2 The average order of $F_0(n)$

We reserve the letter p for a prime variable. As is common, we write $p^i\,\|\,n$ if $p^i\mid n$ and $p^{i+1}\nmid n$ . Let $\mathrm {rad}(n)=\prod _{p\,|\,n}p$ , the squarefree kernel of n. By the Chinese remainder theorem,

$$ \begin{align*} F_0(n)&=\prod_{p^i\,\|\,n}\#\{a\kern-5pt\mod p^i:a^n\equiv1 \pmod{p^i}\}=\prod_{p^i\,\|\,n}p^{i-1}(p-1,n)\\ &=\frac{n}{\mathrm{rad}(n)}\prod_{p\,|\,n}(p-1,n). \end{align*} $$

This formula is to be contrasted with

$$ \begin{align*} \varphi(n)=\frac{n}{\mathrm{rad}(n)}\prod_{p\,|\,n}(p-1). \end{align*} $$

Let $\lambda (n)$ denote the exponent of the multiplicative group $(\mathbb Z/n\mathbb Z)^*$ . Known as Carmichael’s function, we have $\lambda (n)$ equal to the least common multiple (lcm) of the numbers $\lambda (p^i)$ for $p^i\,\|\,n$ . Further, for a prime power $p^i$ , we have $\lambda (p^i)=\varphi (p^i)$ except if $p=2,i\ge 3$ , and then $\lambda (2^i)=\frac 12\varphi (2^i)=2^{i-2}$ . We will use the function

$$ \begin{align*} L(x)=\exp(\log x\log_3x/\log_2x), \end{align*} $$

where $\log _k$ is the k-fold iterate of log. Note that $L(x)=x^{o(1)}$ as $x\to \infty $ , but $L(x)>(\log x)^m$ for any m and x sufficiently large depending on m.

Proof. Let $k=k(n)=\varphi (n)/F_0(n)$ . We consider three cases:

1. (i) $k(n)>L(x)$ ;

2. (ii) $k(n)\le L(x)$ and $\lambda (n)>L(x)^3$ ;

3. (iii) $k(n)\le L(x)$ and $\lambda (n)\le L(x)^3$ .

For $n\le x$ , we have $F_0(n)=\varphi (n)/k(n)\le x/k(n)$ , so that for case (i),

(2.1) $$ \begin{align} \sum_{\substack{n\le x\\k(n)>L(x)}}F_0(n)\le \sum_{n\le x}\frac x{L(x)}\le \frac{x^2}{L(x)}. \end{align} $$

We now assume that $k(n)\le L(x)$ . Let $u=(n,\lambda (n))$ and write $n=uv$ . We have

$$ \begin{align*} k=\frac{\varphi(n)}{F_0(n)}=\prod_{p\,|\,n}\frac{p-1}{(p-1,n)}. \end{align*} $$

Thus, for $p\mid n$ , we have

$$ \begin{align*} p-1=(p-1,n)\frac{p-1}{(p-1,n)} \,\bigg|\, (p-1,n)k \,\bigg|\, nk. \end{align*} $$

Thus, if $p^i\,\|\,n$ , we have $p^{i-1}(p-1)\mid nk$ , which implies that $\lambda (n)\mid nk$ . This then implies that

(2.2) $$ \begin{align} \lambda(n)=(nk,\lambda(n))\mid (n,\lambda(n))k=uk. \end{align} $$

Now, consider those $n\le x$ in case (ii), say there are N of them. That is,

$$ \begin{align*} N:=\sum_{\substack{n\le x\\k(n)\le L(x)\\\lambda(n)>L(x)^3}}1. \end{align*} $$

