Welcome to Tom Roby's Math 3121 homepage! (Fall 1998)

(last updated: 5 November 1998)

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Class Information

Here's a copy of the syllabus in pdf format.

MIDTERM scheduled for Thursday October 29 (during class).
FINAL scheduled for Thurday December 10, 6:00--7:50 p.m.
PRACTICE FINAL SOLUTIONS


#1) See book for def. of center. Say y is in phi[Z(G)], so y=phi(x), some x in Z(G). We claim that y commutes with any elt of G. Let g be in G, then there exists h in G such that phi(h)=g. Hence,
gy=phi(h)phi(x)=phi(hx)=phi(xh)=phi(x)phi(h)=yg.
Hence, y is in Z(G) and we get phi[Z(G)} is a subset of Z(G). Apply the above argument with phi replaced by its inverse to get the reverse inclusion.

#2a) There are two methods. The first is like that problem on the quiz. By Thm. 9.5 G has elts x of order 3 and y of order 5. Then since the group is Abelian, it's easy to see that the order of xy is 15.
Alternatively, use FTFAG to get that any Abelian group of order 45 is isomorphic to Z_9 + Z_5 or Z_3 + Z_3 + Z_5, which contain elements of order 15 (3,1) and (1,0,1) respectively.

#2) [The SECOND number 2] Let E={all elts of order 1 or 2}. It's easy to see that E is a subgroup. By Lagrange's thm, the order of E divides 2^n, so the order of E is even. Hence the set of elts of order 2, i.e., E-{e} has odd order.

#3) Let |G|=n and suppose x^2=y^2. Since n is odd, we can find integers t and s such that 1=gcd(n,2)=2t+ns. Then
x=x^{2t+ns}=x^{2t}=y^{2t}=y^{2t+ns}=y.
If G is Abelian, one can give a simpler proof, based on this map being a homomorphism. (Why isn't it necessarily a homomorphism if G is not Abelian?) The kernel is {x| x^2=e}, which contains elements of order 1 and 2. But a group of odd order has no elements of order 2, so the kernal contains only the identity, so this map is one to one.

#4a) First note that there are none *onto* since the preimage of 1 in Z_8 must have order a multiple of 8 by Thm. 10.1. For maps to Z_8, we consider where the homomorphism sends the generator 1 of Z_20, since that completely determines it. It must be sent to an element whose order divides 20, so not to an element of order 8. It's easy to check that all four maps to elements of order dividing four yield homormorphisms. (1 maps to 0,2,4, or 8.)
b) We can try mapping 1 to any of the four elements of Z_2+Z_2, and all of these extend to a homomorphism. Another way to describe these maps is as follows: x maps to (0,0)
x maps to (x mod 2, 0)
x maps to (0, x mod 2)
x maps to (x mod 2, x mod 2)

5) Not one-to-one unless k=n, since if k is less than n, then n and n-k map to the same reside. It will be onto since the residues 0,1,2,...,k-1 (mod n) map to 0,1,2,...,k-1 (mod k). The kernel is U_k(n) by def.

6) Let a and be be arbitrary elements of G. Then
ab=(ab)^{-1}=b^{-1}a^{-1}=ba
using the Socks-shoes theorem and the given property of G.

7) The subgroup generated by a^5 is A={e,a^5, a^10}. The five cosets are A, aA, a^2A, a^3A, a^4A . (Write out the subsets if you need to.) The factor group is ismorphic to Z_5.

8) In order for beta to the mth to be a 3-cycle, we must have killed the five cycle and the 2 cycle, so m must be a multiple of 5 and of 2, but not a multiple of 3. In other words, m is a mutiple of 10 which is NOT a multiple of 30.

9) Forget about this one. It's on p. 221, #13 of Gallian, but I don't think it makes any sense. (I'll ask him about it later.)



HOMEWORK ASSIGNMENTS

All homework assignments are listed on the date due. (The ones more than two weeks into the future are tentative.)

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