# 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
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
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
ok, I think I understand now, thank you.