0
$\begingroup$

If we have group $G$ whose order $n$ is a prime number, then we know that there are only two subgroups in this group - the trivial subgroup and $G$ itself. Now, we can generate $G$ by $\langle g \rangle$ for some $g \in G$, and $\langle g \rangle$ must be of order $n$.

This should mean that $G$ must be a cyclic group.

However, I'm wondering about two peculiar things:

(1) If $G$ is cyclic and can be represented as $\langle g \rangle = \{g^n\}$, then there must be an element, such as $g^2 \in G$. Then, if we use it as a generator, we'll have $\langle g^2 \rangle = \{1, g^2, g^4, g^6,...\}$, and this set does not appear to include odd powers of $g$. How is this possible? I must be missing something.

(2) Can there exist a non-cyclic group of the same prime order $n$, which is just isomorphic to $\langle g \rangle$, but not cyclic?

Some clarification would be appreciated. I feel that something is not layered properly in my deductions.

$\endgroup$
3
$\begingroup$

Let's just look at an example - I think that will clarify. Consider $\mathbb{Z}_5$. This is a group that meets your criterion above. Now let's say that in some representation we have $\mathbb{Z}_5 = <g>$. We should list the elements generated by $<g^2>$ and see what happens. We have $g^2, g^4, g^6, g^8,....$ What happens with these powers of $g$ that are larger than 5? Well if $g$ was a generator, you know that $g^6=g$, $g^8=g^3$ etc (if this isn't clear, reflect for a moment on why these groups are called cyclic. What fundamental properties do they have?). Now you can see that all the powers are there, since $g^0=g^{10}=e$ in the group.

In fact this happens in general. If $G$ is a cyclic group, all elements of $G$ are powers of a generator $g$ as you noticed. If the order of the group and the power of the element in question are coprime, then this is also a generator of the group. The number of generators of a cyclic group are given by the Euler-function.

In particular, a group with one generator is also always cyclic. This is because the only things you can do with one generator are multiply it by itself and take it's inverses, so you always end up with just powers of that generator, be they finite or infinite.

$\endgroup$
  • $\begingroup$ Thank for your clear answer. I didn't quite understand one thing though: how can there be a generator of an order coprime with the order of G, given that the order of a cyclic group is equal to the order of the generator, and the cyclic group is also a subgroup of G, and thus its order must divide the order of G. $\endgroup$ – sequence Sep 27 '15 at 4:13
  • 1
    $\begingroup$ The power of the element is coprime with the order. Not the order of the element. In the example above, note that 2 is coprime with 5. In a cyclic group of order 10, the subgroup generated by $g^7$ is also the entire group, as is $g^9$ etc. But on the other hand, the subgroup generated by $g^2$ is only the even powers, since 2 and 10 are not coprime. $\endgroup$ – Alfred Yerger Sep 27 '15 at 4:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.