Let $G$ be a group of order $9$.
1) State the possible orders of subgroups and elements in $G$.
2) Find the number of elements of $G$ of order $3$ in the cases where (a) $G$ is
non-cyclic, (b) $G$ is cyclic.
I have a problem with the (b). Here is what I did so far, is it right ?
We know that for G a group of finite order, $\forall H \leq G$ then $\vert H \vert $ divides $\vert G\vert$. $G$ has order $9 =3*3=1*9$, so the possible order of subgroups and elements in $G$ are 3 and 9.
a) If $G$ is not cyclic then there can't be an element $a\in G$ such that $<a>=G$ ie there is not $a\in G$ with $\vert a \vert = \vert G \vert = 9$ so the number of elements in $G$ of order 3 is 2.
b) If $G$ is cyclic then I guess the are none (probably because the only order possible is then 9, but I don't see how to prove it properly.
Thanks
Best Answer
Hints:
(1) For $\;G\;$ cyclic, $\;|G|=n\;$ : for any divisor $\;d\;$ of $\;n\;$, there's exactly one unique subgroup of $\;G\;$ of order $\;d\;$ which, of course, is also cyclic.
(2) A cyclic group of order $\;n\;$ always has $\;\phi(n)\;$ elements of order $\;n\;$ (and thus also generators of the whole group).