4

Since $|\mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_8| = 4\times 4 \times 8$ and $|<(1,2,4)>| = 4$ ,

$(\mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_8) / <(1,2,4)>$ is an abelian group of order 32.

All abelian groups of order $32$ are

$$\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2,$$ $$\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2,$$ $$\mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_2,$$ $$\mathbb{Z}_8 \times \mathbb{Z}_2 \times \mathbb{Z}_2,$$ $$\mathbb{Z}_{16} \times \mathbb{Z}_2,$$ $$\mathbb{Z}_{32}.$$

Let $H = <(1,2,4)>$. Then $H = \{(1,2,4),(2,0,0),(3,2,4),(0,0,0)\}$ and $$|(1,0,0)+H| = 2,$$ $$|(0,1,0)+H| = 4,$$ $$|(0,0,1)+H| = 8.$$

Since $\mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_8$ has no elements of order greater than $8$, and it should have elements of order $2,4,8$, possible abelian groups are $\mathbb{Z}_8 \times \mathbb{Z}_2 \times \mathbb{Z}_2$, $\mathbb{Z}_8 \times \mathbb{Z}_4$.

To classify, we have to find the number of elements of order $4$.

First, $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_8$ has $8$ elements of order $4$, and $\mathbb{Z}_8 \times \mathbb{Z}_4$ has $12$ elements of order $4$.

and $(\mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_8) / <(1,2,4)>$ has $12$ elements of order $4$.

Hence, $(\mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_8) / <(1,2,4)>$ is isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_8$.

For bold part, $(\mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_8) / <(1,2,4)>$ has $12$ elements of order $4$.

It is $12$ because $4$ times addition with elements of $\mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_8$ is smallest positive integers that makes the elements is contained in $H$.

Then in $\mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_8$, we cannot make $(1,2,4)$ $(3,2,4)$ because element in $\mathbb{Z}_4$ with $4$ addition makes it as identity element, $0$.

So it can only be $(0,0,0)$ so we should pick $0$ in $\mathbb{Z}_8$, so there are $16$ cases but we should exclude $(0,0,0)$, $(0,2,0)$, $(2,0,0)$ and $(2,2,0)$.

Hence there are $12$ elements of order $4$.

For this part, should I check that are there anything same in those coset?

I discuss this problem with my friends and he said we should check, but anyway my solution has answer.

ged
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최선웅
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    The answer is correct. This group is, indeed, isomorphic to $\Bbb{Z}_4\oplus\Bbb{Z}_8$. A tool for automating tasks like this is the Smith normal form of integer matrices. Have you covered those in class/ your textbook? Admittedly SNF may be a bit too heavy a hammer for this job, but it is straightforward to apply. Here the simple way would be to observe that each coset of $\langle(1,2,4)\rangle$ has a unique representative with a vanishing first coefficient. The claim follows from that observation rather speedily (representatives of that type actually form a subgroup). – Jyrki Lahtonen Jun 11 '17 at 10:09
  • @JyrkiLahtonen Nope. I didn't have class about SNF, but I'll find for it. Thx. – 최선웅 Jun 11 '17 at 12:26

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