5
$\begingroup$

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.

$\endgroup$
2
  • 3
    $\begingroup$ 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 $n$. $\endgroup$
    – angryavian
    Sep 5, 2020 at 7:14
  • $\begingroup$ 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. $\endgroup$
    – Stahl
    Sep 5, 2020 at 7:31

2 Answers 2

6
$\begingroup$

The usual way to go about this classic problem is the following.

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.

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – user775699
    Sep 7, 2020 at 13:14
  • $\begingroup$ @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. $\endgroup$
    – Evariste
    Sep 7, 2020 at 15:57
  • $\begingroup$ why does in 6 th line of your answer assumption d|n and $G_d\neq 0$ doesn't eliminate some cases(unintentionally)? $\endgroup$
    – user775699
    Sep 19, 2020 at 6:42
  • 1
    $\begingroup$ @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. $\endgroup$
    – Evariste
    Sep 19, 2020 at 12:50
0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.