Math 493 (Fall 2012). Solutions for Problem Set 4
4.1. (a) The assumption of part (a) means that the map ϕ : H × K −→ G given by
ϕ(h, k) = hk is surjective. We need to show that it is a group homomorphism and that it
has trivial kernel. It will then follow that ϕ is a group isomorphism.
Let us first check that H and K commute with each other, i.e., that hk = kh for all
h ∈ H and k ∈ K. Indeed, the identity hk = kh is equivalent to hkh−1 k −1 = e. Writing
hkh−1 k −1 = (hkh−1 )k −1 , we see that hkh−1 k −1 ∈ K since K is a normal subgroup of
G. On the other hand, writing hkh−1 k −1 = h(kh−1 k −1 ), we find that hkh−1 k −1 ∈ H
because H is a normal subgroup of G. So hkh−1 k −1 ∈ H ∩ K = {e}, which proves that
hkh−1 k −1 = e, as asserted. Now it follows that ϕ is a homomorphism:
ϕ (h1 , k1 )(h2 , k2 ) = ϕ(h1 h2 , k1 k2 ) = h1 h2 k1 k2 = h1 k1 h2 k2 = ϕ(h1 , k1 )ϕ(h2 , k2 ),
where we used the identity h2 k1 = k1 h2 we just proved. Finally, if (h, k) ∈ Ker(ϕ), then
hk = e, so h = k −1 , which means that h ∈ H ∩ K = {e}, whence h = e and thus k = e
as well. So the kernel of ϕ is trivial, completing the proof.
(b) Consider again the map ϕ : H × K −→ G given by ϕ(h, k) = hk. Part of the
solution of (a) goes through without the assumption that HK = G and shows that if
H ∩ K = {e}, then ϕ is a group homomorphism with trivial kernel. Hence it is injective,
so |H| · |K| = |H × K| ≤ |G|. The last inequality becomes an equality if and only if ϕ is
also surjective, which is equivalent to the condition that HK = G, proving (b).
4.2. (a) Write G = (Z/pZ) × (Z/qZ) and let f1 : G −→ Z/pZ and f2 : G −→ Z/qZ
denote the two projections, i.e., f1 (x, y) = x and f2 (x, y) = y. Note that f1 and f2 are
both group homomorphisms. If g = (1̄, 1̄), then f1 (g) = 1̄ in Z/pZ and f2 (g) = 1̄ in Z/qZ.
So f1 (g) has order p and f2 (g) has order q. By an earlier discussion of homomorphisms,
the orders of both f1 (g) and f2 (g) must divide the order of g in G, and on the other hand,
the order of g must divide |G| = pq. Since p and q are distinct primes, the order of g
equals pq. In particular, G is cyclic.
(b) By the first Sylow theorem, G has a subgroup H of order p and a subgroup K of
order q. We then have H ∼
= Z/pZ and K ∼
= Z/qZ. Both H and K are normal in G since
G is abelian. Moreover, H ∩ K is a subgroup of both H and K, so |H ∩ K| divides both
p and q, so |H ∩ K| = 1 and H ∩ K is trivial. By the first problem of this homework, we
obtain an isomorphism G ∼
=H ×K ∼
= (Z/pZ) × (Z/qZ), so G is cyclic by part (a).
(c) By the first Sylow theorem, G has a subgroup H of order p and a subgroup K
of order q. If we show that H and K are normal in G, we can repeat the rest of the
argument from part (b) verbatim. Showing that H (resp., K) is normal in G is the same
as showing that H (resp., K) is the only p-Sylow (resp., q-Sylow) subgroup of G. By the
third Sylow theorem, np (G) ≡ 1 mod p and np (G) divides q. Since q is prime, np (G) has
to be 1 or q, and since q 6≡ 1 mod p, we must have np (G) = 1. By the same argument,
nq (G) = 1 (note that q > p, so p certainly cannot be congruent to 1 modulo q), q.e.d.
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4.3. Let G be a group of order p2 , where p is a prime. We claim that if G is not cyclic,
then G ∼
= (Z/pZ) × (Z/pZ), which will solve the problem. Let g ∈ G be any nontrivial
element. Then the order of g is > 1, divides p2 , and is not equal to p2 (as G is not cyclic),
so g has order p. Let K ⊂ G be the subgroup generated by g. Then K ∼
= Z/pZ and K is
normal in G because we already know that G is abelian (from the course notes).
Next choose any h ∈ G such that h 6∈ K and let H ⊂ G be the subgroup generated
by h. By the same argument as above, H ∼
= Z/pZ and H is normal in G. We note that
H ∩ K is a subgroup of H that cannot be equal to H (since h 6∈ K), so H ∩ K = {eG }
since H has prime order. Finally, by construction, |H| · |K| = p2 = |G|. By the first
'
problem of this homework, we obtain an isomorphism (Z/pZ) × (Z/pZ) −→ G.
4.4. We can solve both parts at once. Let f : H × K −→ K be the projection map:
f (h, k) = k. The definition of the group operation on H × K implies that f is a group
homomorphism. It is clear that f is surjective, and by construction, N = Ker(f ). So N
is normal in H × K and by the first isomorphism theorem, (H × K)/N ∼
= K.
4.5. (a) Consider the groups G1 = Z/4Z and G2 = (Z/2Z) × (Z/2Z). These are finite
abelian groups and they are not isomorphic (the first is cyclic, the second is not). We
claim that they are related. Indeed, let N1 = 2Z/4Z ⊂ G1 be the unique subgroup of
order 2 and let N2 ⊂ G2 be the subgroup generated by (1̄, 0̄). Then N2 also has order 2, so
both G1 /N1 and G2 /N2 have order 2, and it follows that N1 ∼
= N2 and G1 /N1 ∼
= G2 /N2 .
