Answer to Question #105098 in Abstract Algebra for Sourav Mondal

Question

Show that if G is a non-cyclic group of order n, then G has no element of order
n. Further, give an example, with justification, of a non-cyclic group with all its
proper subgroups being cyclic.

Answer 1

If $G$ conains element $g$ of order $n$, then $G$ conains cyclic subgroup $<g>$ of order $n$.

Then $|G\setminus<g>|=|G|-|<g>|=n-n=0$ ($G\setminus<g>$ is complement $<g>$ in $G$ ), that is $G=<g>$ is a cyclic group. So if $G$ is a non-cyclic group of order $n$ , then $G$ has no element of order $n$.

$Z_2\oplus Z_2$ is a non-cyclic group with cyclic proper subgroups $\{(0,0)\}, \{(0,0),(0,1)\}, \{(0,0),(1,0)\}, \{(0,0),(1,1)\}$

$Z_2\oplus Z_2$ does not have other proper subgroups, because $\langle (1,0),(0,1)\rangle=\langle (1,0),(1,1)\rangle=\langle (1,1),(0,1)\rangle=$

$=Z_2\oplus Z_2$

So every proper subgroup of $Z_2\oplus Z_2$ is a cyclic group.