I read that a homomorphism is fully determined by where its generators are mapped to.
Let $f\colon\, \Z\oplus \Z\to S_3$ be a group homomorphism.
Generators of $\Z\oplus \Z$ are $(1,0)$ and $(0,1)$. I can map $(1,0)$ to any permutation I want. I just need to make sure that $f(1,0)$ and $f(0,1)$ commute.
Suppose:
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$f(1,0) = id$ (6 options for $f(0,1)$)
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$f(1,0) = (12), (23) \text{ or } (13)$ (2 options each for $f(0,1)$)
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$f(1,0) = (123) \text{ or } (132)$ (3 options each for $f(0,1)$)
Thus, total of $6 + 2×3 + 3×2 = 18$ homomorphisms.
Note: I counted the options by using the fact that the size of centralizer of an element is equal to order of the group divided by size of conjugacy class of the element.
It looks like, this can be generalized. From $\Z\oplus \Z$ to $S_n$, there are $n!\times k$ homomorphisms where $k$ is the number of disjoint conjugacy classes in $S_n$.
Is my conjecture true? I raise this doubt because I assumed that each pair of commuting permutations $f(1,0)$ and $f(0,1)$ determines a homomorphism. I don't know for certain if this is true.
Best Answer
For every group $G$ there is a bijection between $Hom(\mathbb Z^2, G)$ and the set $C_2(G)$ of ordered pairs of commuting elements of $G$. The bijection $\beta$ is given by $$ \beta: \phi\mapsto (\phi(a), \phi(b)), $$ where $a, b$ are fixed generators of $\mathbb Z^2$ and $\phi: \mathbb Z^2\to G$ is a homomorphism. Thus, you are asking about the cardinality $c$ of $C_2(G)$ for finite groups $G$.
We have $$ c= \sum_{h\in G} |Z_G(h)| $$ where $Z_G(h)$ is the centralizer of $h$ in $G$. The group $G$ acts on the conjugacy class $C_h\subset G$ (of $h\in G$) by conjugation; the action is transitive with the stabilizer of every element equal $Z_G(h)$. Thus, by the orbit-stabilizer theorem, we have $$ |G|= |C_h|\cdot |Z_G(h)| $$ for every $h\in G$. Thus, $$ c= \sum_{h\in G} |Z_G(h)|= \sum_{h\in G} \frac{|G|}{|C_h|}= |G| \sum_{h\in G} \frac{1}{|C_h|}. $$ In order to compute $\sum_{h\in G} \frac{1}{|C_h|}$ we pick a representative $h_i$ from every conjugacy class in $G$, $i=1,...,N$, where $N$ is the number of conjugacy classes in $G$. Then $$ \sum_{h\in G} \frac{1}{|C_h|}= \sum_{i=1}^N \sum_{g\in C_{h_i}} \frac{1}{|C_{h_i}|}. $$ The number of summands in each $\sum_{g\in C_{h_i}} \frac{1}{|C_{h_i}|}$ is $|C_{h_i}|$, hence, $$ \sum_{i=1}^N \sum_{g\in C_{h_i}} \frac{1}{|C_{h_i}|}= \sum_{i=1}^N 1= N. $$ Putting it all together, we obtain, $$ |Hom(\mathbb Z^2, G)|= c= |G| \sum_{h\in G} \frac{1}{|C_h|} = |G|\cdot N, $$ where $N$ is the number of conjugacy classes in $G$, confirming your conjecture. For information about computation of the number $N$ for various finite groups, see answers to this Mathoverflow question.