Is every Subgroup of a Cyclic Group itself Cyclic?

[robertjordan]

Homework Statement

Are all subgroups of a cyclic group cyclic themselves?

Homework Equations

G being cyclic means there exists an element g in G such that <g>=G, meaning we can obtain the whole group G by raising g to powers.

The Attempt at a Solution

Let's look at an arbitrary subgroup H= {e, g^k_1, g^k_2, ... , g^k_n}

We know since subgroups are closed that (g^k_i)^t is in H for all integers t, and for all i between 1 and n.

So unless g^k_1, g^k_2,... all have order 1 (which would mean H ={e}), then by the pidgeonhole principle, we have (g^k_i)^x = (g^k_j)^y
for some i,j and some 0<x<ord(g^k_i), 0<y<ord(g^k_2).

WLOG let's say y<x. Then x=yq+r. This implies (g^k_i)^r=(g^k_j).







... I'm stuck

 

[jbunniii]

Hint: if $H$ is a subgroup of $G$, then of course all of its elements are of the form $g^k$. Consider the element of $H$ with the smallest positive exponent. Can you show that it generates $H$?

 

[robertjordan]

Hint: if $H$ is a subgroup of $G$, then of course all of its elements are of the form $g^k$. Consider the element of $H$ with the smallest positive exponent. Can you show that it generates $H$?

Assume BWOC that g^k doesn't generate H. Then there is an element g^t in H such that k doesn't divide t. But that means t=kq+r where r<k. Which means g^r is in the H but that contradicts k being the smallest power of g in H.



Thanks!