Show that all abelian groups of order 21 and 35 are cyclic. Show that all abelian groups of order 21 and 35 are cyclic. I have no idea on how to start. Can anyone give some hints?
 A: All you need is 


*

*the Fundamental Theorem of Finitely Generated Abelian Groups: Every abelian group can be written as the direct product of cyclic groups...etc.

*And $\mathbb Z_{mn}$ is cyclic and isomorphic to $\mathbb Z_m \times \mathbb Z_n$ if and only if $\gcd(m,n) = 1$.

With this, you are equipped to conclude what you need for every abelian group of order 35, 21, respectively.
Note that $\mathbb Z_5 \times \mathbb Z_7 \cong \mathbb Z_{35}$, because $\gcd(5, 7) = 1$.
And it follows that $\mathbb Z_{35}$ is cyclic.
Similarly, you can work with $\mathbb Z_3 \times \mathbb Z_7 \cong \mathbb Z_{21}$ because the $\gcd(3, 7) = 1$. And hence, $\mathbb Z_{21}$ is cyclic
A: You can do this with elementary tools. Here is a possible plan for an abelian group $G$ of order 21.


*

*By Lagrange's theorem, the order of any element is one of 1, 3, 7 or 21.

*Take an arbitrary element $a \neq 1$. Its order is either 3, 7 or 21. Suppose it is 3 (the case when it is 7 is similar, and if it is 21 then we are done).

*The quotient group $G/\langle a \rangle $ has order $7$, so in fact $G/\langle a \rangle \cong \mathbb{Z}_7$. Every element in $\mathbb{Z}_7$ except the identity has order $7$. Then every element in $G - \langle a \rangle$ has order that is divisible by $7$. Then there is an element $b \in G$ of order $7$.

*So, we have elements $a$ and $b$ of orders $3$ and $7$. It is easy to see that then $G$ is a direct product $G = \langle a \rangle \times \langle b \rangle$, so $G \cong \mathbb{Z}_3 \times \mathbb{Z}_7$. It is in fact cyclic, qed.

A: HINT: There are only 2 possible abelian groups of order 21: $\mathbb{Z}_{21}$ and $\mathbb{Z}_3\times\mathbb{Z}_7$. You can show that the latter is cyclic by exhibiting a generator (it's probably the first thing you'll think of); in fact these groups are isomorphic.