For each such n, (2.2) implies that $(n,\lambda (n))\ge \lambda (n)/k>L(x)^2$ . Since $(n,\lambda (n))\le (n,\varphi (n))$ , we have

$$ \begin{align*} \sum_{n\le x}(n,\varphi(n))\ge NL(x)^2. \end{align*} $$

Now, [Reference Erdős, Luca, Pomerance, Granville and Luca1] gives

$$ \begin{align*} \sum_{n\le x}(n,\varphi(n))\le xL(x)^{1+o(1)}, \end{align*} $$

from which we deduce that $N\le x/L(x)^{1+o(1)}$ , so that

(2.3) $$ \begin{align} \sum_{\substack{n\le x\\k(n)\le L(x)\\\lambda(n)>L(x)^3}}F_0(n)\le x^2/L(x)^{1+o(1)}. \end{align} $$

We now consider those $n\le x$ in case (iii), so that $k\le L(x)$ and $\lambda (n)\le L(x)^3$ . For these n, we have $u=(n,\lambda (n))\le L(x)^3$ and from (2.2), $\lambda (n)\mid uk$ . With $n=uv$ , we have $v\le x/u$ . Further, $\lambda (v)\mid \lambda (n)\mid uk$ . For a divisor d of $uk$ , the number of integers $v\le x/u$ with $\lambda (v)=d$ is at most $(x/u)/L(x/u)^{1+o(1)}$ uniformly. Here, we have used [Reference Pomerance and Mollin9, Lemma 5.2]. Since u is small, we have $L(x/u)=L(x)^{1+o(1)}$ , so that our count is bounded above by $(x/u)/L(x)^{1+o(1)}$ . This holds for each $d\mid uk$ , so our count in this case is at most

$$ \begin{align*} \sum_{u\le L(x)^3}\frac{\tau(uk)}{u}\frac x{L(x)^{1+o(1)}}\le \tau(k)\frac x{L(x)^{1+o(1)}}. \end{align*} $$

Now, for each n counted, we have $F_0(n)\le x/k$ , so the contribution of these n is at most

$$ \begin{align*} \sum_{k\le L(x)}\frac{\tau(k)}k\frac{x^2}{L(x)^{1+o(1)}}\le\frac{x^2}{L(x)^{1+o(1)}}. \end{align*} $$

With the estimates (2.1) and (2.3), the proof is complete.

3 The average order of $F_j(n)$ for $j\ne 0,1$

It was shown in [Reference Erdős and Pomerance3] that

(3.1) $$ \begin{align} \sum_{\substack{n\le x\\n\,\mathrm{composite}}}F_1(n)\le \frac{x^2}{L(x)^{1+o(1)}},\quad x\to\infty. \end{align} $$

It is clear that we should restrict to n composite since if n is prime, we have $F_1(n)=~n-~1$ , so that $\sum _{n\le x}F_1(n)\sim \frac 12x^2/\log x$ as $x\to \infty $ . We even have a name for members of $G_1(n)$ when n is composite: these are the bases for which n is a pseudoprime.

When $j>1$ , there is a similar special case. For a prime $p>j$ , we have $F_j(jp)$ a multiple of $p-1$ and a divisor of $\varphi (jp)$ (see (3.2)), so $\sum _{p\le x/j}F_j(jp)\asymp _jx^2/\log x$ . Hence, in this case, we only consider values of n that are not j times a prime.

Theorem 3.1. For each fixed integer $j\ne 0,1$ , we have

$$ \begin{align*} \sum_{\substack{|j|<n\le x\\j\,|\,n\implies n/j\,\mathrm{composite}}}F_j(n)\le \frac{x^2}{L(x)^{1+o(1)}},\quad x\to\infty. \end{align*} $$

Proof. The argument is similar to the proof of (3.1). Fix an integer $j\ne 0,1$ . We have

(3.2) $$ \begin{align} F_j(n)=\prod_{p^i\,\|\,n}\sum_{\substack{a\pmod{ p^i}\\a^{n-j}\,\equiv\,1\pmod{p^i}}}1 =\bigg(\frac n{\mathrm{rad} (n)},j\bigg)\prod_{p\,|\,n}(p-1,n-j). \end{align} $$

