Showing $\Bbb Z_2 + \Bbb Z_3$ is isomorphic to $\Bbb Z_6$.

I want to show that $$\Bbb Z_2 +\Bbb Z_3$$ is isomorphic to $$\Bbb Z_6$$ where '$$+$$' stands for external direct product.

First of all I wrote all the six elements of $$\Bbb Z_2 + \Bbb Z_3$$ and found the orders of each element. I observe that $$\Bbb Z_2 +\Bbb Z_3$$ and $$\Bbb Z_6$$ have same number of elements with number of elements of same order equal. What mapping should I define here in a valid way to show isomorphism?

I already know a theorem that when $$\gcd(m,n)=1$$, $$\Bbb Z_m +\Bbb Z_n$$ is isomorphic to $$\Bbb Z_{mn}$$. Shall I mention this theorem and let it go? Or other way is possible to show this?

• Perhaps you need to construct an isomorphism explicitly. Find a generator of $\mathbb{Z}_2\times\mathbb{Z}_3$. Nov 27 '16 at 3:46
• If you've got a theorem that does the work for you, then by all means use it. It's certainly true that $\gcd(2,3)=1$. Otherwise, you'd want to show an explicit isomorphism. Nov 27 '16 at 4:01
• Ok that means the other way is to show that $Z_2 + Z_3$ is cyclic. Thanks Nov 27 '16 at 4:04
• @Kavita, you can also find out that the order of product in abelian group is the least common multiple of orders of the elements, thus, you can take $(1,1)$, or $(1,2)$ for generator Nov 28 '16 at 3:03

Hint 2: Consider the element $$([1]_2, [1]_3)$$.