what is the order of the quotient group $(\mathbb{Z}\times \mathbb{Z}_{12})/\left \langle (4, 10) \right \rangle$?

Question

If $\left \langle (4, 10) \right \rangle$ is the cyclic subgroup of the abelian group of $\mathbb{Z}\times \mathbb{Z}_{12}$, what is the order of the quotient group $(\mathbb{Z}\times \mathbb{Z}_{12})/\left \langle (4, 10) \right \rangle$?

Any help please. Thank you.

What are your own thoughts about how to solve this? – md2perpe Aug 04 '18 at 08:58 — md2perpe, Aug 04 '18 at 08:58
You may find this comment on a similar question helpful. – Viktor Vaughn Aug 05 '18 at 17:46 — Viktor Vaughn, Aug 05 '18 at 17:46

lhf · Accepted Answer · 2018-08-07T10:26:34.907

1

We have $$ (\mathbb{Z}\times \mathbb{Z}_{12})/ \langle (4, 10) \rangle \cong (\mathbb{Z}\times \mathbb{Z})/ \langle (4, 10), (0,12) \rangle $$ Therefore, the group has $48$ elements, because $$ \begin{vmatrix} 4 & 10 \\ 0 & 12 \end{vmatrix} = 48 $$ If you want to determine which group it is, compute the Smith normal form of this matrix.

edited Aug 07 '18 at 10:26

answered Aug 05 '18 at 11:27

lhf

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what is the order of the quotient group $(\mathbb{Z}\times \mathbb{Z}_{12})/\left \langle (4, 10) \right \rangle$?

1 Answers1