# Prime order of group

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.

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.
• 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. – Alfred Yerger Sep 27 '15 at 4:14