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

Question

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

loser66 · Answer

@oldrin.bataku

anonymous · Answer

https://en.wikipedia.org/wiki/Finite_group#Number_of_groups_of_a_given_order

loser66 · Answer

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

anonymous · Answer

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:

$$\mathbb{Z}_4:\quad\quad\{0,1,2,3\}	ext{ equipped with }+\pmod4\\mathbb{Z}_2	imes\mathbb{Z}_2:\{(i,j):i\in\{0,1\},j\in\{0,1\}\}	ext{ equipped with pairwise addition }+\pmod 2$$

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

anonymous · Answer

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 $n$

loser66 · Answer

(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?

loser66 · Answer

By definition, group needs 2 operators, right?

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

loser66 · Answer

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

anonymous · Answer

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

loser66 · Answer

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

loser66 · Answer

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?

anonymous · Answer

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

anonymous · Answer

$\{3,6,9\}$ 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)

anonymous · Answer

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