Showing that a group of order $21$ (with certain conditions) is cyclic

Question

How can i show that if $o(G)=21$ and if $G$ has only one subgroup of order $3$ and only one subgroup of order $7$, then show that $G$ is cyclic.

Answer 1

Hint: Let $H$ and $K$ be the unique subgroups of order 3 and 7, respectively. Since they are the only subgroups of a given order they must be normal (why?). Conclude that $H, K$ are normal subgroups that intersect trivially, and so their internal direct product $HK \simeq H \times K$.

Answer 2

Use the Sylow Theorems. There exist groups of order 21 that are not cyclic: check the semi-direct products. Consider a group such that $G=\{x,y \mid x^7=y^3=1, yx=x^2y\}$ which is not cyclic but of order 21.

But, since there are only one Sylow-3 subgroup and Sylow-7 subgroup, they are both normal. You are right, these p-subgroups are cyclic, but since they are also normal, the group isomorphic to the direct product $\mathbb{Z}/7 \times \mathbb{Z}/3$ which is cyclic.

Alternative explanation without Sylow: Consider the number of elements. Using Lagrange's Theorem, the order of each element divides the order of the group. Consider the elements that are not in either subgroup, what are their possible orders?

Answer 3

What are the possible orders of elements for this group? One of order 1, two of order 3 (can't have any more as that would mean another subgroup of order 3), six of order 7 (and likewise no more). What of the rest, what must they be?