As discussed in Section 1, $G_j(n)$ is a subgroup of $(\mathbb Z/n\mathbb Z)^*$ ; say it has index $k=k(n)$ , so that $F_j(n)=\varphi (n)/k$ . We have

$$ \begin{align*} \prod_{p\,|\,n}\frac{p-1}{(p-1,n-j)}~\bigg|~k, \end{align*} $$

and so each $p-1$ divides $(n-j)k$ . Thus,

(3.3) $$ \begin{align} \lambda(\mathrm{rad}(n))\mid (n-j)k. \end{align} $$

Since $F_j(n)=\varphi (n)/k(n)< x/L(x)$ for $k(n)>L(x)$ , we may assume that $k\le L(x)$ . Now, assume that $n\le x$ is divisible by a prime $p>kL(x)$ . Write $n=mp$ , where $1\le m\le x/p$ . We have $(mp-j)k\equiv 0\pmod {p-1}$ so that

$$ \begin{align*} m-j\,\equiv\,0\ \ \bigg(\kern-10pt\mod{\frac{p-1}{(p-1,k)}}\bigg). \end{align*} $$

Note that if $j>0$ , we are assuming that $m\ne j$ . Thus, $m\ge j+(p-1)/(p-1,k)$ and m is in a residue class mod $(p-1)/(p-1,k)$ so that the number of choices for m is at most $kx/p(p-1)+|j|k/(p-1)$ . Summing this for $p>kL(x)$ , we get $o_j(x/L(x))$ for the count and, so, $o_j(x^2/L(x))$ for the contribution to the sum in the theorem.

We now may assume that every prime factor of n is bounded above by $kL(x)$ and that $x/L(x)<n\le x$ . Further, if the squarefull part of n is greater than y, then the number of such $n\le x$ is $O(x/\sqrt {y})$ . We apply this with $y=L(x)^2$ and, so, we may assume that the squarefull part of n is at most $L(x)^2$ . Since $n>x/L(x)$ is assumed, we deduce that n has a squarefree divisor d in the interval $I:=(x/kL(x)^2,x/L(x)]$ . For each squarefree integer d in I, we count those $n\le x$ with $n\equiv 0\pmod d$ and $(n-j)k\equiv 0\pmod {\lambda (\mathrm {rad}(d))}$ , using (3.3). Since d is squarefree, the second congruence reduces to $n-j\equiv 0\pmod {\lambda (d)/(\lambda (d),k)}$ . If there are any solutions at all to the two congruences, we must have the greatest common divisor (gcd) of the moduli dividing j.

Then, the number of n in this case is at most $1+(k,\lambda (d))|j|x/d\lambda (d)$ . Letting d run over squarefree numbers in I, using [Reference Pomerance and Mollin9, Lemma 5.2] and partial summation,

$$ \begin{align*} \sum_{d\in I}\frac{(k,\lambda(d))}{d\lambda(d)}=\sum_{l\le x/L(x)}\frac1l\sum_{\substack{d\in I\\\lambda(d)/(k,\lambda(d))=l}}\frac1d\le\sum_{l\le x/L(x)}\frac1l\sum_{u\,|\,k}\sum_{\substack{d\in I\\\lambda(d)=ul}}\frac1d \le\frac{ \tau(k)}{L(x)^{1+o(1)}}. \end{align*} $$

Thus, the number of choices for n in this case is $O_j(\tau (k)x/L(x)^{1+o(1)})$ , so summing over $F_j(n)\le x/k$ , we get $O_j(x^2/L(x)^{1+o(1)})$ . This completes the proof.

4 Lower bounds for the average order

One may wonder how close the expression $x^2/L(x)^{1+o(1)}$ in Theorems 2.1, 3.1 and in (3.1) is to the true average orders. In [Reference Erdős and Pomerance3, Theorem 2.1], it is shown that

(4.1) $$ \begin{align} \frac1x\sum_{\substack{n\le x\\n\,\mathrm{composite}}}F_1(n)> x^{15/23} \end{align} $$

for all large x. The exponent $15/23$ depends on the existence of many primes p with $p-1$ not divisible by a large prime, and it is discussed in [Reference Erdős and Pomerance3] how a certain natural conjecture about these primes leads to the assertion that we have equality in (3.1). For a rigorous lower bound, we have the exponent $15/23$ improving to $0.7156$ using [Reference Lichtman4, Theorem 1.1].

