• Homework #2:

    Find a graph with exactly three automorphisms (not six).
    There are many answers to this. (For example, take one such graph, and add on some other gross thing totally disconnected from it!) Here are a few:

    If you were one of the several people who wrote "the number of automorphisms must always be a multiple of 2", take one (or all) of these graphs and see where your argument breaks down. (At some point you must be making some false statement.)
  • [DF] p21 #1,5,6,9,12,15,27
    1 b,d are associative, not a,c,e. To show associativity, you can just do the (non-abstract) algebra to check (a*b)*c=a*(b*c). To show nonassociativity, it's not enough to get to an equation you don't believe and say "this isn't automatically true". What if it is but you're just not able to prove it? Rather, you must find some numbers to plug in that break the "equation", to show something is not associative.

    Incidentally, you can do 1d if you correspond (a,b) to the matrix
    ba
    0b
    check that multiplication works the same way, and use that matrix mult is associative.

    5. The only element of Z/nZ that could serve as the multiplicative identity is 1, since 1 is the only A such that A * 1 = 1. But for all x, x * 0 = 0. For n>1, 0 and 1 are different, so x * 0 is not 1. Therefore 0 has no multiplicative inverse, and Z/nZ is not a group under multiplication.

    6. a,e but not b,c,d,f. Again, to show they're not subgroups, you must either find that they're empty (though none of these are), or an element without its negative in your "subgroup", or two elements whose sum isn't in your "subgroup". It's usually not very convincing to include a lot of bluster that "obviously, these bad elements exist" -- just write one (or one pair) down, and everyone can agree.

    9a. We need to check that if we add two elements, or negate one element, of this nonempty set that we get another such element. This is easy.

    b. 1/(a + b sqrt{2}) = a/(a^2 - 2b^2) - b/(a^2 - 2b^2) sqrt{2}.

    12.

  • 1 has order 1
  • -1 has order 2
  • 5 has order 2, since 5^2 = 25 = 1 + 2*12 = 1 in Z/12Z
  • 7 has order 2, since 7^2 = 49 = 1 + 4*12 = 1 in Z/12Z
  • -7 has order 2, since -7^2 = 49 = 1 again
  • 13 has order 1, since 13=1 already!

    • 15. This is tautological for n=1. Now for n>1, let's rewrite

    (a_1 a_2 ... a_n)^-1 = ((a_1 a_2 ... a_{n-1}) a_n)^-1 = (g a_n)^-1

    if we let g = a_1 ... a_{n-1}. What can we say about this smaller problem?

    (g a_n)^{-1} (g a_n) = 1
    (g a_n)^{-1} g = a_n^{-1} by multiplying on the right
    (g a_n)^{-1} = a_n^{-1} g^{-1} by multiplying on the right again

    so returning to our previous question,

    (a_1 a_2 ... a_n)^{-1} = a_n^{-1} g^{-1}.

    Now g is a product of n-1 things, so by induction, we already know how to compute its inverse, a_{n-1}^-1 .... a_1^{-1}. Sticking that into our equation gives us the desired result at n too. QED.

    27. We need to check that the identity is in this set (it is, from n=0), that the inverse of any x^n is in this set (it is, at x^{-n}), and that the product of any two x^n and x^m is in this set (it is, at x^{n+m}). Then we're done.