When a group is cyclic

Question

If $G$ is a finite group and has at most one subgroup of an arbitrary order (of course, dividing order of $G$) could we deduce that $G$ is cyclic ?

Answer 1

In fact, something even stronger is true (the original question has already been answered in the comments).

Assume that $G$ is a finite group and let $\pi$ be the set of primes dividing $|G|$.

For each $p\in \pi$, let $n_p$ be the largest power of $p$ that divides $|G|$.

To conclude that $G$ is cyclic, it is enough to assume that $G$ has at most one subgroup of order $p^{n_p}$ for each $p\in \pi$ and that for each $p\in \pi$ there is a natural number $m_p$ with $1\leq m_p\leq n_p - 1$ such that $G$ has at most one subgroup of order $p^{m_p}$. Except that if $2\in \pi$ and $n_2\geq 3$ we need to require $m_2\geq 2$.

Proof: The first condition means that all Sylow subgroups are normal, so it suffices to show that all Sylow subgroups are cyclic.

But if a $p$-group of order $p^a$ has a unique subgroup of order $p^b$ for some $1\leq b\leq a-1$ then it is cyclic unless $p=2$, $b=1$ and the group is generalized quaternion (this is a standard result in $p$-group theory, and a reference can be found in for example Berkovich's book on groups of prime power order). This completes the proof due to our extra requirements when $2\in \pi$.

Answer 2

If you know Lagrange's theorem and the fact that a cyclic group of order$~n$ has exactly one cyclic subgroup of order$~d$ whenever $d$ divides$~n$, then you can prove this by comparing the finite group$~G$ and the cyclic group $C_n$ of order $n=|G|$. For any $d$ one sees as follows that there can be no more elements of order$~d$ in$~G$ than there are in$~C_n$. In order to have any such elements at all in$~G$, one must have $d\mid n$ by Lagrange, and $G$ must have at least one cyclic subgroup of order$~d$ (spanned by such an element); but by hypothesis this is then the unique such cyclic subgroup of $G$, and all elements of order$~d$ are members of it. In the meantime $C_n$ also has a cyclic subgroup of order$~d$; the two cyclic subgroups contain equally many elements of order$~d$.

Now if there were no elements in $G$ of order$~n$, it would have stricly less elements than$~C_n$ (which does have elements of order$~n$); this would contradict $n=|G|$. By a group of order $n$ with an element of order$~n$ is cyclic.

Answer 3

I have found this proof from Keith Conrad’s handout about Sylow applications. Our argument has two steps: verify the theorem for groups of prime-power order and then use Sylow one to derive the general case from the prime-power case. Step 1: Let #G=$p^k$ where p is pr prime,$k≥1$, and assume G has at most one subgroup of each size. To show G is cyclic, let g be an element of G with maximal order. We want G=< g >.Pick any h belong to G, so h is a power of p by Lagrange. Let g have order $p^m$ and h have order $p^n$, so $n≤m$. Then $p^n$ divides $p^m$, so there is a subgroup of the cyclic group < g > with order $p^n$ . Also < h > has order $p^n$, so our hypothesis that G has at most one subgroup per size implies < h > lies in < g >, so h belongs to . Therefore < g > contains G. (This argument was told to me by Trevor Hyde.) Step 2: Let G be group with at most one subgroup per size. Therefore $n_p=1$ for all primes p. For different primes p and q dividing #G, the elements of the p-Sylow and q-Sylow subgroup commute with each other. Any subgroup of G has at most one subgroup of any size (otherwise G itself would have two subgroups of the same size), so by Step 1 the p-Sylow subgroup of G is cyclic. Choose a generator $g_p$ of the p-Sylow subgroup of G. These $g_p$ s commute as p varies, by the previous paragraph, and their orders are relatively prime, so the product of the $g_p $s has order equal to the product of the sizes of the Sylow subgroups of G. This product of sizes is #G, so G is cyclic.