# What is the order of $(3, 1) + \langle(0, 2)\rangle$ in the quotient group $\mathbb{Z}_4 \times \mathbb{Z}_8 /(0, 2)$?

Let $\langle(0, 2)\rangle$ denote the subgroup generated by $(0, 2)$ in $\mathbb{Z}_4 \times \mathbb{Z}_8$.

How I can find the order of $(3, 1) + \langle(0, 2)\rangle$ in the quotient group $\mathbb{Z}_4 \times \mathbb{Z}_8 /(0, 2)$ ?

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Well $$((3, 1) + \langle(0, 2)\rangle)^n = (3, 1)^n + \langle(0, 2)\rangle = (3n, n) + \langle(0, 2)\rangle.$$

The order of $(3, 1) + \langle(0, 2)\rangle$ is the smallest positive value of $n$ such that $((3, 1) + \langle(0, 2)\rangle)^n = \langle(0, 2)\rangle$, the identity of the quotient group $\mathbb{Z}_4\times\mathbb{Z}_8/(0,2)$.

When is $(3n, n) + \langle(0, 2)\rangle = \langle(0, 2)\rangle$? Precisely when $$(3n, n) \in \langle(0, 2)\rangle = \{(0, 0), (0, 2), (0, 4), (0, 6)\}.$$

So we need to determine the smallest positive value of $n$ such that $3n \equiv 0 \operatorname{mod} 4$ and $n \equiv 0, 2, 4, \operatorname{or} 6 \operatorname{mod} 8$. I'll leave that to you.

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Thanks Sir !!! I find that the answer is 4. – ram Jan 3 '13 at 5:54
Simple and nice one +1 – Babak S. Jan 3 '13 at 10:12

Let $G$ be a group, and $H$ a normal subgroup of $G$.

In the quotient group $G/H$, the identity element is the coset $H$. Thus, an element $gH\in G/H$ is the identity if and only if $g\in H$. By definition, the order of $gH\in G/H$ is the smallest $n\geq 1$ such that $$(gH)^n=g^nH=H,$$ which (by our observation above) is the same as the smallest $n\geq 1$ such that $g^n\in H$.

Recall that, by convention, in an abelian group we write the operation as "$+$", so that instead of referring to $g^n=\underbrace{g\cdots g}_{n\text{ times}}$, we write $ng=\underbrace{g+\cdots+g}_{n\text{ times}}$.

Thus, what you should do is determine the elements comprising the subgroup $\langle (0,2)\rangle$, and then compute $g,2g,3g,\ldots$ where $g=(3,1)$ until you find an $ng\in \langle (0,2)\rangle$. The smallest such $n$ will be the order of $g$.

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