Understanding the Cyclic Property of Groups

[PsychonautQQ]

Homework Statement

My online notes stated that it |g| = |G| where g is an element of G then |G| is cyclic.

Can somebody help me understand why this is true?

 

[jbunniii]

I assume $|g|$ means the order of $\langle g \rangle$, the subgroup generated by $g$.

Note that $\langle g \rangle \subseteq G$.

If $|G|$ is finite and $|g| = |G|$, then $\langle g \rangle \subseteq G$ implies $\langle g \rangle = G$. (Do you see why?) What can you conclude?

By the way, the result need not be true if $|G|$ is infinite. For a counterexample, let $G = \mathbb{Q}$, the additive group of rational numbers, and let $g = 1$. Then $\langle g \rangle = \mathbb{Z}$, the additive subgroup of integers. Then $G$ and $\langle g \rangle$ have the same cardinality (countably infinite) but $G$ is not cyclic.

 

[pasmith]

Homework Statement

My online notes stated that it |g| = |G| where g is an element of G then |G| is cyclic.

Can somebody help me understand why this is true?

If |g| = |G| is finite, can there exist any elements of G which are not powers of g?

 

[HallsofIvy]

No. If |g|= |G| then, by definition of "| |", g and G contain the same number of terms. Since <g> is always a subgroup of G, it follows that g is exactly the same as G. Since every member of <g> is a power of g, every member of G is a power of g.

 

[PsychonautQQ]

thanks yall :D