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How many different mappings are GxG -> X, where G is a group of order n and X is a set of n elements?

pauloromero1983

New member
Jul 21, 2020
2
In the context of group theory, there's a theorem that states that for a given positive integer \(n\) there exist finitely different types of groups of order \(n\). Notice that the theorem doesn´t say anything of how many groups there are, only states that such groups exist. In the proof of this statement, they define a map \(f:G\times G \rightarrow X\) where \(X\) is a set with \(n\) elements. Defining a group structure in the same map by means of the product rule \(f(g_{1})f(g_{2})=f(g_{1}g_{2})\), where \(g_{1}, g_{2}\) belong to \(G\) they arrive to the following conclusion: there's an upper bound on the number of different groups of order \(n\), namely: \(n^{n^{2}}\)

My question is how to arrive to such conclusion. Im aware that, for every ordered pair of \(G\times G\) theres \(n\) images (since \(X\) was assumed to have \(n\) elements). For a concrete example, let be \(G\) a group of 2 elements. Then, there are 4 ordered pairs. Each pair has 2 images, so the total number of maps would be 4*2=8. However, by use of the relation \(n^{n^{2}}\) we get \(2^{2^{2}}=16\), i.e, there are 16 different maps, not 8. Im missing something here, but I dont know what exactly what the error is.
 

GJA

Well-known member
MHB Math Scholar
Jan 16, 2013
271
Hi pauloromero1983 ,

In the example you gave for each element of $G\times G$ there are 2 choices in $X$ to which the element can be mapped. Since the total number of elements in $G\times G$ is 4, the total number of possible mappings is $2\times 2\times 2\times 2 = 2^{4} = 16.$

In general, if $A$ and $B$ are finite sets, then there are $|B|^{|A|}$ different mappings/functions from $A$ to $B$. Does this help answer your question?
 

pauloromero1983

New member
Jul 21, 2020
2
ok, I think I understand now, thank you.