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How to prove that $\mathbb{Z/2Z\times Z/2Z}$ is not cyclic?

Here is what I did, what is wrong with it if anything and is there a better way to prove it? A group $G$ is cyclic if $\exists\ a \in G$ such that $\forall g \in G, \exists\ k \in N$ such that $g=a^k$.

Here an element $g$ of $G=\mathbb{Z/2Z\times Z/2Z}$ writes $g=(x,y)$ with $x,y \in \{\bar{0},\bar{1}\}$, ie $g \in \{(0,0),(0,1),(1,0),(1,1)\}$

and there exists no $a \in G$ and $k,k',k'' \in N$ such that $a^k=(0,0), a^{k'}=(0,1)$ and $a^{k''}=(1,0)$

user1729
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astudentofmaths
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  • Should this question maybe be closed as a duplicate of this one? This makes sense given the dates, but this one is slightly more general ("not cyclic" versus "not cyclic of order $4$"). – user1729 Aug 11 '21 at 13:49

2 Answers2

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Because the group is so small, the easiest elementary way is probably finding the order of each nonzero element:

  • $2\cdot(1,0) = (0,0)$
  • $2\cdot(0,1) = (0,0)$
  • $2\cdot(1,1) = (0,0)$

So we can conclude that every element has at most order $2$, so no element has order $4$, therefore no element generates the whole group, therefore the group is not cyclic.

flawr
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  • I don't understand your answer. Why do you multiply by 2 each non-null element of $Z/2Z$ x $Z/2Z$? What does it imply that the product is 0 each time? To show that $a \in G$ is of order $2$ we must show that $a^2=e$ and $a^1 \neq e$, $a^0 \neq e$. I don't understand how it is related with what you wrote. – astudentofmaths Jun 17 '18 at 12:30
  • @astudentofmaths I used additive notation (as it is usually done with commutative groups), and showed exactly what you just wrote. But in the additive notation (in contrast to the multiplicative notation that you just used) the equation $a^2 = e$ becomes $2a = e = 0$. – flawr Jun 17 '18 at 14:14
  • A very simple change in your argument shows that $\mathbb{Z}/312465\mathbb{Z}\oplus\mathbb{Z}/312465\mathbb{Z}$ is not cyclic. – egreg Jun 17 '18 at 14:14
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What you did was fine.

Another way would be to prove that $\mathbb{Z}_2 \times \mathbb{Z}_2$ is not isomorph with $\mathbb{Z}_4$.

Since the latter is the only cyclic group of order $4$ (up to isomorphism), the result follows.

To actually prove they are not isomorph, you can look at the orders of the elements of both groups.

  • Actually I wanted to show that $Z2×Z2$ was not cyclic to prove that it is not isomorph with $Z4$ which is cyclic. What do you mean by "To actually prove they are not isomorph, you can look at the orders of the elements of both groups."? – astudentofmaths Jun 17 '18 at 12:25
  • Isomorphisms preserve orders of elements. So, if one group has an element of order 4 and the other one doesn't they can't be isomorphic. But, a finite group is cyclic iff it contains an element with order the cardinality of the group, so checking that $Z_2 \times Z_2$ does not have an element of order $4$ is equivalent with that the group is not cyclic, and there is actually no need for an isomorphism argument. –  Jun 17 '18 at 12:27
  • Ok with the proof, thanks, but what element in $Z_4$ is of order 4? And: "a finite group is cyclic iff it contains an element with order the cardinality of the group, so checking that Z2×Z2 does not have an element of order 4 is equivalent with that the group is not cyclic" -> how comes $Z_2$ x $Z_2$ is of cardinal 4? I wrote that there were only 3 elements ($(0,0), (0,1),(1,0)$) so it should be of cardinal 3 right? – astudentofmaths Jun 17 '18 at 12:35
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    (1) Look at the element $1$ in $Z_4$. Then $1+1+1+1 = 0$, (2) You forgot $(1,1)$ –  Jun 17 '18 at 12:39
  • (1) I don't understand. The identity $e$ is $1$ as $k*1=k\ \forall k$ and $g \in G$ (G of order $n$) is of order $k$ if $a^k \equiv e (\mod n)$. So 1 should be of order 0 as $1^0=1 \equiv e (\mod n)$ Why do you multiply 1 by 4? Where does that come from? – astudentofmaths Jun 17 '18 at 12:47
  • I see where your confusion comes from. You think of $Z_4$ as a multiplicative group, while it is an additive group. The operation is addition, not multiplication. This means, $* = +$ here. The identity for this additive group is $0$ btw, not $1$. –  Jun 17 '18 at 12:49
  • Let us continue this discussion in chat. – astudentofmaths Jun 17 '18 at 12:50