# Show that $\mathbb{Z}_6 \oplus \mathbb{Z}_6/ \langle (2,3) \rangle$ is or is not cyclic.

I am asked if $\mathbb{Z}_6 \oplus \mathbb{Z}_6/ \langle (2,3) \rangle$ is cyclic or not.

Work:

Well the order of 2 in $\mathbb{Z}_6$ is 3 and the order of 3 in $\mathbb{Z}_6$ is 2. Thus, the order of $(2,3)$ is 6. Since the order of $\mathbb{Z}_6 \oplus \mathbb{Z}_6$ is 36, this means |$\mathbb{Z}_6 \oplus \mathbb{Z}_6/ \langle (2,3) \rangle$|=6.

So the quotient group is of order 6. Any abelian group of order 6 has to be cyclic by the fundamental structure theorem for finitely generated abelin groups.

My question is about this last part: "Any abelian group of order 6 has to be cyclic by the fundamental structure theorem for finitely generated abelin groups."Can someone explain why this is so?

Also, I am asked to find a generator of the group and I have no idea where to start. Any help would be greatly appreciated! Thanks!

By the structure theorem, the options for abelian groups of order $6 = 2 \cdot 3$ are $$\mathbb{Z}_6$$ and $$\mathbb{Z}_2 \times \mathbb{Z}_3$$ The latter group is isomorphic to the former, because $2$ and $3$ are relatively prime. In particular, $([1]_2, [1]_3)$ is a generator of the group (where $[1]_2$ is the congruence class of $1$ modulo $2$, etc.). In fact, any element $(\sigma, \tau)$ with $\sigma, \tau$ being non-identity elements will work.
• Any abelian group of prime order, or of order $pq$ with $p$, $q$ relatively prime prime numbers is cyclic. However, the square / cube / higher power of a prime will always have a non-cyclic group of that order. As far as finding a generator, there are only $6$ elements to work with in $\mathbb{Z}_2 \times \mathbb{Z}_3$; just try computing their powers (in fact, there are only $2$ elements which are reasonable candidates, having non-identity elements in both coordinates). – user296602 Dec 16 '15 at 21:03