# 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, 2020 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, 2020 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.
– user775699
Sep 7, 2020 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, 2020 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)?
– user775699
Sep 19, 2020 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, 2020 at 12:50

I think there is another proof not using the partition of $$G$$ into $$G_d$$.

First we prove the theorem for all finite abelian groups. Let $$G$$ be such a finite abelian groups, which is decomposed as $$G = C_{d_1} \oplus C_{d_2}\oplus\dots\oplus C_{d_m}\,$$, $$d_1|d_2|\dots|d_m$$. If $$m\gt1$$, then there is a subgroup $$D\lt C_{d_2}$$ of order $$d_1$$, contradicting our hypothesis. So $$m = 1$$, $$G=C_{d_1}$$ is cyclic.

It remains to show that any finite group satisfying our hypothesis must be abelian. So let $$G$$ be a finite group. We show first that all Sylow subgroups of $$G$$ are normal, which follows that $$G$$ must be the direct product of its Sylow subgroups.. If $$P\in Syl_p(G)$$ and $$P^g \not= P$$ for some $$g\in G$$ and some $$p \not\mid |G|$$, there is some $$h \in P$$ with $$h \notin P^g$$. $$\langle h \rangle$$ and $$\langle h^g \rangle$$ are two distinct subgroups isomorphic to each other, for otherwise $$h \in \langle h^g \rangle \lt P^g$$.

We have proven that all Sylow subgroups of $$G$$ are normal, so $$G$$ is indeed the direct product of its Sylow subgroups. It follows that if all $$p$$-subgroups of $$G$$ are cyclic, then $$G$$ is abelian, completing the proof. Let $$P$$ be a group of order $$p^n$$ satisfying our hypothesis. We prove by induction on $$n$$. $$P$$ must have a normal subgroup $$Q$$ of order $$p^{n-1}$$, which is already cyclic by the induction hypothesis. Pick an element $$a \in P$$ with $$a \notin Q$$. If $$a$$ has order $$p^n$$, we are done; otherwise, assume $$\langle a\rangle$$ has order $$p^m$$, $$1\leq m \lt n$$. $$Q$$, being cyclic, also has a subgroup $$\langle b \rangle$$ of order $$p^m$$; $$\langle a\rangle \neq \langle b \rangle$$ since $$a \notin Q$$, contradicting the hypothesis. Thus $$P$$ is cyclic.