How do you show $\mathbb{Z}_{3}$ is a proper normal subgroup of $S_{3}$?

Question

I was wondering how to prove that ($\mathbb{Z}_{3}$, +), the group of integers under addition is a proper normal subgroup $S_{3}$, the third group of permutations.

I understand that one can express $\mathbb{Z}_{3} = \{e, 1, 2\}$ and $S_{3} = \{e; (1, 2), (1, 3), (2, 3); (1, 2, 3), (1, 3, 2)\}$ but I am unsure how to go about explicitly proving how $\mathbb{Z}_{3}$ is a proper normal subgroup.

Answer 1

First, you have to see $\mathbb{Z}_3$ as a subgroup of $S_3$. Well, $\mathbb{Z}_3$ is isomorphic to $Z=\{e,(1\ \ 2\ \ 3),(1\ \ 2\ \ 3)\}$.

Now, compute $(1\ \ 2)(1\ \ 2\ \ 3)(1\ \ 2)^{-1}$. Since $(1\ \ 2)^{-1}=(1\ \ 2)$, this is equal to $(1\ \ 3\ \ 2)\in Z$. Next step is to prove that $(1\ \ 3)(1\ \ 2\ \ 3)(1\ \ 3)^{-1}$ and $(2\ \ 3)(1\ \ 2\ \ 3)(2\ \ 3)^{-1}$ belong to $Z$ too.

Since $Z$ is generated by $(1\ \ 2)$, $(2\ \ 3)$, and $(1\ \ 3)$, this proves that $Z$ is a normal subgroup of $S_3$.

Answer 2

Hint: If $(\Bbb Z_3, +_3)=\langle a\mid a^3\rangle$, then map $a\mapsto (1,2,3)$. Use the fact that $\sigma(\tau_1, \tau_2, \tau_3)\sigma^{-1}=(\sigma(\tau_1), \sigma(\tau_2), \sigma(\tau_3))$.

Answer 3

Every group with $3$ elements is isomorphic to $\mathbb{Z}_3$. Now $\{(1),(123),(132)\}$ is the subgroup with $3$ elements in $S_3$. It is normal, because it has index $2$.