MHB A cyclic group with only one generator can have at most two elements

[mathmari]

Hey!

Show that a cyclic group with only one generator can have at most two elements.

I thought the following:

When $a \neq e$ is in the group, then $a^{-1}$ is also in the group.
So, when $a$ is a generator, then $a^{-1}$ is also a generator.

Is this correct?? (Wondering)

But I how can I use this to show that a cyclic group with only one generator can have at most two elements??

Or should I use something else to prove this?? (Wasntme)

 

[Euge]

Hey!

Show that a cyclic group with only one generator can have at most two elements.

I thought the following:

When $a \neq e$ is in the group, then $a^{-1}$ is also in the group.
So, when $a$ is a generator, then $a^{-1}$ is also a generator.

Is this correct?? (Wondering)

But I how can I use this to show that a cyclic group with only one generator can have at most two elements??

Or should I use something else to prove this?? (Wasntme)

Hi mathmari,

By having an $a\in G$ different from $e$, you are assuming $G$ has more than one element. It's true that $a$ and $a^{-1}$ will generate $G$, but this does not follow from the fact that $a^{-1}$ is in the group. Instead, it follows from the fact that any element in a group has the same order as its inverse. Now, knowing that $a$ and $a^{-1}$ generate $G$ and $G$ has only one generator, you deduce that $a = a^{-1}$, i.e., $a^2 = e$. So $G$ has order two, with elements $e$ and $a$.

Of course, if $G = {e}$, then $G$ is cyclic with one generator. So all cases have been covered.

 

[mathmari]

Hi mathmari,

By having an $a\in G$ different from $e$, you are assuming $G$ has more than one element. It's true that $a$ and $a^{-1}$ will generate $G$, but this does not follow from the fact that $a^{-1}$ is in the group. Instead, it follows from the fact that any element in a group has the same order as its inverse. Now, knowing that $a$ and $a^{-1}$ generate $G$ and $G$ has only one generator, you deduce that $a = a^{-1}$, i.e., $a^2 = e$. So $G$ has order two, with elements $e$ and $a$.

Of course, if $G = {e}$, then $G$ is cyclic with one generator. So all cases have been covered.

Have we proven in that way that a cyclic group with only one generator can have at most two elements??

 

[Euge]

Certainly. Either $G$ has one element (which is cyclic with only one generator) or it has more than element, in which case $G$ has only two elements.

 

[Deveno]

Have we proven in that way that a cyclic group with only one generator can have at most two elements??

Yes, but it might be hard for you to SEE.

Suppose that we have $b \in G$ with $b \neq a,e$. Since $a$ generates $G$, it must be that $b = a^k$ for some integer $k$. However, since $a^k = e$ if $k$ is even, and $a^k = a$ if $k$ is odd, there is no such $b$.