Why is $\Bbb Z_3$ cyclic?

Question

I want to prove that $\Bbb Z_3$, defined as the only group of order $3$, is cyclic.

The identity element $e$ needs to be there, so $\Bbb Z_3=\{a,b,e\}$. Now, $ab$ can't be nor $a$ or $b$, otherwise one of them would be $e$, so $ab=e \implies b=a^{-1}$. Now, the next step would be proving that $a^2=a^{-1}$, but why is that necessary? Couldn't the multiplication table be like this? $$ab=ba=e, \quad a^2=a, \quad ({a^{-1}})^2=a^{-1}$$

Answer 1

From $a^2 = a$ we can multiply both sides with $a^{-1}$ and thus obtaining $a=e$. This is a contradiction(since $a$ and $e$ are two different elements), so you cannot have the multiplication table stated.

Answer 2

$a^2$ could be $1$ or $a$ or $b$.

• In the first case $ab\ne aa=1$ and $ab\ne a1=a$ and $ab\ne 1b=b$.
• In the second $a=a1\ne aa=a$

Answer 3

By Lagrange's theorem, the order of $a$ is divisor of $3$. So either it is $1$, which means $a=e$, which is excluded, or it is $3$, i.e. $a^3=e$, which is equivalent to $a^2=a^{-1}$.

Answer 4

Are you asking why we must prove it? Or why it must be true.

Why we must prove it: We need to prove any 3 element group is cyclic. that means there is an element $a \ne e$ and the three elements are precisely $e,a$ and $a^2$.

So if you have that the elements are $e,a, a^{-1}$ you haven't proven that yet. You have to prove the third element is $a^2$. It's not enough to prove it is $a^{-1}$.

that's why you have to prove it.

Why must it be true: Well, .... that's what you have to prove, isn't it?

You ask if $a^2 = a$ isn't possible. $a^2 = \color{red}a\color{blue}a \ne \color{red}a$ for the exact same reason $\color{red}a\color{blue}b \ne \color{red}a$.

.... which come to think of it you didn't give a good reason.

$e$ is the identity element but there is nothing in the definition of a group that says that $e$ is the only identity, nor does it mean if $ab = b$ that $a=e$. Maybe there are two elements $e$ and $a$ where $ab=b$ and $eb=b$

But note that if $ab = a$ then $a^{-1}ab= a^{-1}a$. But $a^{-1}ab = eb =b$ and $a^{-1}a = e$ so that can only happen if $b=e$.

Likewise if $a^2 = a$ then $a^{-1}a^2 = a^{-1}a$. But $a^{-1}a^2 = a^{-1}aa =ea =a$ and $a^{-1}a = e$ so $a^2 = a\implies a = e$. (That's true for all groups).