphic, that the mapping between them must be a bijection on $\{a,b,c\}$. Fortunately for you, there are only six such mappings:

The identity map (if they are not the same table, we can rule this out),

Three transpositions: swap $(a,b)$, swap $(a,c)$ or swap $(b,c)$

Two "cyclical" maps: $a \to b \to c \to a\dots$ ,

$a \to c \to b \to a\dots$

But there is a faster way: deduce how many elements $x$ have the property that:

$x\ast x = e$ (this amounts to finding out how many times $e$ (or its analogue) occurs on the diagonal of the table).

This is the same as counting the number of elements of order $2$.

While having the same number of elements of order $2$ does not guarantee two groups are isomorphic; what is true, is that having differing numbers of elements of order $2$ proves two groups are not isomorphic. This is something worth remembering, as it often proves to be a useful short-cut.

And in the special case of groups of order $4$, it indeed settles the matter (because $4$ has a limited set of divisors, being a prime power).

If your two tables have the same number of elements of order $2$, and they are indeed group tables (it turns out showing associativity is "hard" from just the table information), they will be isomorphic. But I caution you here, the isomorphism between them will not be unique, in fact you should be able to find at least two such permutations (bijections) of $\{a,b,c\}$ that will work.

This underscores one of the inherent difficulties with Cayley tables: it is possible to have several distinct tables that represent "the same" (that is, isomorphic) group. One kind of group of order $4$ can have six tables, the other kind can have two. So in this sense Cayley tables are "inefficient conductors of information".

If we list the orders of the respective elements, there are only two lists (when ordered in increasing order):

$\{1,2,2,2\}$

$\{1,2,4,4\}$

It turns out for groups of order $4$ (but the proof of this is too long for this post) that this is all there is, and that any two groups of order $4$ of the same "order type" are isomorphic. I would recommend you try to prove this for yourself, it's interesting.

If the two groups are of the first order type, any permutation of $(a,b,c)$ yields an isomorphism. If the two groups are of the second type, there are only two possibilities, the identity, or a swap of the two elements of order $4$ (assuming you use the same "alphabet" for your two tables).

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  • $\begingroup$ I have seen many answers referring to "order" when discussing group isomorphisms - at this point in the text the concept of order isn't introduced. I understand it, but it feels like cheating to use. Thank you very much for your answer. $\endgroup$ – Alex Van de Kleut Sep 4 '17 at 2:24

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