How "at most one" cyclic group means "exactly one " cyclic groups $?$

Question

$G$ is a finite group of order $n$ and the equivalence relation on $G$ is $$x\equiv y\quad \text{iff}\quad (x)=(y)$$

Given that for any divisor $d$ of $n$ , $G$ has at most one cyclic subgroup of order $d$ . Let the equivalence class of $x$ be denoted by $C(x)$. Since equivalence relation $$|G|=\sum |C(x)|$$ where the sum runs over all the elements of $G$.

Since $$|C(x)| \le \phi(d) \ \ \ \ \text{and} \ \ \ \ n=\sum_{d|n} \phi(d)$$ we can write $$n=\sum |C(x)| \le \sum_{d|n} \phi(d) =n $$ i.e. $$\sum |C(x)|= \sum_{d|n} \phi(d)$$

Now what I do not get is how this last equation can imply that number of cyclic subgroups , that was said to be "at most one" is actually "exactly one" . Thanks for any help.

We have shown there exists an element of order $n$ in $G$, so that $G$ is cyclic. Hence it has exactly one subgroup of order $d$ for all divisors $d\mid n$. See also here. — Dietrich Burde, Sep 13 '15 at 21:29
$G$ has at least $\phi(d)$ elements of order $d$ whenever $d$ divides $n$. So it has elements of order $n$. Hence $G$ is cyclic. — Dietrich Burde, Sep 13 '15 at 21:35
@DietrichBurde : How can you say that ? I read exactly this proof in a book and cannot understand how the conclusion came? — , Sep 13 '15 at 21:37
See the answer of Pierre-Yves Gaillard at the link above. Isn't it exactly the conclusion you want ? — Dietrich Burde, Sep 13 '15 at 21:38

score 1 · Answer 1 · answered Sep 14 '15 at 06:26

1

As seen, $|C(x)|\le\phi(d) $ whenever $x$ is of order $d$. If there were any $x$ and $d$ with $|C(x)|<\phi(d)$ (where of course $d\mid n$), then you'd have $\sum |C(x)|<\sum_{d\mid n}\phi(d)=n$

answered Sep 14 '15 at 06:26

Hagen von Eitzen

374,180

score 0 · Answer 2 · answered Sep 14 '15 at 06:15

A cyclic group is a group which contains elements of its own order.

By the assumption $G$ has at most $\phi(d)$ elements of order $d$ whenever $d$ divides $n$ (here n is the order of the group G)

But we know that $G$ has atleast $\phi(d)$ elements of order d whenever $d$ divides n.

By the above two statements we can say that $G$ has elements of order $n$. Hence $G$ is cyclic(as I stated earlier).

How "at most one" cyclic group means "exactly one " cyclic groups $?$

2 Answers2