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?

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    $\begingroup$ Perhaps you need to construct an isomorphism explicitly. Find a generator of $\mathbb{Z}_2\times\mathbb{Z}_3$. $\endgroup$ Nov 27 '16 at 3:46
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    $\begingroup$ 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. $\endgroup$ Nov 27 '16 at 4:01
  • $\begingroup$ Ok that means the other way is to show that $Z_2 + Z_3$ is cyclic. Thanks $\endgroup$
    – Kavita
    Nov 27 '16 at 4:04
  • $\begingroup$ @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 $\endgroup$ Nov 28 '16 at 3:03

Hint: Use the Chinese Remainder Theorem.

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


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