# showing group $G$ is cyclic if every number $m$ that divides $(G:1)$, there are atmost $m$ elements such that … [duplicate]

The following problem is from J.S. Milne book

Let $G$ be a finite Abelian group. If $G$ has at most $m$ elements of order dividing $m$ for each divisor of $m$ of $(G:1)$, show that $G$ is cyclic.

I do not understand the notation $(G:1)$. Also I would highly appreciate any hints (need not solve the problem as it coincides with my assignment) for solving the problem.

We are given $x^n=e$ has at most $n$ solutions for $n|o(G)$. Let $F(d)$ denote # of elements of order $d$ where $d|n$.
Claim : $F(d)\le \phi(d)$ $\forall d|o(G)$. Suppose there does not exist any element of order $d$ then $F(d)=0\le\phi(d)$. If there exists an element of order $d$ say $g$ Then $\{{g^i :i=1,2,...,d}\}$ are all the solutions of $x^d=e$. Hence we have there can exist at most $\phi(d)$ elements of order $d$. Now we have $$o(G)=\sum_{d|o(G)}F(d)$$ since every element has some order $d$. again we have $$\sum_{d|o(G)}\phi(d)=o(G)$$ This implies $F(d)=\phi(d)$ $\forall d|o(G)$ In particular $F(o(G))=\phi(o(G))$$\gt$$0$ and this proves $G$ is cyclic.
• Is this true for additive group too? because taking $\Bbb Z_{12}$, we know $6 \mid 12$ and $\phi(6) = 2$ but in $\Bbb Z_6$ there are $6$ elements. – Santosh Linkha May 12 '17 at 4:37
• Seems that I was confused about $F(d)$. I thought it were roots of $x^d = e$. It was set generators of group that contains the roots. – Santosh Linkha May 14 '17 at 14:47
$(G:1)$ is the index of the trivial subgroup, which coincides with the order of $G$.