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:
$\displaystyle 125^2 \equiv 0 \text{ (mod 625)} \not\in G$
CB
$\displaystyle 125^2=(5^3)^2=5^6=5^2\!\!\!\!\pmod{625=5^4}\neq 0\!\!\!\!\pmod{625}$ , but the explanantion's still correct, of course: $\displaystyle 25^2=5\cdot 125=0\!\!\!\!\pmod{625}\notin \{1,5,25,125\}$ and then this set isn't closed under multiplcation modulo 625.
Tonio
Again no: $\displaystyle 5^5=3125=621=-5\!\!\!\!\pmod{626}\notin G$ ...
If you want to make sure to get a MULTIPLICATIVE subgroup modulo n, for some integer n, take any INVERTIBLE element $\displaystyle x\in\mathbb{Z}_n$ (i.e., any element $\displaystyle x\in\mathbb{Z}_n\,\,\,s.t.\,\,\,(x,n)=1$ and then take $\displaystyle G:=<x>=\{1,x,x^2,\ldots,x^{r-1}\}\,,\,\,with\,\,\,r=ord_n(x)$ .
Tonio