1

Let $G$ be finite. Suppose that $\left\vert \{x\in G\mid x^n =1\}\right\vert \le n$ for all $n\in \mathbb{N}$. Then $G$ is cyclic.

What I have attempted was the fact that every element is contained in a maximal subgroup following that cyclic iff not a union of cyclic subgroups, the order of elements of a cyclic group, and Sylow-$p$ subgroups......But none of them seems helpful.

: ) It’s very kind of you to give me some hints to push me further. Thanks!

3 Answers3

2

Here are some hints of a possible way to prove this:

1) Show that for all $n$ dividing $|G|$, there is at most one cyclic subgroup of cardinal $n$ in $G$. Call it $H_n$ (when it exists).

2) Look at the map $\Psi: G\to \{\text{cyclic subgps} \}$, $x\mapsto \langle x \rangle$. Compute $|\Psi^{-1}(H_n)|$.

3) Write an equation giving $|G|$ in terms of $|\Psi^{-1}(H_n)|$, and compare with a famous formula involving Euler's totient function.

user1729
  • 31,015
user120527
  • 2,129
  • 9
  • 17
  • Thanks for your time. But I’m still confused about how to do with (3). Could you give me more details? –  Feb 23 '18 at 14:12
  • 1
    write $m=|G|$. Let $E$ be the set of divisor $n$ of $m$ such that there is a cyclic subgroup of order $n$. Then $$m=\sum_{n\in E} \varphi(n)=\sum_{n|m} \varphi(n),$$ so $E$ is the full set of divisors of $m$. – user120527 Feb 23 '18 at 14:19
  • 1
    See also https://math.stackexchange.com/a/1615033/589 – lhf Feb 23 '18 at 14:55
0

I have another idea, please have a look.

For each prime $p\mid |G|$ we have $H$, a Sylow $p$-subgroup(uniqueness comes from the given condition) s.t. $|H|=p^n$(say). Then $G=H_1 \oplus \cdots \oplus H_n$. Now each $H_i$ has at least one normal subgroup of order $p^a$ s.t. $a\mid n_i$ for all $0 \leq a \leq n_i$.

Claim: Each $H_i$ is cyclic.

Suppose $H_1$ is not cyclic then $\exists P_1\neq P_2$ subgroups of $H_1$ s.t. $|P_1|=|P_2|=p^b$ for some $0\leq b\leq n_1$ then the given condition would break.

Ri-Li
  • 9,038
  • How does that $H_i$ is cyclic imply $G$ is cyclic? –  Feb 26 '18 at 12:23
  • 1
    Because the orders of $H_i$ are relatively prime. – Ri-Li Feb 26 '18 at 12:36
  • "Now each $H_i$ has at least one normal subgroup of order $p^a$ s.t. $a\mid n_i$ for all $0 \leq a \leq n_i$" might seem useless to me, the rest may be sufficient. Did I miss anything? –  Feb 26 '18 at 14:05
  • That statement would be useful to argue the claim. The existence of at least a subgroup. – Ri-Li Feb 26 '18 at 14:07
  • Subgroup of $H_i$? How would that help? –  Feb 26 '18 at 14:09
  • To show that $H_i$ is cyclic, you have to say that it has a unique subgroup of the divisor of the order of $H_i$. Then you start with an existence right? – Ri-Li Feb 26 '18 at 14:12
  • Yes, but I'm so sorry that I still have one more question. When $H_1$ is non-cyclic, I want to know how you make sure there must $\exists P_1\neq P_2$? –  Feb 26 '18 at 15:02
  • Okay, this is an exercise for you to check that a group $G$ is cyclic iff for every divisor of $|G|$ you have a unique subgroup of that order. – Ri-Li Feb 26 '18 at 16:23
  • I was thinking about the completely same thing...... –  Feb 26 '18 at 16:24
  • When $G$ is cyclic, it’s well-known. The other side would be easy for me if $G$ is $p$-group, but I got confused when $G$ is not a $p$-group. –  Feb 26 '18 at 16:30
  • https://math.stackexchange.com/questions/1302635/a-finite-group-which-has-a-unique-subgroup-of-order-d-for-each-d-mid-n – Ri-Li Feb 26 '18 at 16:35
  • And the way you are thinking https://math.stackexchange.com/questions/1108316/finite-group-with-unique-subgroup-of-each-order – Ri-Li Feb 26 '18 at 16:35
  • Wow, Thanks A Lot! Actually it’s where my confusion exactly came from, that to prove this statement I probably have to use the very problem we’ve been discussing. Anyway, my problem is solved, thanks : ) –  Feb 26 '18 at 16:50
0

Although the duplicate supposes $G$ to be abelian, Ihf’s answer doesn’t. It’s a perfect answer and please see an even more complete one here. By the way, the answers given here are also very nice!