Express the elements of $\mathbb Z_{\times mod 5} = \{1,2,3,4 \}$ as permutations in cycle form. Hence show that $\mathbb Z_{\times mod 5}$ is cyclic

Question

Am I expected to list all the permutations of $ \{ 1,2,3,4 \} $, 24 in total? And if so, how would I show that its cyclic?

I realised I can show that its cyclic by letting any number from the set $ \{ 2,3,4 \} $ be a generator for the group hence it's cyclic but I wonder what is the original expectation.

Answer 1

You consider the group of unities in the ring $\mathbb{Z}/(5)=\mathbb{Z}_5$ which is $\mathbb{Z}_5^{*}=\{1,2,3,4\}$ where $1,..,4$ denote the nonzero residue classes modulo $5$ then You get a homomorphism into the symmetric group on four elements by the multiplication tables, e.g. $1$ maps to $()$ and $2$ to $(1243)$, $3$ to $(1342)$ and $4$ to $(14)(23)$. Can you take it from here?

Answer 2

Consider Cayley's embedding $\theta\colon\mathbb{Z}_5^{\times}\to \operatorname{Sym}(\mathbb{Z}_5^{\times})(=S_4)$, $i\mapsto (j\mapsto ij \pmod 5)$. Then:

$$\theta_1(1)=1, \space \theta_1(2)=2, \space \theta_1(3)=3, \space \theta_1(4)=4$$ $$\theta_2(1)=2, \space \theta_2(2)=4, \space \theta_2(3)=1, \space \theta_2(4)=3$$ $$\theta_3(1)=3, \space \theta_3(2)=1, \space \theta_3(3)=4, \space \theta_3(4)=2$$ $$\theta_4(1)=4, \space \theta_4(2)=3, \space \theta_4(3)=2, \space \theta_4(4)=1$$

or, in disjoint cycle notation:

$$\theta_1=(), \space \theta_2=(1243), \space \theta_3=(1342), \space \theta_4=(14)(23)$$

So we get: $\mathbb{Z}_5^\times \cong \Theta:=\{\theta_i, i=1,2,3,4\}=\langle\theta_2\rangle=\langle\theta_3\rangle$, and $\mathbb{Z}_5^\times$ is cyclic with generators $2$ and $3$.