Index of $G^n$ in $G$, where $G=(\mathbb{Z}/p\mathbb{Z})^*$

Question

Consider the group $G=(\mathbb{Z}/p\mathbb{Z})^*$ and for a fixed $n\in \mathbb{N}, G^n:=\{x^n:x\in G\}.$

Question: Is it true that Index of $G^n$ in $G$, $|G:G^n|\leq n.$

Answer 1

I shall use the notation $G^{\cdot n}$ for $\big\{x^n\,\big|\,x\in G\big\}$ for any group $G$, instead. (The notation $G^n$ is usually understood as the direct product of $n$ copies of $G$.) For a unital ring $R$, I shall denote by $R^\times$ the group of units of $R$ under multiplication (since some people use $R^*$ for the set of nonzero elements of $R$). You can compute the index $$f(m,n):=\Big[(\mathbb{Z}/m\mathbb{Z})^\times:\big((\mathbb{Z}/m\mathbb{Z})^\times\big)^{\cdot n}\Big]$$ explicitly for any positive integers $m$ and $n$. You can directly obtain $f(p,n)=\gcd(p-1,n)$ for a prime natural number $p$ without knowing the general form of $f(m,n)$, but I leave this as an exercise. If you use the hint given by the user named random, you can show that $f(p,n)\leq n$ without knowing the value of $f(p,n)$ at all.

First, factorize $$m=2^t\,p_1^{k_1}\,p_2^{k_2}\,\cdots \,p_r^{k_r}\,,$$ where $r$ and $t$ are nonnegative integers, $p_1,p_2,\ldots,p_r$ are pairwise distinct prime numbers greater than $2$, and $k_1,k_2,\ldots,k_r$ are positive integers. Then, $$(\mathbb{Z}/m\mathbb{Z})^\times\cong H_t\times C_{p_1^{k_1-1}(p_1-1)}\times C_{p_2^{k_2-1}(p_2-1)}\times C_{p_r^{k_r-1}(p_r-1)}\,,$$ where $$H_t:=\left\{\begin{array}{ll}C_1\,,&\text{if }t\in\{0,1\}\,,\\ C_2\,,&\text{if }t=2\,,\\ C_2\times C_{2^{t-2}}\,,&\text{if }t\in\{3,4,5,\ldots\}\,. \end{array}\right.$$ Here, $C_d$ denotes the cyclic group of order $d\in\mathbb{Z}_{>0}$. That is, $$\big((\mathbb{Z}/m\mathbb{Z})^\times\big)^{\cdot n}\cong H_t^{\cdot n}\times C_{s_1}\times C_{s_2}\times\ldots\times C_{s_r}\,,$$ where $$s_j:=\frac{p_j^{k_j-1}(p_j-1)}{\gcd\left(p_j^{k_j-1}(p_j-1),n\right)}$$ for $j=1,2,\ldots,r$. Furthermore, $$H_t^{\cdot n}\cong\left\{\begin{array}{ll} H_t\,,&\text{if }n\text{ is odd}\,,\\ C_{2^{\max\left\{0,t-2-v_2(n)\right\}}}\,,&\text{if }n\text{ is even}\,.\\ \end{array} \right.$$ Here, $v_p(l)$ is the largest integer $\mu\geq 0$ such that $p^\mu$ divides $l\in\mathbb{Z}_{>0}$, where $p$ is a prime natural number. That is, $$f(m,n)=\small\left\{ \begin{array}{ll} \prod\limits_{j=1}^r\,\gcd\left(p_j^{k_j-1}(p_j-1),n\right)\,,&\text{if }n\text{ is odd}\text{ or }t\in\{0,1\}\,,\\ 2^{t-1-\max\left\{0,t-2-v_2(n)\right\}}\,\prod\limits_{j=1}^r\,\gcd\left(p_j^{k_j-1}(p_j-1),n\right)\,,&\text{if }n\text{ is even}\text{ and }t\in\{2,3,4,\ldots\}\,. \end{array} \right.$$ In particular, for a prime natural number $p$, $$f(p,n)=\gcd(p-1,n)\leq n\,,$$ where the equality case is $n\mid p-1$.