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

Question

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.

Answer 1

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.

Answer 2

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 $\{id,(12)(34),(13)(24),(14)(23)\}$.