Homework #4, due Feb 26th:

1.7 #8,15,16,17,23

2.1 #2,5,6,9,10,11

1.7 #8. Let B = k-element subsets of A, and let S_A act on B in the obvious way.

a. To show it's an action, we need to check two things:

1 . {a_1,...,a_k} = {1.a_1, ..., 1.a_k} = {a_1,...,a_k}. So the identity does act as the identity.

(gh) . {a_1,...,a_k} = {(gh).a_1, ..., (gh).a_k} = {g.h.a_1, ...} = g.{h.a_1,...} = g.h.{a_1,...}

Both of them come from the fact that the action of S_A on A is indeed an action. (So in fact the proof generalizes to any group G acting on A, not just G=S_A.)

b. The cycle decomposition of (12) on the 2-element subsets of {1,2,3,4} is
( 11 22 ) ( 13 23 ) ( 14 24 )
whereas for (123) it's
( 11 22 33 ) ( 14 24 34 )
where I've abbreviated {i,j} as ij.

#15. Show that g.a = a g^{-1} does define a left action of G on G.

A. We have to check two things:

1 . a = a 1^{-1} = a. Good

(gh) . a = a (gh)^{-1} = a h^{-1} g^{-1} = (h.a) g^{-1} = g.(h.a)

Note that the second one would get screwed up if we didn't put the inverse there.

#16. Same question for conjugation.

A. We have to check two things:

1 . a = 1 a 1^{-1} = a. Good

(gh) . a = gh a (gh)^{-1} = g h a h^{-1} g^{-1} = g (h.a) g^{-1} = g.(h.a)

#17. Prove that for each g, a |-> g a g^{-1} is an automorphism.

A. We have to check that it's a homomorphism, and that it's invertible. For the first:
(g.a)(g.b) = (g a g^{-1}) (g b g^{-1}) = g a b g^{-1} = g.(ab) Good
For the second, note that conjugating by g^{-1} is the inverse operation to conjugating by g. (It's not good enough to prove that conjugation by g is 1:1, or that it's onto, because G might not be a finite group!) Very often the easiest way to prove something's invertible is to write down the inverse.

#23. Let {top,bot,left,right,front,posterior} be the six faces of the cube, so that our set is the 3-element set {{t,b}, {l,r}, {f,p}} (abbreviating). To write down a rotation, it's enough to say where all faces go, so for example one of the 90 degree rotations is (l p r f).

The kernel is given by all the 180 degree rotations plus the identity, namely {1, (l r)(f p), (l r)(t b), (f p)(t b)}. It's nontrivial, so this action is not faithful.

2.1 #2.

(12)(23) = (1 2 3) not a 2-cycle

f * fr = r not a reflection (here f is some flip)

if n=ab, a nontrivial factorization, and g of order n, then g^a is of order b which is not 1 or n

1+1 is not odd

1 + sqrt(2) does not square to a rational

#5. If |G|>2, then |G| can't be a multiple of |G|-1. (If you really want to see a proof, let d = |G| / (|G|-1). Then d|G| - d = |G|, so d = (d-1)|G| > 2(d-1), so 2 > d, but then d=1, so |G| - 1 = |G|, contradiction.) Now apply Lagrange's theorem.

#6. This was a hard question, in that one needed to invent an infinite nonabelian group for the purpose. Probably the only one you are familiar with is NxN invertible matrices. Let f =
[-1 0]
[ 0 1]
which flips across the y-axis, and g = f times rotation by 1 radian (or some other irrational times 2 pi). Then g and f are both flips, so order 2, but fg = just the rotation, so infinite order.

#9. Need to check three things:

det 1 = 1. Sure.

If det A = 1, then det A^{-1} = (det A)^{-1} = 1^{-1} = 1. Good.

If det A = det B = 1, then det AB = det A det B = 1*1 = 1. Done.

#10. It's enough to do (b). Let X be the set of subgroups being intersected.

For each H in X, 1 is in H, therefore 1 is in the intersection of all.

If g is in H is in X, then g^{-1} is in H too. So if g is in H for all H in X, then g^{-1} is in the intersection of all.

Same line of reasoning for products -- for each one it's there, therefore it's in the intersection.

#11. In all three it's obvious that the identity, (1,1), is there. Let's check multiplication:

(a,1) * (a',1) = (a a',1)

(1,b) * (1,b') = (1, b b')

(a,a) * (a',a') = (a a', a a')

Checking that these subsets are closed under inverses is much the same.