## Loser66 one year ago For which integers n such that $$3\leq n\leq 11$$ is there only one group of order n (up to isomorphism)? 1) For no such integer n 2) For 3, 5, 7, and 11 only 3) For 3, 5,7,9, and 11 only 4) For 4,6,8, and 10 only 5) For all such integers n Please, help. I don't understand the question

1. Loser66

@oldrin.bataku

2. anonymous
3. Loser66

so, the correct one is 2) 3, 5, 7 , and 11 only. I would like to know how to find it out without using that table. Please

4. anonymous

the order of the group is the number of its elements; there can be groups of the same order that are not isomorphic in that their elements are 'linked' in a different way. consider the following groups of order 4: $$\mathbb{Z}_4:\quad\quad\{0,1,2,3\}\text{ equipped with }+\pmod4\\\mathbb{Z}_2\times\mathbb{Z}_2:\{(i,j):i\in\{0,1\},j\in\{0,1\}\}\text{ equipped with pairwise addition }+\pmod 2$$ it just happens that every group of order 4 is isomorphic to either of these -- they have the only 'structures' (up to relabeling elements) that satisfy the group axioms possible for a set of four elements

5. anonymous

anyways, for every positive integer $$n$$, the additive group of integers mod $$n$$ is a cyclic group of order $$n$$, so there's always at least one finite group of order $$n$$

6. Loser66

(mod 4) That is what I want!!! so, if we have this kind of problem, we put it in mod , right? One more question: To group, when define the order of elements, we have to use multiplication,right?

7. Loser66

By definition, group needs 2 operators, right?

8. anonymous

so now we're interested in finding which of these $$n$$ permit more than just this type of cyclic group

9. anonymous

@Loser66 no, a group is a set S equipped with a single associative, invertible operation under which S is closed

10. anonymous

we're talking about the order of the group itself, not a particular element; the order of an element $$g$$ in a particular group is the size of the cyclic subgroup it generates

11. Loser66

Why did you delete it, I didn't get it yet!!

12. anonymous

hmm, this problem is far from trivial in the general case, but you know the answer must include at least $$3,5,7,11$$

13. Loser66

if I choose n =3, then the group is {3,6,9} and its order is 3, right?

14. Loser66

I don't see the logic on it :( if n =5, then the group starts at 5 must have order 5, but how?? if using addition {5,10} just have 2 elements, how its order is 5?

15. anonymous

that works, but the more obvious choice of a cyclic group is the mod 3 additive group on $$\{0,1,2\}$$

16. anonymous

$$\{3,6,9\}$$ under addition mod 9 is isomorphic to the above cyclic group, though; it's in fact a cyclic subgroup of order $$3$$ of what is equivalent to $$\mathbb{Z}_9$$, namely $$\{1,2,3,4,5,6,7,8,9\}$$ under addition mod 9 as well where we use 9 instead of 0 (but they are the same equivalence class mod 9 so it doesn't matter)

17. anonymous

for $$n=5$$ the cyclic group is usually written in the form $$\{0,1,2,3,4\}$$ with addition mod $$5$$