Solutions to Test #1 (Oct 26, 2011)
1) There are six elements mod 9 that are rel prime with 9, and there are
also six elements mod 7 that are rel prime with 7. Thus G has 6*6=36
elements. If G were cyclic, then a generator would have order 36.
However, the sixth power of every element in G is the identity (1,1) in G.
Hence G is not cyclic.
2) (1235)(426) = (435126) has order 6.
3) The eight elements of order 20 are g^(6k), where k=1,3,7,9,11,13,17,19.
4) Say the answer is yes. Then (15)(24) equals a product of 37
transpositions, so the Identity equals a product of 39 transpositions.
That contradicts the fact that the Identity cannot be written as a product
of an odd number of transpositions. Thus the answer is no.
5) Say a and b have orders m and n respectively.
Since G is abelian, ab^(mn) equals a^(mn) * b^(mn) = 1*1 = 1.
Thus ab has finite order (dividing mn).
6) hH is contained in H, by closure of H. Also, H is contained in hH,
since every element c in H equals hd for some d in H, namely d=h'c,
where h' denotes the inverse of h in H. (Note d=h'c is in H by closure.)
7) Write n = mq + r, where the remainder r is a nonnegative integer < m.
We have 1 = g^n = g^(mq+r) = g^(mq) * g^r = 1 * g^r = g^r.
If r is positive, this contradicts the definition of order, since r < m.
Thus r=0, so m divides n.
8) The only way an element of S_5 can have order 3 is if it is a 3-cycle.
There are 5*4*3 = 60 ways to choose the three elements of a 3-cycle.
However, this counts each 3-cycle thrice, since for example
(123)=(231)=(312). Thus there are 60/3 = 20 elements of order 3.