Is $\mathbb{Z}_{4} \oplus \mathbb{Z}_{6} \cong \mathbb{Z}_{2} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{4}$?

Question

Is $\mathbb{Z}_{4} \oplus \mathbb{Z}_{6} \cong \mathbb{Z}_{2} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{4} $?

In my opinion, this statement is correct because the maximal order of element in each side is $12$, but is it enough to prove it by stating it?

In general, should I show a specific isomorphism between the groups? If I should, how do I find one?

Answer 1

Since $gcd(2,3) = 1$, we have that $$\mathbb{Z}_2 \oplus\mathbb{Z}_3 \cong \mathbb{Z}_6$$ and so the claim follows.

The fact that the maximal order of an element of two groups is equal is not sufficient to show that they are isomorphic.

Answer 2

Page 166, Chapter 8 Corollary 2, Contemporary Abstract Algebra, 8th ed. Joseph A. Gallian

Let m = n₁n₂.....n_k. Then $\mathbb{Z}$_m is isomorphic to $\mathbb{Z}$_n1 $\oplus$$\mathbb{Z}$_n2$\oplus$....$\oplus$$\mathbb{Z}$_nk if and only if n_i and n_j are relatively prime when i $\ne$ j.

So, $\mathbb{Z}$₆ $\approx$ $\mathbb{Z}$₂$\oplus$$\mathbb{Z}$₃ as 2 and 3 are relatively prime.

Therefore, $\mathbb{Z}$₄$\oplus$$\mathbb{Z}$₆ $\approx$ $\mathbb{Z}$₂$\oplus$$\mathbb{Z}$₃$\oplus$$\mathbb{Z}$₄

Answer 3

This might not be the most efficient way to do it, but my thought process tends towards "an isomorphism between any two structures is an invertible homomorphism".

Since the function $X\oplus Y\mapsto Y\oplus X$ is trivially an isomporphism, you really only need to determine if $\Bbb{Z}_2\oplus\Bbb{Z}_3\cong\Bbb{Z}_6$, since the rest follows from $X \oplus Y \cong W \implies X\oplus Y \oplus Z\cong W \oplus Z$. It suffices to show that a homomorphism $f:\Bbb{Z}_2\oplus\Bbb{Z}_3\to\Bbb{Z}_6$ is invertible.

While there are certainly much better methods for any particular example (including this one, see answers and comments), this is the only method I can think of which might work in all cases.