if $\displaystyle a$ is generator of a cyclic group, show that $\displaystyle a^-1$ is also the generator of that group.

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- Oct 21st 2009, 10:03 AM #1

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- Oct 21st 2009, 10:51 AM #2

- Oct 21st 2009, 11:01 AM #3
I believe you forgot the word "Finite"...

So, assuming we are in finite groups:

The question is essentially asking you "does $\displaystyle o(a)=o(a^{-1})$ hold?", where $\displaystyle o(g)$ is the order of the element $\displaystyle g \in G$. Can you see why these two things are equivalent?

If you can, then the proof becomes much easier. Assume they have different orders, and w.l.o.g. let $\displaystyle o(a)=n<m=o(a^{-1})$. Then plug this in and if you can't find a contradiction just say and someone will give more help.

I should point out that*by definition*, $\displaystyle <X>=<X^{-1}>=<X,X^{-1}>$ where $\displaystyle X^{-1}=\{x^{-1}:x \in X\}$. That is to say, when we talk about generating elements of a group then the inverse of the elements are also taken to be in the generating set. For instance, $\displaystyle (\mathbb{Z},+)=<1>$, however clearly no sum of 1's will give you the element $\displaystyle -1$.

Also, try to post fewer new threads in such a short space of time. It make people think you just haven't tried. At the very least, put them all into one thread...

- Oct 21st 2009, 03:48 PM #4

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- Oct 21st 2009, 04:56 PM #5

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How is $\displaystyle \mathbb{Z}$ generated by $\displaystyle 1$ and not by $\displaystyle -1$? If we use additive notation then $\displaystyle <1>= \{ n1 : n \in \mathbb{Z} \} =<-1>$. Please explain

Actually, doesn't this work for arbitrary group $\displaystyle G$: since $\displaystyle a^{-1} \in <a>$ and $\displaystyle a \in <a^{-1}>$ (since they're both groups) we have $\displaystyle <a^{-1}> \subseteq <a> \subseteq <a^{-1}>$ and so $\displaystyle <a>=<a^{-1}>$

- Oct 21st 2009, 11:10 PM #6
In reply to aman_cc and Jose27, essentially it comes down to how we define $\displaystyle <X>$ where $\displaystyle X \subseteq G$.

Technically, it is defined to be,

$\displaystyle <X> = \displaystyle\bigcap_{X \subset H \leq G}H$.

Basically, what Jose27 said is what I mean - you take powers of integers not natural numbers. Although this is correct it is, perhaps, less correct than saying $\displaystyle <X>:=<X \cap X^{-1}>$. For instance, when you construct the free group over the alphabet $\displaystyle X$ you would say it in this way (as in, you take an element of your alphabet and you have to place it's inverse in by hand).

Also, when working in a finite group you would just use powers of natural numbers, as you get the inverses for free. Thus, people don't think about the inverses.

Finally, if we talk about the generator of a group as a semigroup then you*cannot*take integer powers, you have to use the natural numbers. So, for instance, as semigroups a finite cyclic group is just $\displaystyle <a>$ for some $\displaystyle a \in G$, but $\displaystyle \mathbb{Z}=<1, -1>=<11, -123>$

In short, $\displaystyle \mathbb{Z}=<1>$, and answering the question the way I did it stops it becoming a non-question and is probably because I play to much with semigroups...