# Is every Subgroup of a Cyclic Group itself Cyclic?

1. Mar 20, 2013

### robertjordan

1. The problem statement, all variables and given/known data

Are all subgroups of a cyclic group cyclic themselves?

2. Relevant 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.

3. The attempt at a solution

Let's look at an arbitrary subgroup H= {e, gk_1, gk_2, ... , gk_n}

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

So unless gk_1, gk_2,... all have order 1 (which would mean H ={e}), then by the pidgeonhole principle, we have (gk_i)x = (gk_j)y
for some i,j and some 0<x<ord(gk_i), 0<y<ord(gk_2).

WLOG let's say y<x. Then x=yq+r. This implies (gk_i)r=(gk_j).

... I'm stuck

2. Mar 20, 2013

### 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 $H$?

3. Mar 20, 2013

### robertjordan

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!