using Lagrange's Theorem and cylic geoups Suppose |G| = 21. If G has exactly one subgroup of order 3 and exactly one subgroup of order 7, prove that G is cyclic.
So far, I only know that by using LaGrange's theorem, the order of every element in G must factor 21. So my options are 1, 3, 7, or 21.
Does it have anything to do with the fact that 3 and 7 are primes or am I reaching a bit too far?
Also, do I have to assume (in the proof) that the subgroups have those specific orders?
 A: To show that $G$ is cyclic is equivalent to showing that $G$ has an element of order $21$. 
Let $H$ be the unique subgroup of order $3$.
Then if any element $x\in G$ has order $3$, then $\langle x\rangle$, the group generated by $x$ is a subgroup of order $3$. So $\langle x \rangle = H$. In particular, there can only be $2$ elements of $G$ of order $3$, which are exactly the non-identity elements of $H$.
In a similar way, how many elements of order $7$ are there?
And how many elements do we have left? What can their orders be?
A: Let $x$ be an element of $G$ of order $3$; let $y$ be an element of $G$ of order $7$. Define $H := \langle x\rangle$ and $K := \langle y\rangle$. These subgroups of $G$ intersect trivially since any $g \in H \cap K$ must have order a divisor of $|H| = 3$ and $|K| = 7$, and $3$ and $7$ are relatively prime. Since $H$ is the unique Sylow 3-subgroup of $G$ and $K$ is the unique Sylow $7$-subgroup of $G$, $H$ and $K$ are normal in $G$. Now $xyx^{-1}y^{-1} = (xyx^{-1})y^{-1} \in K$ since $xyx^{-1} \in K$ (by normality) and $y^{-1} \in K$. Also, $xyx^{-1}y^{-1} = x(yx^{-1}y^{-1}) \in H$ as $x \in H$ and $yx^{-1}y^{-1} \in H$ by normality. Therefore, $xyx^{-1}y^{-1} \in H \cap K = \{e\}$. So $xy = yx$, and consequently, $G \approx H \times K$. Since $H \approx \Bbb Z_3$, $K \approx \Bbb Z_7$, and $\Bbb Z_3 \times \Bbb Z_7 \approx \Bbb Z_{21}$, it follows that $G$ is cyclic.
