# Characterizing cyclic group of order $n=p_1^{n_1}p_2^{n_2}\cdots p_r^{n_r}$, by Lattice isomorphisms

Given a finite cyclic group $G$ with order $n=p_1^{n_1}p_2^{n_2}\cdots p_r^{n_r}$, where $p_i$s are distinct prime numbers $n_i>1$ for all $i$. Let $H$ be any group.Assume that $L(G)≅L(H)$and Aut$(G)≅$Aut$(H)$. Is it true that $G≅H$?

• Can you remind us what is L(G)? Commented Dec 23, 2015 at 6:11
• It was defined in mathoverflow.net/questions/226683 by the previous incarnation of the author. Commented Dec 23, 2015 at 7:41
• L(G) is the lattice of subgroups of G. Aut (G) is the group of automorphisms of G Commented Dec 23, 2015 at 8:40

Assume that $G$ is finite cylic, $G$ and $H$ have isomorphic lattices, and that $Aut(G)\approx Aut(H).$ Let the unique prime factorization of $\lvert G\rvert$ be $\lvert G\rvert= p_1^{a_1}\cdots p_n^{a_n}$ where $a_i>1$ and $p_i <p_{i+1}$ for all $i.$ Also, let the unique prime factorization of $\lvert H\rvert$ be $\lvert H\rvert= q_1^{b_1}\cdots q_n^{b_m}$ where $b_i>1$ and $q_i <q_{i+1}$ for all $i$.
Roland Schmidt's Subgroup Lattices of Groups Corollary 1.2.8 states that if $G$ is finite cyclic and $H$ is any group, then $L(G)\approx L(H)$ only if $H$ is also cyclic and $n=m$. Since $G$ and $H$ are cyclic, $\phi(\lvert Aut(G) \rvert)=\phi(\lvert Aut(H)\rvert)=p_1^{a_1-1}(p_1-1)\cdots p_n^{a_n-1}(p_n-1)$= $q_1^{b_1-1}(q_1-1)\cdots q_n^{b_n-1}(q_n-1)$ were $\phi(n)$ is the Euler's totient function.
I claim that $p_n=q_n$. If not, suppose $p_n>q_n$. Note that $p_n\nmid [q_1^{b_1-1}(q_1-1)\cdots q_n^{b_n-1}(q_n-1)]$ which is a contradiction.
So, $p_1^{a_1-1}(p_1-1)\cdots p_n^{a_n-1}(p_n-1)=q_1^{b_1-1}(q_1-1)\cdots p_n^{b_n-1}(p_n-1).$ We also have $a_n=b_n$ since $p_n=q_n$ is the largest prime divisor.
So, $p_1^{a_1-1}(p_1-1)\cdots p_n^{a_n-1}(p_n-1)=q_1^{b_1-1}(q_1-1)\cdots p_n^{a_n-1}(p_n-1).$ After cancelling we have $p_1^{a_1-1}(p_1-1)\cdots p_{n-1}^{a_{n-1}-1}(p-1)=q_1^{b_1-1}(q_1-1)\cdots q_{n-1}^{b_{n-1}-1}(q_{n-1}-1).$
Repeating this process gives $p_i=q_i$ and $a_i=b_i$ for all $i$. Thus $G$ and $H$ are cyclic of the same order, and therefore, $G\approx H$. Edit: Originally, I missed the hypothesis that the exponents were bigger than 1. This is a different answer.