# cosets of $\langle1_3\rangle\times\langle 0_2\rangle\times \langle0_4\rangle\trianglelefteq\mathbb{Z}_3\times\mathbb{Z}_2\times\mathbb{Z}_4$

Determine the left and right cosets of $H$ in $G$ and determine $(G:H)$. $G=\mathbb{Z}_3\times\mathbb{Z}_2\times\mathbb{Z}_4, H=\langle1_3\rangle\times\langle 0_2\rangle\times \langle0_4\rangle$

I cannot find a generator of $\mathbb{Z}_3\times\mathbb{Z}_2\times\mathbb{Z}_4$, I only know in direct product how to compute the order of a cyclic group in which I know the generator(finding the order through least common multiple). On this case I think $|\mathbb{Z}_3\times\mathbb{Z}_2\times\mathbb{Z}_4|=4\times3\times2=24$ and $ord(\langle1_3\rangle\times\langle 0_2\rangle\times \langle0_4\rangle)=3$ . So $(G:H)=\frac{|G|}{|H|}=\frac{24}{3}=8$

So we would have 8 right or left cosets.

However I do not know how to find the cosets.

Question:

How can I find the cosets(equivalence classes)? How can I write them down?

Do you know what the subgroups $\langle 1_3 \rangle, \langle 0_2 \rangle, \langle 0_4 \rangle$ actually are? Not their order or anything else, but exactly what these three groups are.

Then can you figure out what the elements in their product are? They are not hard to list.

Now what does it say about two elements $(x_1, y_1, z_1), (x_2, y_2, z_2) \in \Bbb Z_3 \times \Bbb Z_2 \times \Bbb Z_4$ that there difference lies in $\langle 1_3 \rangle \times \langle 0_2 \rangle \times \langle 0_4 \rangle$?

After answering above, you should be able to list the cosets.

Edit to answer the question in the comment.

$\langle 1_3\rangle = \Bbb Z_3 = \{0_3, 1_3, 2_3\}\\\langle 0_2\rangle = \{0_2\} \subset \Bbb Z_2\\\langle 0_4\rangle = \{0_4\} \subset \Bbb Z_4$

So \begin{align}\langle 1_3 \rangle \times \langle 0_2 \rangle \times \langle 0_4 \rangle &= \{0_3, 1_3, 2_3\}\times \{0_2\} \times \{0_4\} \\&= \{(0_3,0_2,0_4),(1_3,0_2,0_4),(2_3,0_2,0_4)\} \end{align}

• I think $\langle 1_3 \rangle\times \langle 0_2 \rangle\times \langle 0_4 \rangle$ is $\mathbb{Z}_3$, right? However I only found this equivalent class $\{(2_3,1_2,1_4),(1_3,1_2,1_4)\}$. I think only the first element can vary in this case $2_3$. However I am not seeing the other cosets and hoe they are going to be 8. Thanks for your answer! – Pedro Gomes Dec 30 '17 at 12:10
• $\langle 1_3\rangle = \Bbb Z^3, \langle 0\rangle = \{0\}$, so $$\langle 1_3 \rangle \times \langle 0_2 \rangle \times \langle 0_4 \rangle =\{(0,0,0),(1,0,0),(2,0,0)\}$$. Now, what does it say about $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ if their difference is in $\langle 1_3 \rangle \times \langle 0_2 \rangle \times \langle 0_4 \rangle$? (By the way, your proffered equivalence class is missing a member.) – Paul Sinclair Dec 30 '17 at 15:32
• I cannot understand this equality $\langle 1_3 \rangle \times \langle 0_2 \rangle \times \langle 0_4 \rangle =\{(0,0,0),(1,0,0),(2,0,0)\}$. What does it mean? How did you derive it? I am having a hard time understanding this exercise. – Pedro Gomes Dec 30 '17 at 15:58
• Could you provide me the solution please? That would help me out figuring out how I would solve this exercise. – Pedro Gomes Dec 30 '17 at 16:25
• I could not solve the question. Could you provide me an answer? – Pedro Gomes Dec 30 '17 at 19:11