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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

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  1. Loser66
    • one year ago
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    @oldrin.bataku

  2. anonymous
    • one year ago
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    https://en.wikipedia.org/wiki/Finite_group#Number_of_groups_of_a_given_order

  3. Loser66
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    (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
    • one year ago
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    By definition, group needs 2 operators, right?

  8. anonymous
    • one year ago
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    so now we're interested in finding which of these \(n\) permit more than just this type of cyclic group

  9. anonymous
    • one year ago
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    @Loser66 no, a group is a set S equipped with a single associative, invertible operation under which S is closed

  10. anonymous
    • one year ago
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    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
    • one year ago
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    Why did you delete it, I didn't get it yet!!

  12. anonymous
    • one year ago
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    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
    • one year ago
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    if I choose n =3, then the group is {3,6,9} and its order is 3, right?

  14. Loser66
    • one year ago
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    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
    • one year ago
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    that works, but the more obvious choice of a cyclic group is the mod 3 additive group on \(\{0,1,2\}\)

  16. anonymous
    • one year ago
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    \(\{3,6,9\}\) 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
    • one year ago
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    for \(n=5\) the cyclic group is usually written in the form \(\{0,1,2,3,4\}\) with addition mod \(5\)

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