Homework #7, due March 18th
1. If A,B are normal in G and have intersection {1}, show they commute.
Ans. We want to know that for all a in A and b in B, that ab=ba.
All we're given is a way of guaranteeing that some group element is 1.
So let's massage our equation into something being 1:
a b a^-1 b^-1 = 1, that's what we want.
If we group as (a b a^-1) b^-1, we see that this is an element of B.
If we group as a (b a^-1 b^-1), we see that this is an element of A.
Therefore it's 1.
2. If H is a subgroup of G, what is the kernel of the action of
G on G/H?
By definition, it's those {g in G : gC = C for all C in G/H}.
Unwinding the definition of G/H, it's {g in G : gkH = kH for all k in G}.
We can rewrite this as [k^-1 g k H = H]. But this is the same as k^-1 g k in H.
So now it looks like {g in G : k^-1 g k in H for all k in G},
or {g in G : g in k H k^-1, for all k in G}.
That's the intersection over all k in G, of k H k^-1.
So what? Why should this be preferable, as an answer, to our first line?
4.3 #5. The size of the conjugacy class through b
is the number of cosets of G by the centralizer of b. (This is just our
usual |orbit| = |G/stabilizer|, applied to the conjugation action.)
Since the centralizer contains the center (follow the definitions),
the partition of G by cosets of the center -- into n parts -- is a refinement
of the partition of G by cosets of the centralizer. So the conjugacy class
has at most n elements.
4.3 #8. Let pi be a nonidentity permutation (a b ...) (...) ... (...),
i.e. the first cycle listed has length > 1. Let c be different from
a and b = pi(a), which uses n at least 3. We now claim that pi does not
commute with (b c).
Proof: pi (b c) a = pi a = b, whereas (b c) pi a = (b c) b = c.
Note that we need c different from b to learn that these two answers
are different, and c different from a to learn that (b c) a = a.
Anyway, the upshot is that we have a witness (b c) to the fac that
pi doesn't commute with everyone. So it's not in the center.
Only the identity is.
4.3 #10. sigma = (12345), sigma^2 = (13524),
sigma^-1 = (15432), sigma^-2 = (14253).
Find taus conjugating sigma to these powers. All you have to do is
match up elements.
4.3 #13. If G has two conjugacy classes, then one of them is the identity,
and the other is of size n-1 (where G is size n). But then n-1 | n.
That forces n=2. So G is necessarily of order 2, and every group of order
2 has two conjugacy classes.
If you're disappointed by this answer, you should try three conjugacy
classes, and show the only groups are of size 3 and 6. I don't know the
answer for 4. I also don't know if the original question has more
infinite examples.
4.3 #35. For a permutation to be of order k, the lcm of the cycle lengths
should be k. If k is a prime p, this means that all the cycles are of
length 1 or p. So our permutation is either a p-cycle and all 1s,
or two p-cycles, or three, or ... up to floor(n/p), the maximum number
of p-cycles we could cram in n elements. That's the greatest integer
less than or equal to n/p.