# Isomorphism Type of $\mathbb{Z_8}\times\mathbb{Z_6}\times\mathbb{Z_4} /\langle (2,2,2) \rangle$

Determine the isomorphism type of $\mathbb{Z_8}\times\mathbb{Z_6}\times\mathbb{Z_4}/\langle (2,2,2) \rangle$. Give two proofs: one using elementary analysis of orders of elements, and the other using the First Isomorphism Theorem.

The first part was fairly straight forward. The factor group will have 16 elements and will be abelian, so we can apply the Fundamental Thm of Finitely Generated Abelian Groups. The possible groups that this factor group could be isomorphic to are:

1. $\mathbb{Z_{16}}$

2. $\mathbb{Z_{8}}\times\mathbb{Z_{2}}$

3. $\mathbb{Z_{4}}\times\mathbb{Z_{4}}$

4. $\mathbb{Z_{4}}\times\mathbb{Z_{2}}\times\mathbb{Z_{2}}$

5. $\mathbb{Z_{2}}\times\mathbb{Z_{2}}\times\mathbb{Z_{2}}\times\mathbb{Z_{2}}$

I calculated the orders of each of the 16 cosets. The highest order I found was 4, thus the two possible isomorphism types that our factor group can be are $\mathbb{Z_{4}}\times\mathbb{Z_{4}}$ or $\mathbb{Z_{4}}\times\mathbb{Z_{2}}\times\mathbb{Z_{2}}$. Since our factor group also has 7 elements order 2, this group must be isomorphic to the latter.

For the second proof using the First Isomorphism Theorem, I think I need to find an onto mapping

$\phi$: $\mathbb{Z_{8}}\times\mathbb{Z_{6}}\times\mathbb{Z_{4}}$ $\rightarrow$ $\mathbb{Z_{4}}\times\mathbb{Z_{2}}\times\mathbb{Z_{2}}$ whose kernel, $\ker\phi$ = $\langle (2,2,2) \rangle$

If I achieve this criterion, then we know by the First Isomorphism Thm that

$\mathbb{Z_{8}}\times\mathbb{Z_{6}}\times\mathbb{Z_{4}}/\langle (2,2,2) \rangle$ $\cong$ $\mathbb{Z_{4}}\times\mathbb{Z_{2}}\times\mathbb{Z_{2}}$

as the mapping is onto and thus the im$\phi=\mathbb{Z_{4}}\times\mathbb{Z_{2}}\times\mathbb{Z_{2}}$

However, no matter what map I try, I can't get the kernel correct. I'll be able to get all the elements of $\langle(2,2,2) \rangle$ in the kernel, but I also get other elements as well.

For example, I was playing around with the map that sends $(a,b,c)$ $\rightarrow$ $(2a$ mod $4, b, c)$, but with this map, the $(2,0,0) \in$ $\ker\phi$, but $(2,0,0) \notin\langle (2,2,2) \rangle$.

Any help on how to construct this mapping would be most appreciated.

• What happens if you change the last coordinate of your example attempt from $c$ to $a+c$? – Greg Martin Oct 20 '15 at 23:11