Download Solution to Worksheet 6/30. Math 113 Summer 2014.

Solution to Worksheet 6/30. Math 113 Summer 2014.
These problems are intended as supplementary material to the homework exercises and
will hopefully give you some more practice with actual examples. In particular, they may be
easier/harder than homework.
1. Let H = 4Z = {4k | k ∈ Z}. Show that H is a subgroup of Z and determine the cosets
of H.
Solution: H contains 0 since 0 = 4 · 0; H is closed under addition because 4k + 4m =
4(k + m) ∈ H; H is closed under additive inverses because −(4k) = 4(−k) ∈ H. The
cosets of H are
H = {... , −4, 0, 4, 8, ...}
1 + H = {... , −3, 1, 5, 9, ...}
2 + H = {... , −2, 2, 6, 10, ...}
3 + H = {... , −1, 3, 7, 11, ...}.
Note: This example shows that even in an infinite group, subgroups may have only finitely
many cosets.
2. Recall that the group Q of quaternions consists of elements ±1, ±i, ±j, ±k, with i 2 =
j 2 = k 2 = −1, and with multiplication of i, j and k defined by
ij = k, jk = i, ki = j, ji = −k, kj = −i, ik = −j.
(a) Find a subgroup of Q of order 2 and list all of its left cosets.
(b) Prove that Q can be generated1 by the elements i and j.
(c) Is there a nontrivial homomorphism f : Q → Z/15Z?
(d) (harder, possibly) Is there a homomorphism from Q → D8 which sends i to s and
j to r (here s is reflection across the x-axis and r a 90 degree counterclockwise
rotation)?
Solution:
(a) The only subgroup of Q which has order 2 is H = {±1}. The reason for this is that
all the other elements have order 4, so as soon as a subgroup contains one of them,
its order must be at least four. Anyway, the four left cosets of H in Q are:
H = {±1}
iH = {±i}
jH = {±j}
kH = {±k}
(b) This is because 1 = i 0 j 0 , −1 = i 2 , −i = i 3 , −j = j 3 , k = ij, and −k = ji. Thus all
the other elements can be writen using only i and j.
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A group can be generated by a set S of elements if all the other elements in the group can be written as
products of elements in S.
(c) No. Suppose f is such a homomorphism. Then ker f is a subgroup of Q, so its
order must be 1, 2, 4, or 8, by Lagrange’s theorem. It can’t be 8, or else the
homomorphism is trivial (since then ker f = Q, so f sends everything to zero).
It can’t be 1, because then f would be injective, so =f would have 8 elements,
and this cannot be since im f is a subgroup of Z /15Z but 8 doesn’t divide 15.
Now suppose | ker f | = 2. Then f is 2-to-one, meaning that each output has
exactly two corresponding inputs (see problem 4 for an justification). But then
since there are eight inputs, there would be 4 distinct outputs, so |im f | = 4, which
is impossible because 4 doesn’t divide 15. Similarly, if | ker f | = 4, then f is 4-to-one,
so |im f | = 2, which is impossible.
(d) No, there is no such map. Suppose for contradiction that there were. Then f (−1) =
f (i 2 ) = f (i)2 = s 2 = e. Also ij = −ji in Q, so applying f we would get sr =
f (i)f (j) = f (ij) = f (−ji) = f (−1)f (j)f (i) = e · r · s = rs in D8 . But the relation
sr = rs i false in D8 , so this is a contradiction.
3. Find a subgroup H in W3 and g ∈ W3 such that gH 6= Hg .
Solution: If you’ve done the HW, you maybe already realize we should look for a subgroup
of index greater than 2. Thus we try a subgroup of order 2 (hence index 3, by Lagrange),
say H = {I I I, X I}. If we take g = I X, then
gH = {g , c},
Hg = {g , c 2 },
where c is the unique diagram with vertex 2 attached to vertex 1, and vertex 3 attached
to vertex 2. As c 6= c 2 we have Hg 6= gH.
4. Let f : G → G 0 be a homomorphism and set H = ker f . Pick g ∈ G and set h = f (g ).
Prove that the coset gH is equal to the pre-image f −1 ({h}) = {a ∈ G | f (a) = h} of h.
Solution: We need to show that gH ⊆ f −1 ({h}) and f −1 ({h}) ⊆ gH. For the first
inclusion, pick gx ∈ gH, with x ∈ H. Then f (gx) = f (g )f (x) = f (g )e = f (g ) = h,
since x ∈ H = ker f . So gx ∈ f −1 ({h}). For the other inclusion, pick y ∈ f −1 ({h}).
Then f (y ) = h = f (g ). Thus f (yg −1 ) = f (y )f (g )−1 = e, so yg −1 ∈ H, call it
yg −1 = b ∈ H. Then y = gb ∈ gH.
5. Let g be an element of a cyclic group G , where |G | = n. Prove that an element g k is a
generator of G if and only if k is coprime to n.
Solution: Since it’s an “if and only if” statement, we split it into two directions.
(⇒) Suppose that x = g k is a generator of G . Then, x has order n, by definition of the
order of an element and generator. If k is not coprime to n then gcd(k, n) = m > 1. Let
r = n/m, s = k/m. Then,
x r = g kr = g ns = (g n )s = eG .
However, r < n, so that o(x) 6= n and x is not a generator of G , which is absurd. Hence,
we can’t find such an m above so that k and n must be coprime.
(⇐) Suppose that gcd(k, n) = 1, and k 6= ±1. Then, by the Fundamental Theorem
of Arithmetic (for example), we have that lcm(k, n) = kn. Let x = g k and suppose
o(x) = m, for some mdividing n. Thus,
eG = x m = g mk .
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As n = o(g ) is the least positive integer such that g n = eG , we must have n < mk (we
can’t have n = mk else gcd(n, k) = k. There exists b, r , with 0 ≤ r < n such that
mk = bn + r =⇒ eG = g mk = g bn+r = (g n )b g r = g r =⇒ r = 0.
Hence, bn = mk; call this integer y . It is a common multiple of k and n, but the least
such multiple is kn, so that m ≤ n ≤ m. Hence, n = m = o(x), so that x is a generator
of G .
6.
∗
Let G = GL2 (R), B = {upper triangular matrices in G }. Show that B is a subgroup
of G and
1 0
0 1
G /B =
B|a∈R ∪
B .
a 1
1 0
Show that there is a bijection
{lines in R2 } ←→ G /B
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