Daniel Homework 1

Comments: Overall, good job. But try not to overthink the problems and just use the definitions of groups.

2.2 duh. 2.8. The key idea is that a group need not be commutative: so x*y need not equal y*x. This is an honest mistake since school has never taught any non-commutative multiplications. So to prove t is an inverse r, we just have to show that t*r = e. So to show (xy)^-1 = y^-1 * x^-1, simply take xy and multiply it on the right by y^-1 * x^-1. To see it in detail, you can check out Julian or Jame's proof.

3.1. Good, you have key steps down. But for a first homework, it is better to write a checklist and write everything out. Or at least copy and paste : ).

3.2. Good, you checked the inverse was in the group (i.e, in the same form). For future note, since we are working on the computer, please write out parentheses.

3.3 Dude, you wayyyy over thought this one. No need to do binomial expansion on a +bi (though it is correct). Its like killing a fly with an atomic bomb. Ten sec simple proof: z^n = 1, y^n = 1. Thus, z^n * y^n = 1 i.e, (zy)^n = 1. Associativity from def of multiplication of complex numbers. 1/z is the inverse and since (1/z)^n = 1/(z^n)=1, it is in the group. 1 is the identity and 1^n = 1. Your solution was more of an analytical one. But we want to use algebra to solve these problems (simply looking at the given properties of the object).

3.10. Ah, but remember we are working with a MODULO group. division, in a sense, does exist. Specifically multiplication by inverses. For example, if we are in the group mod 5, the element 3 has the inverse 2 since 3*2 = 1 mod 5. And how do we know the inverse always exist? Not by pigeonhole principle since there is no guarentee. It is by the identity we learned in Number theory, Bezout's identity.

4.2. Good, but for grading ease, please separate groups.

4.4 Good, unique is the key word. In the future (college exam), please write out the simple immediate step before a=g: Suppose ax = gx. Since the inverse exists by definition of a group, multiply both sides by x^-1 on the right. Then a = g.

4.5 Inverses are unique, yes. But e also has a unique inverse: it's e. But I understand absolutely why you removed it from consideration since it is anomalous (the order of e is 1), and the counting argument is correct. Good job.

4.6 First observation from squares are definitely correct. Second and third part good! Dude, you wrote up to the the third semicolon correct and this was the key idea needed in 2.8. But I do not see how you guy y x = x inverse y inverse part. Did you mean y x = y inverse x inverse part, which you demonstrated in the first sentence?

4.8 YESSSSS!!!! Very Pretty Trick!!!

5.5. Yes, yes, you clearly understand the key word, "finite." And unlike the other two homeworks I got, you remembered to mention why associativity still hold. You also remembered this was an if and only if, GOOD! If you want to sound more mathy though, and less English, instead of other way around, say "conversely." It is like math secret code.

5.7. Remember to check closure: x^n = e, y^m = e, so (xy)^mn = x^m * y^n (since xy=yx)=e*e=e. Also, check inverse is in the set (though you got the correct inverse): (x^(n-1)). This is by simple commutivity of exponents: (x^(n-1))^n= (x^n)^(n-1)=e^(n-1)=e.

5.10. Yep!

6.1. Sorry to make you do this one, but every math major got to write out a multiplication table once in their life.

6.3　good! you just beat out like 20 Math 109 students who thought the answer was 6! Good job!.

6.11. Hm..right you asked me about permutations. Watch my video http://www.youtube.com/watch?v=8eflPEqlrUU and ask me the specifics. Or we can talk via phone.