The same goes for $F_0(n)$ , namely

$$ \begin{align*} \frac1x\sum_{n\le x}F_0(n)>x^{0.7156} \end{align*} $$

for all large x and, conjecturally, this average is equal to $x/L(x)^{1+o(1)}$ . See the final paragraph of [Reference Luca, Pomerance, Landman, Nathanson, Nešetřil, Nowakowski, Pomerance and Robertson6] where the distribution of those n with $\lambda (n)\mid n$ is discussed. Note that $\lambda (n)\mid n$ if and only if $F_0(n)=\varphi (n)$ .

For fixed $j\ne 0,1$ , we have the same lower bound estimates for

$$ \begin{align*} \frac1x\sum_{\substack{|j|<n\le x\\ j\,|\,n\implies n/j\,\mathrm{composite}}}F_j(n). \end{align*} $$

To see this, one merely replaces the number 1 in [Reference Erdős and Pomerance3, (2.6)] with j.

5 The normal order

For each positive integer n, let

$$ \begin{align*} g(n)=\sum_{d\,|\,n}\frac{\Lambda(d)}{\varphi(d)}, \end{align*} $$

where $\Lambda $ is the von Mangoldt function. We have the following theorem.

Theorem 5.1. For each integer j, there is a set $S_j\subset \mathbb N$ of asymptotic density $1$ such that

$$ \begin{align*} \frac{\log F_j(n)}{\log\log n} =g(n-j)+o(1) \end{align*} $$

for $n\in S_j$ and $n\to \infty $ .

This theorem for $j=1$ is [Reference Erdős and Pomerance3, Theorem 4.1]. To generalise to $j\ne 1$ , one need only check that we have $\log (n/\mathrm {rad}(n))/\log \log n =o(1)$ as $n\to \infty $ in a set of asymptotic density 1.

As in [Reference Erdős and Pomerance3], we have the corollary that for each j, $F_j(n)\le (\log x)^{\psi (x)}$ for all but $o(x)$ integers $n\le x$ , where $\psi (x)\uparrow \infty $ arbitrarily slowly. In fact, from the Erdős–Wintner theorem, for each positive real number u, the asymptotic density $D_j(u)$ of the set of n with $F_j(n)\le (\log n)^u$ exists, with $D_j(u)$ continuous, strictly increasing, tending to 0 as $u\downarrow 0$ and tending to 1 as $u\uparrow \infty $ .

6 The number of subgroups

For a finite abelian group $\mathcal G$ , written multiplicatively, and m a positive integer, the sets

$$ \begin{align*} {\mathcal G}_m=\{g\in {\mathcal G}:g^m=1\} \end{align*} $$

are subgroups of $\mathcal G$ ; call them power kernels. Let $\lambda (\mathcal G)$ be the exponent of $\mathcal G$ . It is easy to see that

$$ \begin{align*} g^m=1 \quad\hbox {if and only if}\quad g^{(m,\lambda(\mathcal G))}=1, \end{align*} $$

so that $\mathcal G_m=\mathcal G_d$ , where $d={(m,\lambda (\mathcal G))}$ . Applying this to the groups $G_j(n)$ , we have the following result, where $\tau $ is the divisor function.

It is possible to consider the count $\tau (\lambda (n))$ statistically. Its logarithm has the normal order $\tfrac 12\log 2 (\log \log n)^2$ and there is a Gaussian distribution (see [Reference Erdős and Pomerance2]). One may also consider the total number of nonisomorphic subgroups of $(\mathbb Z/n\mathbb Z)^*$ . This was considered in [Reference Martin and Troupe7] and the result is similar to what we have for $\tau (\lambda (n))$ in [Reference Erdős and Pomerance2]. (The paper [Reference Martin and Troupe7] also considered the larger statistic, where one counts subsets that are subgroups.) One can also consider $\tau (\lambda (n))$ on average; for this see [Reference Luca and Pomerance5].