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Math Help - Cayley table inverse

  1. #1
    Junior Member froodles01's Avatar
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    Cayley table inverse

    Surely it can't be this simple.

    I have a group G = {5,10,15,20,25,30} & have constructed a Cayley Table for this group

    x35 5 10 15 20 25 30
    5 25 15 5 30 20 10
    10 15 30 10 25 5 20
    15 5 10 15 20 25 30
    20 30 25 20 15 10 5
    25 20 5 25 10 30 15
    30 10 20 30 5 15 25

    Would 15 be the identity.
    I'd like to find the inverse, now. Surely it can't be the elements all negative, can it?

    (x35) 5 10 15 20 25 30
    5 -25 -15 -5 -30 -20 -10
    10 -15 -30 -10 -25 -5 -20
    15 - 5 -10 -15 -20 -25 -30

    Surely this is too simple
    Last edited by froodles01; May 29th 2012 at 04:52 AM. Reason: Adding info
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  2. #2
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    Re: Cayley table inverse

    cayley tables are stupid. it's much simpler to just list the rule by which a product is evaluated.

    in this case:

    a*b = ab (mod 35). note that a*b = b*a, since the integer ab = ba, so reducing either of these mod 35 gives the same result.

    to find an identity (if there is one) we need to find an element e such that:

    a*e = e*a = a, for all a in G. since e*a = a*e (see above), we only need to check for a right-identity.

    for example 5 can't be the identity, since 5*5 = 25 ≠ 5 (take either 5 to be "a").
    10 can't be the identity since 5*10 = 15 ≠ 5.
    5*15 = 5, so it's possible 15 could be the identity. we could check the rest of the products (there's only 5 more to check), but let's work smart, not hard.

    note that all of our elements are multiples of 5, so we can write them as 5x, x = 1,2,3,4,5,6.

    now (5x)(15) = 75x = 70x + 5x.

    hence (5x)(15) = (2x)(35) + 5x, so:

    75x = 5x (mod 35).

    this shows that (5x)*(15) = 5x for each x = 1,2,3,4,5,6, so that 15 is indeed our identity (why do case-by-case, when you can do all at once?).

    now we turn to inverses (if we have them).

    for every a, we need to find a b so that:

    a*b = 15 (since * is commutative, this will mean b*a = 15).

    well, we COULD examine our table, and find that 5-1 = 10, etc.

    let's see if we can do this "algebraically".

    suppose (5x)*(5y) = 15 (mod 35).

    this means 25xy = 15 (mod 35).

    note that this means:

    25xy = 35k + 15, so we can divide by 5 to get:

    5xy = 7k + 3, that is:

    5xy = 3 (mod 7)

    also: note that (3)(5) = 15 = 1 (mod 7) (because 15 = 2(7) + 1)

    so if we multiply by 3, we get:

    xy = 15xy = 9 = 2 (mod 7)

    therefore y = 2x-1 (mod 7).

    now we need to know the inverses of each x (mod 7). this is somewhat easier to compute:

    1 has inverse 1 (duh!)
    2 has inverse 4 (since 8 = 1(7) + 1)
    3 has inverse 5 (we saw this above)
    4 has inverse 2 (since 2 has inverse 4)
    5 has inverse 3 (because....)
    6 has inverse 6 (because that's the only inverse left...or: because 36 = 5(7) + 1, take your pick).

    ok, let's put this back together for G:

    5 = 5(1), and 1 has inverse 1 mod 7, and 2*1 = 2 (mod 7), so y = 2, so 5 has inverse 5(2) = 10.
    10 = 5(2), and 2 has inverse 4 mod 7, and 2*4 = 8 = 1 (mod 7), so y = 1, so 10 has inverse 5.
    15 should be its own inverse, but we'll check, just for laughs:

    15 = 5(3), and 3 has inverse 5 mod 7, and 2*5 = 10 = 3 (mod 7), so y = 3, so 15 has inverse 15. hey, it works!

