How many elements $a$ with $o(a)\mid 5$ in $\mathbb{Z}/360\mathbb{Z}$

Question

I am trying to understand why there are exactly $5$ elements of order dividing $5$ in $\mathbb{Z}/360\mathbb{Z}$.

Hello and welcome to math.stackexchange! What are the possible values $o(a)$? Do you know an element whose order divides 5? What could be the order of the other four elements? — Hans Engler, Dec 29 '14 at 14:49

score 3 · Answer 1 · edited Apr 13 '17 at 12:19

3

Hint: $\mathbb{Z}/360\mathbb{Z}$ is a cyclic group. So there is exactly one subgroup of order $5.$ So...?

$\bf{EDIT:}$ This link (kindly provided by Bill Dubuque) says that, if $G$ is a cyclic group of order $n$ and if $d$ is +ve integer dividing $n$, then there is a unique subgroup of $G$ order $d.$ Applying this result in this particular case $(n = 360, d=5),$ we get that $\mathbb{Z}/360\mathbb{Z}$ has a unique subgroup $H$ of order $5.$ Surely for any $a \in H, o(a)|5$ (why?). Now if there is an element of order $5$ in $G$ which is not in $H$ then that element will generate a subgroup of order $5$ which is different from $H.$ But his can't happen because the above mentioned result says that $H$ is unique.

edited Apr 13 '17 at 12:19

Community

1

answered Dec 28 '14 at 20:46

Krish

7,102

See e.g. this question for proofs of said theorem. – Bill Dubuque Dec 28 '14 at 21:15
i am still lost, why does this help me? – Tom Turner Dec 29 '14 at 13:31
@IdoBen-Shaul: see the Edit. let me know if you have any doubt. – Krish Dec 29 '14 at 13:43

How many elements $a$ with $o(a)\mid 5$ in $\mathbb{Z}/360\mathbb{Z}$

1 Answers1