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My first exercise on Group Theory

  1. Oct 1, 2015 #1
    EDIT: I've just realised this is the 'Calculus and beyond' subforum - I saw 'beyond' and thought, "Well I've done all my calculus, and now I'm doing group theory, so this thread must go here!". But now I realise it surely belongs somewhere else. Sorry about that. Mods feel free to shift it to the right place...

    I'm new to group theory and am sure this is a very simple question as I'm only on Week 1 of my Group Theory module at university. I may even have got the correct answer (apart from one bit I'm stuck on), but I'd like to check it all before next week's lectures come around, just to be sure I'm on the right track with the various definitions and concepts.

    1. The problem statement, all variables and given/known data


    Write out the multiplication table for the group G = ℤ7*
    List all the inverses of the elements of G.
    Find an element of G such that every member of G is a power of that element

    2. Relevant equations


    3. The attempt at a solution

    So we were told that a group ℤn* is defined such that it contains all integers except 0, and that all values in the group can only exist if they have no common factor with n. (e.g. ℤ4* = {1,3} . 2 is omitted because it is a factor of 4)

    Hence the group G = ℤ7* = G{1,2,3,4,5,6,7}

    Also, I understand a numerical group such as this uses the modulo form of multiplication, whereby any multiple resulting in something greater than n is simply treated as a the remainder. Hard to explain but I'm sure you all know what I mean. e.g. 5*3 with modulo 6 = 3

    • So... The Multiplication table for ℤ7*:

    0 1 2 3 4 5 6 (ignore the 0 - I needed to add it to keep the table aligned)
    1 1 2 3 4 5 6
    2 2 4 6 1 3 5
    3 3 6 2 5 1 4
    4 4 1 5 2 6 3
    5 5 3 1 6 4 2
    6 6 5 4 3 2 1

    I'm hoping this is correct.

    • Next, the inverse of each element of G:
    1→6
    2→4
    3→5
    4→2
    5→3
    6→1

    Again I'm hoping this is correct.

    • And the final part: "Find an element of G such that every member of G is a power of that element"

    Well, this is the bit I'm really unsure about. What does it mean by "every member of G"..? I assumed it was just another word for element, but still I can't see a clear, specific answer. I was thinking the answer is 1, because 1 to any power will give another member of G (assuming a member is also an element)..? I just don't know to be honest. Hopefully it's something blindingly simple.

    Thanks for any help, particularly regarding that final part at the end.

    :smile:
     
  2. jcsd
  3. Oct 1, 2015 #2

    pasmith

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    Homework Helper

    Yes.

    1 is the identity; 1 to any power gives 1. This doesn't generate the whole group.
    Look at the powers of 2: 2 -> 4 -> 1 -> 2. This doesn't generate the whole group either. Keep looking.
     
  4. Oct 1, 2015 #3
    Ahhh I think I get it now. I forgot that I should keep in mind the use of the modulo, so that, for example, if I take 32 I get 9, which is therefore 2 in ℤ7*

    The answer appears to be 3.
    31 mod 7 = 3
    32 mod 7 = 2
    33 mod 7 = 6
    34 mod 7 = 4
    35 mod 7 = 5
    36 mod 7 = 1
     
  5. Oct 1, 2015 #4

    FeDeX_LaTeX

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    Gold Member

    I don't agree with the bolded. Is the product of 1 and 6 the identity element?

    Correct. Such an element is called a generator of G.
     
  6. Oct 1, 2015 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, it doesn't. As you say just below, it contains only those positive integers, less than n, that are relatively prime to n.

    Again, no. Z7 does not include "7". Surely, that was a typo?

    It would have been better to use something like "X".
    Yes, that is correct.
    No, the first and last are wrong. 6*1= 6 not 1. 1*1= 6 so 1 is its own inverse. Similarly 6*6= 1 so 6 is its own inverse.
    Yes, "member" and "element" are the same thing. However, 1n= 1 for all n so powers of 1 do not give "all members" Also, since 6*6= 1, powers of 6 can only give 6 and 1.
    21= 2, 22=2*2= 4, 23= 2*4= 1, 24= 2*1= 2 and then it repeats- powers of 2 can give only 2, 4, and 1.
    31= 3, 32=3*3= 2. 33= 3*2= 6, 33= 3*6= 4, 34= 3*4= 5, 35= 3*5= 1. We get all members. Check 4, 5, and 6 also.

    (Notice that when a power is 1 we start repeating- we get all elements as powers only when the last, n-1, power gives 1.)
     
  7. Oct 1, 2015 #6
    Ahhh of course, my mistake.

    1 is the identity element, 1*6 = 6 ≠ 1

    1*1 = 1

    It should be this, I believe?
    1→1
    2→4
    3→5
    4→2
    5→3
    6→1
     
  8. Oct 1, 2015 #7

    FeDeX_LaTeX

    User Avatar
    Gold Member

    Almost. Are you sure the inverse of 6 is 1?

    Can two elements have the same inverse?
     
  9. Oct 1, 2015 #8
    Hahaha christ I even saw and thought, "Why is that in bold as well?", then looked at the multiplication table to verify that the inverse of 6 was definitely 1.

    I now see that the inverse of 6 is 6.

    Thanks!
     
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