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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
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

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0https://en.wikipedia.org/wiki/Finite_group#Number_of_groups_of_a_given_order

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0so, 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
 one year ago
Best ResponseYou've already chosen the best response.0the 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0anyways, 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
 one year ago
Best ResponseYou've already chosen the best response.0(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
 one year ago
Best ResponseYou've already chosen the best response.0By definition, group needs 2 operators, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so now we're interested in finding which of these \(n\) permit more than just this type of cyclic group

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Loser66 no, a group is a set S equipped with a single associative, invertible operation under which S is closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we'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
 one year ago
Best ResponseYou've already chosen the best response.0Why did you delete it, I didn't get it yet!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm, this problem is far from trivial in the general case, but you know the answer must include at least \(3,5,7,11\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0if I choose n =3, then the group is {3,6,9} and its order is 3, right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0that works, but the more obvious choice of a cyclic group is the mod 3 additive group on \(\{0,1,2\}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\{3,6,9\}\) 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
 one year ago
Best ResponseYou've already chosen the best response.0for \(n=5\) the cyclic group is usually written in the form \(\{0,1,2,3,4\}\) with addition mod \(5\)
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