What is the order of subgroup $\langle5\rangle \bigoplus\langle3\rangle$ of the group $\mathbb{Z}_{30} \bigoplus \mathbb{Z}_{12}$

Question

What is the order of subgroup $\langle5\rangle \bigoplus\langle3\rangle$ of the group $\mathbb{Z}_{30} \bigoplus \mathbb{Z}_{12}$

I think the order will be $LCM(6,4)=12$ Because $\langle5\rangle$ has order $6$ in $\mathbb{Z}_{30}$ and $\langle3\rangle$ has order $4$ in $\mathbb{Z}_{12}$

Answer 1

Hint: the underlying set structure of a direct sum $G \oplus H$ is $G \times H$, so $|G \oplus H| = |G||H|$. What are the sizes of $\langle 5 \rangle \triangleleft \mathbb{Z}_{30}$ and $\langle 3 \rangle \triangleleft \mathbb{Z}_{12}$?.

Answer 2

You're notation looks like you are making the subgroups $\langle 5 \rangle \triangleleft \mathbb{Z}_{30}$ and $\langle 3 \rangle \triangleleft \mathbb{Z}_{12}$ and only after that taking the sum. This gives you 24. But what you wrote is instead $\langle (5,3) \rangle \triangleleft \mathbb{Z}_{30} \bigoplus \mathbb{Z}_{12}$. Where you put the brackets matters.

For $\langle (5,3) \rangle$, you are okay. Let's list them.

(5,3) (10,6) (15,9) (20,0) (25,3) (0,6) (5,9) (10,0) (15,3) (20,6) (25,9) (0,0)