# Show that order of the element $(1,1)$ of $\mathbb{Z}/2 \times \mathbb{Z}/3$ is $6$

I need to show that the order of the element $(1,1)$ of $\mathbb{Z}/2 \times \mathbb{Z}/3$ is $6$. I am not sure if the binary operation is addition modulo or multiplication modulo. Both cases can be tried.

So, basically I need to show that $(1,1)^6 = (1,1)$ (if binary operation is multiplication modulo) or $(1,1)^6 = (0,0)$ if binary operation is addition modulo.(please correct me here if I am wrong). But what will be the product after $(1,1)*(1,1)$ where $*$ can be any one of the two binary operations.

Any kind of help is appreciated.

• Hint $\ n(\color{#c00}1,\color{#0a0}1) = (\color{#c00}n,\color{#0a0}n) = (\color{#c00}0,\color{#0a0}0)\,$ in $\, \Bbb Z/\color{#c00}2\times \Bbb Z/\color{#0a0}3\iff \color{#c00}{2\mid n},\,$ and $\,\color{#0a0}{3\mid n}\ \$ Commented Sep 7, 2016 at 14:52

The usual operation in $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/3\mathbb{Z}$ as groups is the addition, component-by-component. In this case, it is easy to show that $$(1,1)^{\oplus 6}=(1,1)+(1,1)+(1,1)+(1,1)+(1,1)+(1,1)=(6 \mod 2, 6\mod 3)=(0,0).$$

As rings there is a product component by component, but in this case $(1,1)$ is the neutral element.

As mathmax pointed out in his comment, you also have to show that $(1,1)^{\oplus k}\neq (0,0)$ for $i=1,2,3,4,5$. Here there are the calculations: \begin{align*} &(1,1)^{\oplus 1}=(1,1) & & (1,1)^{\oplus 2}=(2,2)=(0,2) & &(1,1)^{\oplus 3}=(3,3)=(1,0)\\ &(1,1)^{\oplus 4}=(4,4)=(0,1) & & (1,1)^{\oplus 5}=(5,5)=(1,2) & &(1,1)^{\oplus 6}=(6,6)=(0,0) \end{align*}

• By neutral element you mean identity element? Commented Sep 7, 2016 at 13:59
• Yes, I meant the identity element (that element $a$ such that $x\cdot a=x$ for all $x$ in the ring) Commented Sep 7, 2016 at 14:02
• Actually, your computation only shows that $6$ is an upper (resp. a multiple) bound for the order of $(1,1)$. One has to note further that $(1,1)^k\neq (0,0)$ for $k=1,2,3,4,5$.
– user149890
Commented Sep 7, 2016 at 14:22
• @mathmax that's true. I initally intended to calculate each of the powers, but I forgot! Commented Sep 7, 2016 at 14:23