Download Math 120: Problem Set 4 Solutions 3.1.1. First, φ(1 G)=1H (since φ is

Math 120: Problem Set 4 Solutions
3.1.1. First, φ(1G ) = 1H (since φ is a homomorphism) and 1H ∈ E (since E is a subgroup
of H), so 1G ∈ φ−1 (E). Second, if x and y are in φ−1 (E), then φ(x) and φ(y) are in E,
so φ(x)φ(y) ∈ E. But φ(x)φ(y) = φ(xy), so φ(xy) ∈ E, so xy ∈ φ−1 (E). Thus φ−1 (E) is
closed under multiplication. Finally, if x is in φ−1 (E), then φ(x) is in E. Since E is closed
under inverses, φ(x)−1 is in E. But φ(x)−1 = φ(x−1 ). So φ(x−1 ) ∈ E, which means x−1
is in φ−1 (E). Thus φ−1 (E) is closed under inverses.
We have shown that φ−1 (E) is nonempty and is closed under multiplication and inverses.
Therefore it is a subgroup.
3.1.22(a). Recall that a subgroup N of G is normal if and only if gng −1 ∈ N for every
g ∈ G and n ∈ N . We know from earlier work that H ∩ K is a subgroup, so we just have
to show that it is a normal subgroup of G. Let x be in H ∩ K and g be in G. Then x ∈ H
and x ∈ K, so gxg −1 ∈ H (by normality of H) and gxg −1 ∈ K (by normality of K), so
gxg −1 ∈ H ∩ K.
3.1.24. As in 22(a), we already know that N ∩ H is a subgroup, so we only have to show
that it is normal in H. Thus let h be an element of H and x be an element of N ∩ H. Then
of course h ∈ G and x ∈ N , so hxh−1 ∈ N (by normality of N ). However, hxh−1 is also
an element of H since h and x are both elements of H. So hxh−1 is in both N and H,
which implies that it is in N ∩ H. We have proved that h ∈ H and x ∈ N ∩ H implies
hxh−1 ∈ N ∩ H. Thus N ∩ H in normal in H.
3.2.4. Let Z = Z(G). Suppose G is NOT abelian, so Z is not all of G. Let x be an element
of G that is not in Z. Let H be the set of all elements of G that commute with x. (In
other words, H is the centralizer of x.) Then Z is a subgroup of H and H is a subgroup
of G. Furthermore, Z is not all of H since x is in H but not in Z. Also, H is not all of
G since x is not in the center. Thus (by Lagrange’s theorem, |Z| is a proper factor of |H|
and |H| is a proper factor of G. The only proper factors of |G| = pq are 1, p, and q, but 1
has no proper factors. So |H| must be p or q, and |Z| must be 1.
Here is a slightly different proof: By Lagrange’s theorem, the order of Z = Z(G) can only
be 1, p, q, or pq. If it’s 1 or pq, we’re done. Thus suppose |Z| = p (of course the case
|Z| = q is handled the same way.) Now Z is normal, so G/Z is a group, and it’s order is
|G|/|Z| = (pq)/p = q. Since q is prime, G/Z must be cyclic. (Corollary 10 on page 91.)
By problem 3.1.36, this means G is abelian.
[Here is the solution to 3.1.36, i.e. the proof that if G/Z is cyclic, then G is is abelian.
Since G/Z is cyclic, by assumption it’s generated by some element gZ. That is, we get
all the cosets by taking powers of gZ: (gZ)k = g k Z (where k can be any integer). This
means every element x of G can be written as g k z for some integer k and some element
z ∈ Z (k and z will depend on x). Thus any two elements x and y of G can be written as
g k z and g m z 0 for suitable integers k and m and suitable elements z and z 0 of Z. Now g k
1
2
and g m commute with each other, and and z and z 0 commute with everything, so all four
of g k , g m , z, and z 0 commute with each other. This implies that x and y commute. Since
this is true for all x and y in G, G is abelian.]
3.2.5. (a) If A is any subset of G and g is an element of G, then f (a) = ga defines a
bijection from A to gA. (It is surjective by definition of gA, and it is injective because
ga = ga0 implies g −1 ga = g −1 ga0 which implies a = a0 .) So A and gA have the same size.
Likewise A and Ah have the same size. Thus for any subset A and any elements g and h
of G, A, gA, and gAh = (gA)h all have the same size. In particular, H and gHg −1 have
the same order.
Thus we only have to show that gHg −1 is a subgroup. It is certainly nonempty since H is
nonempty. Suppose x and y are elements of gHg −1 . Then x = gag −1 and y = gbg −1 for
some elements a and b of H. Then
xy = gag −1 gbg −1 = gabg −1 ,
so xy ∈ gHg −1 . Also, a−1 ∈ H and
x−1 = (gag −1 )−1 = (g −1 )−1 a−1 g −1 = ga−1 a−1
so x−1 ∈ gHg −1 . We have shown that gHg −1 is nonempty and is closed under multiplication and inverses, so it is a subgroup.
3.2.8. Let n be the order of intersection. By Lagrange, n divides |H| and n divides |K|.
Thus n divides their greatest common factor, which is 1. So n = 1.
3.2.14. Proof that S4 has no normal subgroup of order 8: Suppose N were a normal
subgroup of order 8. Then G/N is ∼
/ N , then xN is not the identity
= Z/3Z. Thus if x ∈
element of G/N , so it has order 3 (as an element of G/N ), so
(xN )k = N if and only if 3|k
i.e., xk ∈ N if and only if 3|k. In particular, if xk = 1, then 3|k. So |x| is divisible by 3.
By writing an element σ ∈ S4 as a product of disjoint cycles, we see that the only possible
orders are 1, 2, 4, and 3. So if |x| is divisible by 3, then |x| = 3. Furthermore, there are
exactly 8 elements of order 3, namely the 3-cycles. (There are 8 ways to pick a 3-cycle in
S4 : 4 choices for which number i will not be in the 3-cycle, and then, having chosen i, two
ways to cyclically order the remaining three numbers.)
Combining these two paragraphs: if x ∈ G and x ∈
/ N , then x has order 3. Since there
are 8 such elements, this means there are at most 8 elements of G not in N . But that is
impossible since G has 24 elements and N has 8 elements.
Proof that S4 has no normal subgroup N of order 3. Suppose N was such a subgroup. Let
x be an element of S4 not in N . Then the order of xN (as an element of G/N ) divides
3
|G/N | = |G|/|N | = 24/3 = 8. Thus (xN )k 6= N if n is odd, which means xn 6= N if n is
odd. In particular, xn 6= 1 if n is odd. We have shown: if x ∈
/ N , then x has even order.
Thus all the elements of odd order belong to N . But there are more than 3 such elements.
(Indeed, as we saw above, there are exactly 8 such elements.)
Remark: we can generalize the principle we used:
Lemma. Suppose G is a nite group, N is a normal subgroup, and x ∈ G. If |x| and
|G|/|N | are relatively prime, then x ∈ N .
Additional problems: check back later....