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MATH 370 - Assignment 1
Solutions provided by Matthew
Problem 1:
For a, b G, multiply by a1 on the left:
a1 (a2 b2 ) = ab2 = a1 (ab)2 = a1 abab = bab = ab2 = bab
Now, multiply on the left by b1 :
(ab2 )b1 = ab = (bab)b1 = ba = ab = ba
Therefore, G is abel
Assignment 7
MATH 370, Honours Algebra 3
Prof. Pichot
Solutions provided by Spencer
December 3, 2012
1. Find out and prove for which n the dihedral group Dn is nilpotent, and for which n it is solvable.
Proof. We claim that the only n for which Dn is nilp
MATH 370 - Assignment 4
Solutions provided by Costin
[1] Let A be in the center of GLn (k ) and let B GLn (k ) be a diagonal matrix. We have that
n
( AB)ij =
Ai Bj = Aij Bjj
=1
n
( BA)ij =
Bi Aj = Bii Aij
=1
AB = BA implies that Aij Bjj = Bii Aij . If
1. Let K be a proper maximal subgroup of Q. Since Q is abelian, K is a normal subgroup.
By the fourth isomorphism theorem, there is a bijection between subgroups of Q containing
K and subgroups of Q/K. Since K is maximal, it follows that Q/K has no proper
1. Let G be a nilpotent group and let H be a maximal subgroup of G. Since G is nilpotent,
H < NG (H). As H NG (H), the maximality condition implies that NG (H) = G; i.e. H is normal
in G.
Consider the quotient group G/H. By the fourth isomorphism theorem,
1. We have that x = 1x and y = 1y, so 1 | x and 1 | y. Suppose d | x and d | y for some d Q[x, y].
Then there exist p, q Q[x, y] such that x = pd and y = qd. Since Q is an integral domain,
Q[x] and Q[x, y] are integral domains as well, so degy (x) = degy
Assignment 2
MATH 370, Honours Algebra 3
Prof. Pichot
Solutions provided by Spencer
October 17, 2012
1. Find a generating set for Sn containing only two elements. Prove that the subset is generating
the whole group.
Proof. We claim that Sn is generated by
Assignment 3
MATH 370, Honours Algebra 3
Prof. Pichot
Solutions provided by Costin
October 17, 2012
[0]
See p6 below.
[1] Center of D2n
When n = 1, D2 is a group with two elements. D2 is abelian, since the identity always
commutes with every other element
1. (a) Z[i] is a principal ideal domain and 1 + i is an irreducible, hence prime, element in Z[i].
Since prime ideals are maximal in a principal ideal domain, it follows that (1 + i) is a maximal ideal
in Z[i]. Therefore, Z[i]/(1 + i) is a eld.
Let z Z[i]
4
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Normal approximation to binomial distribution. We saw that when X bin(n, p) for
a large n and a small p, the binomial distribution can be approximated with the Poisson
distribution with mean = np.
What if n is large but p is not small ?
In this case we ca