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Prove that a group with $3$ elements is cyclic.

I tried the case where $G=\{e,a,b\} $ and I kept trying multiplication and finally I found that $a^2$ must equal to $b$ and $b^2$ must equal to $a$. Then $a^3=e$.

Are there any other methods ? I have another question :

Prove that a group with $4$ elements may or may not be cyclic.

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    $\begingroup$ Another method is to use the Lagrange's Theorem. Using Lagrange Theorem, you can prove that any group of prime order is cyclic. $\endgroup$ – Thomas Shelby Jan 3 at 17:23
  • $\begingroup$ I think that it doesn't take a lot of work to prove that any group of order $4$ (note that I omitted the word “finite”; are you aware of some infinite group of order $4$?) may by cyclic or not. $\endgroup$ – José Carlos Santos Jan 3 at 17:31
  • $\begingroup$ I recommend to read my answer here math.stackexchange.com/questions/1570641/…. A similar method works for a group of 4 elements. $\endgroup$ – Nicky Hekster Jan 3 at 17:52
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    $\begingroup$ @ThomasShelby: you mean Cauchy's theorem, right? Lagrange's theorem only gives $o(g)\mid o(G)$, while Cauchy's theorem ensures $o(G)=p\Longrightarrow\exists g:o(g)=p$. $\endgroup$ – Jack D'Aurizio Jan 3 at 20:27
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    $\begingroup$ @ThomasShelby: all right, I get it. Since the identity is unique and $o(g)\mid o(G)$, in a group with prime order all the elements except the identity are generators, fine. $\endgroup$ – Jack D'Aurizio Jan 4 at 1:22
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According to Lagrange's theorem, a group with 3 elements can have only the trivial subgroups. Because 3 is prime only 1 and 3 divide it. So, only possible subgroups are $\{e\}$ with one element and $G$ with three. For $x \in G$ that is not $e$ a group generated with this element,
$<x>$ must be whole $G$. So $G$ is cyclic.

A group with 4 elements can have untrivial subgroups as 4 isn't prime. For instance $\mathbb{Z_4} = \{0, 1, 2, 3\}$ with $+$ is cyclic. It is generated with the element $1$. But group $\mathbb{Z_2} \times \mathbb{Z_2} = \{(0, 0), (1, 0), (0, 1), (1, 1)\}$ with $+$ defined as $(a, b) + (a', b') = (a + a', b + b')$ is not cyclic.

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  • $\begingroup$ can you explaine in more terms " For x∈G that is not e a group generated with this element, <x> must be whole G. So G is cyclic." $\endgroup$ – El Mouden Jan 8 at 12:23
  • $\begingroup$ We can take any element (that is not the unit $e$) $x \in G$ and by multiplying it with itself, we will get every other element of this group. So let $x \in G$. $x x$ is another element that must be in the group but is different from $x$ since we said $x$ is not a unit. So, now we have three distinct elements: $e, x, x^2$. We also know, that since our group can only have either one or three elements, $x^2$ can not be the unit. Now we have the whole $G$. $\endgroup$ – Coupeau Jan 9 at 17:49
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Instead of “keeping multiplying” it's easier to fill the Cayley diagram: in every row and column every element must appear.

\begin{array}{c|ccc} & e & a & b \\ \hline e & e & a & b \\ a & a & \\ b & b & \\ \end{array} In the slot corresponding to $a^2$ you cannot put $e$, because otherwise the slot for $ab$ would contain $b$. Hence $a^2=b$ \begin{array}{c|ccc} & e & a & b \\ \hline e & e & a & b \\ a & a & b\\ b & b & \\ \end{array} and now the diagram has a unique completion \begin{array}{c|ccc} & e & a & b \\ \hline e & e & a & b \\ a & a & b & e \\ b & b & e & a\\ \end{array} Then $a^3=a^2a=ba=e$.

You can try your hand with a four element group and see that the diagram admits different completions.

\begin{array}{c|cccc} & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & \\ b & b & \\ c & c & \end{array} In the slot for $a^2$ we can put any of $e$, $b$ or $c$. Let's try with $e$: \begin{array}{c|cccc} & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & e\\ b & b & \\ c & c & \end{array} Then we are forced to put in the following \begin{array}{c|cccc} & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c\\ c & c & b \end{array} For $b^2$ we can have either $e$ or $c$. Let's try $e$: \begin{array}{c|cccc} & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & e\\ c & c & b & \end{array} Now we can complete: \begin{array}{c|cccc} & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & e & a\\ c & c & b & a & e \end{array} It would be very tedious to verify that this diagram indeed produces a group. It does and is the product of two cyclic groups of order two; it is also the subgroup of $S_4$ given by $\{id,(12)(34),(13)(24),(14)(23)\}$.

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  • $\begingroup$ I need another proof, without using the Cayley diagram $\endgroup$ – El Mouden Jan 8 at 12:25
  • $\begingroup$ @ElMouden Then using Lagrange's theorem is what you need. $\endgroup$ – egreg Jan 8 at 12:59
  • $\begingroup$ Oh Thanks Finally i did it $\endgroup$ – El Mouden Jan 9 at 14:07

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