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Q. Show that $\mathbb{Z_3 x Z_4}$ is a cyclic group.

So my question is there a faster way besides listing all the elements and besides knowing the theorem.

Since the process I am doing is:

I know: $\mathbb{Z_3 x Z_4}=\{(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3),(2,0),(2,1),(2,2),(2,3)\}$

and I've been taking each element and figuring out the order such as:

$$(1,0): (1,0); (1,0)+(1,0)=(2,0);(2,0)+(1,0)=(0,0)$$ which is order 3.

and I believe I have to keep doing this until I find the element that gives me order 12.

but I realize that is kinda tedious. So my question is, is there a simpler way or would be it suggested to continue on the method I am doing.

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    $\begingroup$ try $(1,1)$, strategy: don't pick elements with zeros.. $\endgroup$ – MAM Mar 14 '15 at 23:00
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You can calculate the order of an element without listing every element. For example, you can show that $(1,0)$ has order $3$ and $(0,1)$ has order $4$.

So $(1,1)$ has order divisible by $3$, since the first element has order $3$. Similarly its order must be divisible by $4$. Hence, it must have order $12$ since $3,4$ are coprime and the order of the group is $12$.

This is a group theoretic way of phrasing the Chinese Remainder theorem.

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  • $\begingroup$ how can we know that $(1,0)$ has order $3$ and $(0,1)$ has order 4? $\endgroup$ – JOJO Sep 12 at 11:30
  • $\begingroup$ @JOJO calculate $(0, 1), (0, 1) + (0, 1), (0, 1) + (0, 1)+(0, 1)$ etc, and note when it's first equal to $(0, 0)$ $\endgroup$ – Mathmo123 2 days ago
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To find an element of order $12$ don't check elements randomly: think about which elements might possibly be of order $12$. For example, if for $(x,y)\in \mathbb{Z}_3\times\mathbb{Z}_4$ one of the components is not a generator for it's group, you won't possible generate all $12$ different elements. So why don't you look at a pair $(x,y)$ where $x$ is a generator for $\mathbb{Z}_3$ and $y$ is a generator for $\mathbb{Z}_4$.

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Taking $(1,1)$ in $\mathbb{Z_3 x Z_4}$, we have $12(1,1)=0$ and $n(1,1)\neq0$ if $n<12$. That is, it is cyclic.

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This is the abstract version of the Chinese Remainder Theorem::

If $m$ and $n$ are two coprime integers, then $\mathbf Z/m\mathbf Z\times \mathbf Z/n\mathbf Z\simeq \mathbf Z/mn\mathbf Z$.

In the present case, consider the canonical morphism: \begin{align*} \varphi \colon\mathbf Z\ &\rightarrow \mathbf Z/3\mathbf Z\times \mathbf Z/4\mathbf Z\\ x&\mapsto (x\bmod 3,x\bmod 4) \end{align*} Bézout's identity for $3$ and $4$ is simply $4-3=1$. It is easy to see $\varphi$ is surjective. Indeed, if $(\bar a, \bar b)$ is an element of $\mathbf Z/3\mathbf Z\times \mathbf Z/4\mathbf Z$, we have: $$\varphi(4a-3b)=(4a-3b\bmod 3, 4a-3b\bmod 4)=(a\bmod 3, b\bmod 4)$$ since $4\equiv 1\mod 3$, $-3\equiv 1\mod 4$. Now $x\in \ker\varphi$ if and only if $x$ is divisible by both $3$ and $4$, i.e. $x$ is divisible by $12$, whence an isomorphism; $$\bar\varphi\colon \mathbf Z/12\mathbf Z\rightarrow \mathbf Z/3\mathbf Z\times \mathbf Z/4\mathbf Z. $$

This isomorphism proves $ \mathbf Z/3\mathbf Z\times \mathbf Z/4\mathbf Z$ is cyclic, of order $12$.

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The fastest "minimal computation" way:

$6(1,1) = (0,2) \neq (0,0)$

$4(1,1) = (1,0) \neq (0,0)$. Therefore, $(1,1)$ has order $12$ (thanks to Bill Dubuque).

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    $\begingroup$ Note: even if you didn't know to test $(1,1)$, you could just test the $6$ elements $(a,b)$ with $a,b \neq 0$. So, $12$ computations at most (actually less, since we have $4$ elements of order $12$, so even if we hit the "bad" $2$ elements first, we'd hit success on the $3$rd try, so $6$ computations). $\endgroup$ – David Wheeler Mar 15 '15 at 1:37

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