How do I show that a finite group $G$ of order $n$ is cyclic if there is at most one subgroup of order $d$ for each $d\mid n$?

This particular question was asked in masters exam for which I am preparing and I could not solve it.

Question:

(a) Prove that if $$G$$ is a finite group of order $$n$$ such that for integer $$d>0$$, $$d\mid n$$, there is no more than one subgroup of $$G$$ of order $$d$$, then $$G$$ must be cyclic .

(b) Using (a) prove that multiplicative group of units in any finite field is cyclic.

For (a), I thought that as $$n\mid n$$ and there is only one subgroup of $$G$$ of order $$n$$ and order of a subgroup is order of element so, there exists an element $$a$$ such that $$|a|=n$$. But same argument can be used if statement says that there are more than one subgroup of order $$d$$ for each $$d \mid n$$. So, what mistake I am making? and kindly tell right approach.

For (b), the number of elements of in group is $$p^{n} -p^{n-1}$$. I don't know how can I show that there exists an element equal to order of the group.

• If $n$ is the order of group $G$, then there is one subgroup of order $n$ (the whole group $G$). This is true of any finite group. This does not imply the existence of an element $a$ of order $n$. Sep 5 '20 at 7:14
• Check your calculation of the number of elements in the multiplicative group of a finite field -- it's not $p^n - p^{n-1}.$ Also, your argument will need to involve more than the order of the group, you'll need to use the fact that you're in a field; there are non-cyclic groups $G$ such that the $\#G$ is the same as the number of elements in some finite field. Sep 5 '20 at 7:31

Let $$G=\cup G_d$$ where $$G_d$$ is the set of elements of $$G$$ of order $$d$$ for each $$d|n$$.
Since there's at most one subgroup of order $$d$$, $$|G_d|\leq\varphi(d)$$
However, $$\sum_{d|n}\varphi(d)=n$$ and $$|G|=\sum_{d|n}|G_d|$$, therefore it must be that $$|G_d|=\varphi(d)$$ for all $$d|n$$, and in particular, there is an element of order $$n$$, so $$G$$ is cyclic.
Now let $$G$$ be the multiplicative group of units of a finite field. Assume $$d|n$$ and $$G_d \neq \emptyset$$. Since any element of $$G_d$$ generates a cyclic group of order $$d$$, there must be at least $$\varphi(d)$$ such elements. However, the elements of the cyclic group are roots of $$X^d-1=0$$ which has at most $$d$$ roots in a field, so the cyclic group is the set of its roots. So $$G_d$$ is entirely contained in the cyclic group and $$|G_d|=\varphi(d)$$. Once again, since $$\sum_{d|n}\varphi(d)=n$$, it must be that $$G_d \neq \emptyset$$, so in particular there is an element of order $$n$$ and $$G$$ is cyclic.
• how in last paragraph of your answer you are sure that that there exists a d| n such that $G_{d} \neq 0$ ? Kindly explain. Sep 7 '20 at 13:14
• @Tim If $G_d=\emptyset$ for some $d|n$, then $|G|=\sum_{d|n}(\text{#elements of order d})<n=\sum_{d|n} \varphi(d)=|G|$, which is a contradiction. Sep 7 '20 at 15:57
• why does in 6 th line of your answer assumption d|n and $G_d\neq 0$ doesn't eliminate some cases(unintentionally)? Sep 19 '20 at 6:42
• @Tim We always have $d|n$. Now either $G_d = \emptyset$, or $G_d \neq \emptyset$. I have shown that if $G_d \neq \emptyset$, then $|G_d|=\varphi(d)$. Obviously if $G_d = \emptyset$ then $|G_d| = 0$. So $|G_d| \leq \varphi(d)$ in all possible cases. However, since $\sum |G_d| = n$, and $\sum \varphi(d) = n$, it must be that $G_d = \varphi(d)$ always, since otherwise we'd get that $|G|=n<n$ which is a contradiction. So it is always the case that $G_d \neq \emptyset$ and I am not missing any cases. Sep 19 '20 at 12:50