Find a Subgroup of $\mathbb{Z}_{12} \oplus \mathbb{Z}_{18}$ Isomorphic to $\mathbb{Z}_9 \oplus \mathbb{Z}_4$ This is what I have so far:
$$\mathbb{Z}_{12} = \{0, 1, 2,\cdots, 11\}$$
$$\mathbb{Z}_{18} = \{0, 1, 2,\cdots, 17\}$$
$$\mathbb{Z}_9 = \{0, 1, 2,\cdots, 8\}$$
$$\mathbb{Z}_4 = \{0, 1, 2, 3\}$$
Using that fact that isomorphisms must contain the same number of elements, we must find a subgroup in $\mathbb{Z}_{12}\oplus \mathbb{Z}_{18}$ with the same number of elements as in $\mathbb{Z}_{9}\oplus \mathbb{Z}_{4}$.
Now, let's determine the number of elements in $\mathbb{Z}_{9}\oplus \mathbb{Z}_{4}$:
$$|\mathbb{Z}_{9}\oplus \mathbb{Z}_{4}| = \text{lcm}(|\mathbb{Z}_{9}|, |\mathbb{Z}_{4}|) = \text{lcm}(|9|, |4|) = 36$$
Next, let's find a subgroup in $\mathbb{Z}_{12}\oplus \mathbb{Z}_{18}$ with order $36$. First, $\langle 3 \rangle$ in $\mathbb{Z}_{12}$ gives us the group $\{0, 3, 6, 9\}$ which has order $4$. Second, $\langle 2\rangle$ in $\mathbb{Z}_{18}$ gives us the group $\{0, 2, 4, 6, 8, 10, 12, 14, 16\}$ which has order $9$. Then, the group generated by $(3, 2)$ has order $36$ (because $\text{lcm}(9,4) = 36$)
My question is, does this properly show that there exists a subgroup in $\mathbb{Z}_{12}\oplus \mathbb{Z}_{18}$ that is isomorphic to $\mathbb{Z}_{9}\oplus \mathbb{Z}_{4}$? Or do I have to show that the mapping is one-to-one, onto, and preserves the group operation? If so, what is the mapping function to show one-to-one, onto, and operation preservation?
Thanks!
 A: Hint: Is there an element in $\mathbb{Z}_{12}$ that doesn't generate all of $\mathbb{Z}_{12}$ (hence is a factor of $12$) but generates $\mathbb{Z}_{9}$ or $\mathbb{Z}_4$? Then is there an element in $\mathbb{Z}_{18}$ not relatively prime to $18$ that generates $\mathbb{Z}_9$ or $\mathbb{Z}_4$? 
Call these two elements $a$ and $b$. Then look at the element $a \oplus b \in \mathbb{Z}_{12} \oplus \mathbb{Z}_{18}$. Create an 'obvious' mapping from this element into $\mathbb{Z}_9\oplus \mathbb{Z}_4$. Prove that this mapping must be $1:1$, onto, and preserves the group operation. 
But makes things easy on yourself! Focus on the element $a \oplus b$, it should generate $\mathbb{Z}_9 \oplus \mathbb{Z}_4$ (why?) then those three things are easy to prove. (again, think about why).
EDIT: I'll expand a bit more on my hint then. You already looked at a few elements in $\mathbb{Z}_{12}$. Look closely at the subgroup $\langle 3 \rangle=\{0,3,6,9\}$. Could we map that to $\mathbb{Z}_4$ with a mapping that preserves the operation, say $\varphi$? Now look at $\langle 2 \rangle =\{0,2,4,6,8,10,12,14,16\}\subset \mathbb{Z}_{18}$. Can we take those elements and map it to $\mathbb{Z}_9$, say the mapping is $\theta$, in a way that preserves the operation? Now since clearly $\mathbb{Z}_{12} \oplus \mathbb{Z}_{18} \cong \mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$, would the 'mapping' 
$$\psi((a,b))=\varphi(a) \oplus \theta(b)$$
for $(a,b) \in \mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$ be an isomorphism to $\mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$? 
FINAL EDIT: With your last comment you essentially have the idea. It is clear that $\mathbb{Z}_{12} \oplus \mathbb{Z}_{18} \cong \mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$ (just take the map that sends $(a,b)$ to $(b,a)$). So we just need to find a subgroup of $\mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$ isomorphic to $\mathbb{Z}_9 \oplus \mathbb{Z}_4$. Your idea is right. We make a mapping $\varphi$ that takes each element in $\mathbb{Z}_{18}$ mod $2$ and a mapping $\theta$ that takes each element in $\mathbb{Z}_{12}$ mod $3$. Then show that $\varphi \oplus \theta$ is an isomorphism. Now here is where you need to be careful. You said it would be an isomorphism of $\mathbb{Z}_4 \oplus \mathbb{Z}_9$ to $\mathbb{Z}_{18} \oplus \mathbb{Z}_12$. This is wrong. It's an isomorphism from $\mathbb{Z}_{4} \oplus \mathbb{Z}_9$ to $\langle 2 \rangle \oplus \langle 3 \rangle$ in $\mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$ (so a subgroup of that group). Since this group is isomorphic to $\mathbb{Z}_{12} \oplus \mathbb{Z}_{18}$, we would be done since it would be clear what the subgroup was. But there is an easier way to do this.
Let's look at the subgroup $\langle 2 \rangle \oplus \langle 3 \rangle$ of $\mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$. Let's create a mapping $\varphi$ of $\langle 2 \rangle$ in $\mathbb{Z}_{18}$ into $\mathbb{Z}_9$. Notice that $1$ generates $\mathbb{Z}_9$. Moreover, $\langle 2 \rangle$ generates the subgroup (itself) in $\mathbb{Z}_{18}$. So just map $2 \in \mathbb{Z}_{18}$ to $1 \in \mathbb{Z}_{9}$. Similarly, we can map $3 \in \mathbb{Z}_{12}$ to $1 \in \mathbb{Z}_4$. Since we are mapping generators to generators, it is clear that we preserve the operations, it is $1:1$, and onto. So we have the required isomorphism. But all the same, if this shorter way isn't as clear to you then you can can always show it by the preceding paragraph (which was what you stated in your comment, I just cleaned up a bit of the language). Hope that helped finally clear things up.
