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

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.

• Another method is to use the Lagrange's Theorem. Using Lagrange Theorem, you can prove that any group of prime order is cyclic. – Thomas Shelby Jan 3 at 17:23
• 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. – José Carlos Santos Jan 3 at 17:31
• I recommend to read my answer here math.stackexchange.com/questions/1570641/…. A similar method works for a group of 4 elements. – Nicky Hekster Jan 3 at 17:52
• @ThomasShelby: you mean Cauchy's theorem, right? Lagrange's theorem only gives $o(g)\mid o(G)$, while Cauchy's theorem ensures $o(G)=p\Longrightarrow\exists g:o(g)=p$. – Jack D'Aurizio Jan 3 at 20:27
• @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. – Jack D'Aurizio Jan 4 at 1:22

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,
$$$$ must be whole $$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.

• 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." – El Mouden Jan 8 at 12:23
• 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 $G$. – Coupeau Jan 9 at 17:49

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

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