1. ## Group theory

Hi,

G={1,5,25,125} is a cyclic group under modulo 625. Is this correct?

I had a course on Group theory recently. Based on the answer to the above question I want to know whether I had understand the concepts well.

Thank you.

2. Originally Posted by Sudharaka
Hi,

G={1,5,25,125} is a cyclic group under modulo 625. Is this correct?

I had a course on Group theory recently. Based on the answer to the above question I want to know whether I had understand the concepts well.

Thank you.
First you have to show that G is a group under the operation specified, which I will assume is multiplication modulo 625.

Then you have to show that there is an element that generates the entire group.

G fails to be a group as it is not closed under multiplication modulo 625 since:

$125^2 \equiv 0 \text{ (mod 625)} \not\in G$

CB

3. Thank you very much for your reply. Yes you are correct the group was'nt closed under multiplication modulo 625. Now then suppose,

G={1,5,25,125} is a cyclic group under modulo 626. Is this correct?

4. Originally Posted by CaptainBlack
First you have to show that G is a group under the operation specified, which I will assume is multiplication modulo 625.

Then you have to show that there is an element that generates the entire group.

G fails to be a group as it is not closed under multiplication modulo 625 since:

$125^2 \equiv 0 \text{ (mod 625)} \not\in G$

CB

$125^2=(5^3)^2=5^6=5^2\!\!\!\!\pmod{625=5^4}\neq 0\!\!\!\!\pmod{625}$ , but the explanantion's still correct, of course: $25^2=5\cdot 125=0\!\!\!\!\pmod{625}\notin \{1,5,25,125\}$ and then this set isn't closed under multiplcation modulo 625.

Tonio

5. Originally Posted by Sudharaka
Thank you very much for your reply. Yes you are correct the group was'nt closed under multiplication modulo 625. Now then suppose,

G={1,5,25,125} is a cyclic group under modulo 626. Is this correct?

Again no: $5^5=3125=621=-5\!\!\!\!\pmod{626}\notin G$ ...

If you want to make sure to get a MULTIPLICATIVE subgroup modulo n, for some integer n, take any INVERTIBLE element $x\in\mathbb{Z}_n$ (i.e., any element $x\in\mathbb{Z}_n\,\,\,s.t.\,\,\,(x,n)=1$ and then take $G:==\{1,x,x^2,\ldots,x^{r-1}\}\,,\,\,with\,\,\,r=ord_n(x)$ .

Tonio

6. Originally Posted by tonio

$125^2=(5^3)^2=5^6=5^2\!\!\!\!\pmod{625=5^4}\neq 0\!\!\!\!\pmod{625}$ , but the explanantion's still correct, of course: $25^2=5\cdot 125=0\!\!\!\!\pmod{625}\notin \{1,5,25,125\}$ and then this set isn't closed under multiplcation modulo 625.

Tonio
According to my calculator:

$125^2=5^6=5^4 \times 25=625 \times 25\equiv 0 \text{ (mod 625)}$

CB

7. Thanks so much to all of you. Now for the last time,

G={1,5,25,125} under multiplication modulo 624 is a cyclic group. Is'nt it?

8. Originally Posted by CaptainBlack
According to my calculator:

$125^2=5^6=5^4 \times 25=625 \times 25\equiv 0 \text{ (mod 625)}$

Of course you're right and even more: the above's exactly the calculation I wrote, but for some reason hidden deep in my brains and kept jealously by my 3 working neurons, I did that calculation ADDITIVELY modulo 625...a shame.

Tonio

CB
.

9. Originally Posted by Sudharaka
Thanks so much to all of you. Now for the last time,

G={1,5,25,125} under multiplication modulo 624 is a cyclic group. Is'nt it?

This time yes, since $(5,624)=1\,\,\,and\,\,\,5^4=1\!\!\!\!\pmod{624}$ , so that $G=<5>\!\!\!\!\pmod{624}$ , a group of order 4 (since $625=1\!\!\!\!\pmod{624}$)

Tonio

10. Thank you tonio. The problem is cleared. I appreciate the help given by all of you. It's nice to know that such people like you are out there willing to help at any time.