    20 = 5(4), and 4 has inverse 2 mod 7, and 2*2 = 4 (mod 7), so y = 4, so 20 has inverse 20 (yes, a non-identity element CAN be its own inverse).
    25 = 5(5), and 5 has inverse 3 mod 7, and 2*3 = 6 (mod 7), so y = 6, so 25 has inverse 30.
    30 = 5(6), and 6 has inverse 6 mod 7, and 2*6 = 12 = 5 (mod 7), so y = 5, so 30 has inverse 25.

    in fact, there's a reason why (mod 7) figured so prominently in this.

    our group G is really "the same" (has the same cayley table) as the group {1,2,3,4,5,6} under multiplication mod 7:

    15 <---> 1 = 36 mod 7
    25 <---> 2 = 32 mod 7
    5 <---> 3
    30 <---> 4 = 34 mod 7
    10 <---> 5 = 35 mod 7
    20 <---> 6 = 33 mod 7

    what happens is that 3k (mod 7) gets sent to 5k (mod 35):

    52 = 25 mod 35
    53 = 125 = 20 mod 35
    54 = 625 = 30 mod 35
    55 = 3125 = 10 mod 35
    56 = 15625 = 15 mod 35
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  3. #3
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    Re: Cayley table inverse

    You seem to be misunderstanding the idea of "inverse" element. If your operation is ordinary addition of numbers, so that 0 is the identity, then -x is the inverse of x. If your operation is ordinary multiplication of positive numbers, so that 1 is the identity, then 1/x is the inverse of x.

    Given a group, so that we are given the operation, o, with identity, i, the inverse of x is the member of the group, y, such that xoy= i.

    I disagree with Deveno that "Cayley tables are stupid". They can be a good way of showing the operation and can be used where there is no simple "rule" for the operation. (Though they may be tedious to write out, here we are given the table and should make use of it.)

    Here, yes, it is true that the operation "multiplication modulo 35" but we can see immediately by looking at the third row that the identity is "15". Further, looking at the first two rows we see that 5*10= 10*5= 15 so 5 is the inverse of 10 and vice versa. 15, being the identity is, of course, its own inverse. From the fourth row, we see that 20*20= 15 so 20 is its own inverse. Finally, from the last two rows, 25*30= 30*25= 15 so 20 is the inverse of 20 and vice versa.
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  4. #4
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    Re: Cayley table inverse

    have you ever tried to manually verify that an operation on a set S with 12 elements is associative? the table itself has 144 products, and there are 1,728 possible "3-term" combinations, each of which has to be computed 2 different ways, and then each of these pairs compared. assuming you can look up each entry in the table in around 2 seconds, that's 8 seconds for each comparison, for a total of 13,824 seconds. if you work non-stop, error-free, that's just under 4 hours.

    furthermore, there are 12144 possible such tables to investigate. determining which of these might be associative using tables takes more time than you or i possess in a lifetime.

    contrast this with the fact that an abstract understanding of groups allows us to determine there are only 5 isomorphism classes of groups of order 12.

    cayley tables are only practical for groups of very small order (even a group of order 12 is pushing the limits). sure they CAN be illustrative, but i have had conversations with people who don't feel they actually have an isomorphism until they have "the same cayley table", or who think that the uniqueness of the identity and inverses somehow derive from the properties of cayley tables.

    as a pedogogical device, cayley tables are over-used. and they actually obscure the concept of isomorphism, as one can have two different tables on the same underlying set, which nevertheless represent the same group (for example, just exchange the rows/columns corresponding to 1 and 3 for Z4, or more properly invoke the automorphism x→x-1). perhaps my statement "cayley tables are stupid" sounds a bit extreme. it's meant to sound extreme enough to provoke thought.

    groups are a beautiful and elegant concept. algebra is more than combinatorics.
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  5. #5
    Junior Member froodles01's Avatar
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    Re: Cayley table inverse

    Thank you all very much for the contributions.
    I have finished this for now, but may ask another self explanatory question later, when I revise for the final exam.
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