Sachchidanand Prasad

22-05-2024

    Problem: Let $S_3$ denotes the group of permutation on three letters (symmetric group). Find the number of group homomorphisms
    • from $S_3$ to $\mathbb{Z} _3 \left( \cong \mathbb{Z} / 3\mathbb{Z} \right) $,
    • from $\mathbb{Z} _3$ to $S_3$.
    abstract-algebra group-theory group-homomorphism
Solution: We need to find number of group homomorphisms between $S_3$ and $\mathbb{Z} _3$ and vice versa. let us write \[ S_3 = \{ e, \sigma = (1,2,3), \sigma ^2, \tau = (1,2), \sigma \tau, \sigma ^2\tau\}. \]
  • Let $f: S_3 \to \mathbb{Z} _3$ be a homomorphism. If $f$ is not trivial, then it must be surjective. Therefore, from the first isomorphism theorem we have \[ S_3 / \ker f \cong \mathbb{Z} _3 \implies \left\vert S^3 /\ker f \right\vert = \left\vert \mathbb{Z} _3 \right\vert \implies \left\vert \ker f \right\vert = 2. \] We also know that $\ker f$ is normal subgroup. Thus if $f$ is nontrivial, then $\ker f$ is of order $2$ normal subgroup of $S_3$. From here we can can get contradiction in two ways.
    • If $N \leqslant G$ is a normal subgroup of order $2$ and $G / N$ is cyclic, then $G$ is abelian. (See problem for 24th April 2024.)
    • Look at the order $2$ subgroups in $S_3$. Let $N = \{ e, \tau \} $. Consider \begin{align*} \sigma N \sigma ^{-1} = \{ e, \sigma \tau \sigma ^{-1} \} = \{ e, \sigma ^2\tau \} \neq N. \end{align*} Similarly, if you take the other two subgroups, then also it is easy to see that they are not normal. Therefore, there does not exist any normal subgroup of order $2$.
    Therefore, $\ker f$ can not be of order $2$ and hence, $f$ must be the trivial homomorphism. Thus, there is only one homomorphism from $S_3$ to $\mathbb{Z} _3$.
  • Now we need to find the number of homomorphisms from $\mathbb{Z} _3$ to $S_3$. Since $\mathbb{Z} _3$ is cyclic, any homomorphism is determined by its generator. That is, if $f : \mathbb{Z} _3 \to S_3$ is a homomorphism, then it is determined by $f(1)$. If $o(g)$ denotes the order of $g$, then we know that $o(f(1))$ must divide $o(1) = 3$. This implies $o(f(1))$ is either $1$ or $3$. Look at the following possibilities. \begin{align*} & o(f(1)) = 1 \implies f(1) = e \implies f \text{ is trivial}. \\ & o(f(1)) = 3 \implies f(1) \in \{\sigma ,\sigma ^2\}. \end{align*} Thus, the total number of homomorphisms are $3$. (The number of nontrivial homomorphisms are precisely the order $3$ elements in $S_3$.)
The second part of the problem can be generalized as the number of nontrivial group homomorphisms from $\mathbb{Z} _p$ to $S_n$, where $p$ is a prime, is the number of order $p$ elements in $S_n$.