Let $h :\Bbb Z_{5}\rightarrow S_{z_{5}}$ be the left-regular representation of $\Bbb Z_5$? For each $n$ in $\Bbb Z_5$, compute $h(1)(n)$and $h(3)(n)$.
With this question I am confused with so many terms Like left-regular representation, I have use this term while proofing cayley’s theorem but I can’t relate this question. Also when question ask find $h(1)(n)$, so does $(n)$ represents $1$-cycle (identity mapping) or it’s something else. If so then how we find $h(1)$.
I know about symmetric groups and isomorphism but not able to proceed with this question.
By left regular representation is meant Cayley's embedding of a group into its symmetric group, namely the embedding by left multiplication used in the proof of Cayley's theorem: $h\colon G\to \operatorname{Sym}(G)$, $a\mapsto(g\mapsto ag)$.
In your particular case (the additive group of the integers modulo $5$), we get: $h(1)(n)=1+n, \forall n\in \Bbb Z_5$, and hence:
\begin{alignat}{1} h(1)(0)=1 \\ h(1)(1)=2 \\ h(1)(2)=3 \\ h(1)(3)=4 \\ h(1)(4)=0 \\ \end{alignat}
and likewise for $h(3)(n)$.
Extra. $\Bbb Z_5$ has $5$ elements and hence any bijection $f\colon \{1,2,3,4,5\}\to \Bbb Z_5$ induces an isomorphism $\varphi_f\colon S_{\Bbb Z_5}\to S_5$, defined by $\varphi_f(\alpha):=f^{-1}\alpha f$. You can prove that under the bijection $\tilde f\colon i\mapsto i-1$, we get: $\varphi_{\tilde f}(h(1))=(12345)$ and $\varphi_{\tilde f}(h(3))=(14253)$.