let $G$ is group,that for all $n \in\mathbb{N}$, it has exactly $\phi(n)$ elements of order $n$. Then Every finite subgroup of $G$ is cyclic.

Asked Jun 14 '19 at 07:17

Active Jun 14 '19 at 07:37

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let $G$ be a group having property that for all $n \in\mathbb{N}$, it has exactly $\phi(n)$ elements of order $n$. Show that every finite subgroup of $G$ is cyclic.

let $H$ is subgroup of order $m$ and $H$ is not cyclic. then $x \in H$ imply $o(x)| m \ $ and $o(x )< m$ .

But then $\sum_{d|m} \phi(d)= m$ will not longer be true, which is contradiction.

is this approach wright ?

or any better suggestion ?

Thanks

asked Jun 14 '19 at 07:17

Eklavya

2,671

2

Yes. Your approach is right, that's a correct proof. – Giorgio Mossa Jun 14 '19 at 07:37
2

Why wouldn't it be true? I don't see it. – Jun 14 '19 at 08:53
Maybe to be clear, show why $\sum_{d|m} \phi(d)= m$ will no longer be true. – Wang Kah Lun Jun 14 '19 at 15:36

let $G$ is group,that for all $n \in\mathbb{N}$, it has exactly $\phi(n)$ elements of order $n$. Then Every finite subgroup of $G$ is cyclic.

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