Question

True or false: If every proper subgroup of a group G is cyclic, then G is cyclic. Justify your answer.

Short answer

The statement is false.

Step-by-step solution

Consider the given statement

If every proper subgroup of a group G is cyclic, then G is cyclic.

Check whether the statement is true or false 

Consider the group as follows:

$G={\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$

The subgroups of this group are$\left\{\left(0,0\right)\right\},\left\{\left(0,0\right),\left(0,1\right)\right\},\left\{\left(0,0\right),\left(1,0\right)\right\},\left\{\left(0,0\right),\left(1,1\right)\right\}$, and $\left\{\left(0,0\right),\left(1,0\right),\left(0,1\right),\left(1,1\right)\right\}$. Therefore, all proper subgroups of ${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$are cyclic but ${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$is not cyclic.

Assume that there exits an element a from${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$ such that it generates ${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$but this is not possible because each element generates proper subgroups written above.

For example, element (1,0) generates subgroup {(0,0),(1,0)} ,which is a proper subgroup of ${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$.

Therefore, the statement is false.