Proof's about cyclic groups

Question

Let $(G,+)= \langle g \rangle$ be a cyclic group with $\vert G \vert = n < \infty $

(i) Show that for every divisor $d\geq 1$ of n exists exactly one subgroup with cardinality $d$ of $G$, namely $U_d := \langle \frac{n}{d}g \rangle$.

(ii) For Divisors $d, e$ of $n$ applies: $U_d \subseteq U_e \leftrightarrow d \ \vert \ e $

(iii) For $k_1,...,k_r \in \mathbb{Z}$ is

$\vert \langle k_1g,...,k_rg\rangle \vert = \frac{n} {gcd(k_1g,...,k_r,n)}$.

I would really appreciate if anyone could provide me with proof's of the above statements as I can't seem to figure them out by myself.

Thanks alot in advance.

What knowledge of cyclic groups do you have? And what have you tried? In the i) part there are the groups given so you only have to show, that the group $U_d$ given there has cardinality $d$ and is the only one. You can show that by considering any subgroup with cardinality $d$ and showing that it has to be generated by $\frac{n}{d}g$. The other parts work quite similarly. — Verdruss, Jul 18 '17 at 14:07
@MarianoSuárez-Álvarez But have you spelt prural correctly? :) — Derek Holt, Jul 18 '17 at 15:25

score 0 · Answer 1 · answered Jul 18 '17 at 14:07

0

Hint: Consider $H_d=\{ x \in G : dx = 0 \}$.

answered Jul 18 '17 at 14:07

lhf

1 Answers1