(b) Take G1 = S3 and G2 = Z/6Z. Let N1 = A3 (the alternating subgroup of S3 )
and let N2 = 2Z/6Z ⊂ G2 be the unique subgroup of order 3. Then N1 and N2 are both
isomorphic to Z/3Z, and G1 /N1 and G2 /N2 are both isomorphic to Z/2Z.
(c) Take G1 = S4 and G2 = A4 × (Z/2Z). Note that both S4 and A4 are noncommutative (by an easy inspection), so G1 and G2 are noncommutative. We have obvious normal
subgroups N1 = A4 ⊂ G1 and N2 ⊂ G2 that are both isomorphic to A4 , and the quotients
G1 /N1 and G2 /N2 are both isomorphic to Z/2Z. Finally, G1 and G2 are not themselves
isomorphic, because G2 has nontrivial center while the center of S4 is trivial.
4.6. (a) If x ∈ X, then eG · x = x, so x ∼ x. Next, suppose that x, y ∈ X and x ∼ y, so
that g · x = y for some g ∈ G. Then
g −1 · y = g −1 · (g · x) = (g −1 g) · x = eG · x = x,
so y ∼ x. Finally, suppose that x, y, z ∈ X are such that x ∼ y and y ∼ z. Then y = g · x
and z = h · y for some g, h ∈ G. Hence z = h · (g · x) = (hg) · x, so x ∼ z.
(b) Since eG · x = x, we have eG ∈ StabG (x). Next, if g, h ∈ StabG (x), then
(gh) · x = g · (h · x) = g · x = x,
so gh ∈ StabG (x). Finally, if g ∈ StabG (x), then g −1 · x = g −1 · (g · x) = (g −1 g) · x =
eG · x = x, so g −1 ∈ StabG (x). Thus StabG (x) is a subgroup of G.
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4.7. The given action is not free, for example because G has 6 elements and G/H only
has 3 elements (if the action were free, the size of any G-orbit would be equal to 6).
Now let us prove that the action is faithful. Suppose that g ∈ G is such that g ·x = x for
all x ∈ G/H. This means that gg 0 H = g 0 H for all g 0 ∈ G. Equivalently, for every g 0 ∈ G
there exists h ∈ H with gg 0 = g 0 h, i.e., g = g 0 hg 0−1 . We conclude that g ∈ g 0 Hg 0−1 for all
g 0 ∈ G, i.e., g belongs to the intersection of all G-conjugates of the subgroup H. But this
intersection is trivial (so g = e, as required): indeed, if g 0 6∈ H, then g 0 Hg 0−1 6= H (by an
easy inspection of the multiplication table of S3 ), which means that H ∩ g 0 Hg 0−1 = {e}
because H already has order 2.
4.8. We have |S4 | = 4! = 24 = 23 · 3, so if p is a prime with p ≥ 5, then the only p-Sylow
subgroup of S4 is the trivial subgroup {e} ⊂ S4 .
We also see that the 3-Sylow subgroups of S4 are the subgroups of order 3. Any such
subgroup must be cyclic and hence generated by an element of order 3. In S4 every
element of order 3 is a 3-cycle. Moreover, if σ ∈ S4 is a 3-cycle, then σ and σ −1 both
generate the same 3-Sylow subgroup of S4 . Suppose we fix an integer 1 ≤ j ≤ 4. By
removing j from {1, 2, 3, 4} we get a set {1, 2, 3, 4}\{j} with 3 elements. There are exactly
two 3-cycles, that are inverse to each other, that permute these three elements. If we let
Kj be the subgroup of S4 consisting of these three cycles together with e, we see that
Kj is a 3-Sylow subgroup of S4 , and that S4 has exactly four 3-Sylow subgroups, namely,
K1 , K2 , K3 , K4 (with the notation that we just introduced).
Finally, let us describe all 2-Sylow subgroups of S4 . These are the subgroups of order
8. I will first state the answer, and then explain how the answer can be obtained.
For each pair of distinct elements i, j ∈ {1, 2, 3, 4}, write τi,j ∈ S4 for the transposition
that interchanges i with j. There is a total of 4 · (4 − 1)/2 = 6 such transpositions.
By a straightforward inspection, the subset N = {e, τ1,2 τ3,4 , τ1,3 τ2,4 , τ1,4 τ2,3 } is a normal
subgroup of S4 of order 4. By the second Sylow theorem, N is conjugate to a subgroup of
some 2-Sylow subgroup of S4 , but because N is normal and because all 2-Sylow subgroups
of S4 are conjugate, N is contained in every 2-Sylow subgroup of S4 .
Now for each of the transpositions τi,j introduced earlier, consider the subgroup Hi,j
of S4 generated by N and by τi,j . As we will soon see, this is a subgroup of S4 of order
8 (hence it is a 2-Sylow subgroup), and moreover, every 2-Sylow subgroup of S4 can be
obtained in this way. Note that this implies that S4 has exactly three 2-Sylow subgroups:
even though S4 contains six distinct transpositions, there are some overlaps between the
−1
subgroups Hi,j we just described. For example, τ1,2 τ3,4
= τ1,2 τ3,4 ∈ N , which implies
that τ1,2 and N together generate the same subgroup as τ3,4 and N , that is, H1,2 = H3,4 .
Similarly, H1,3 = H2,4 and H1,4 = H2,3 . The upshot is that with the notation above, the
2-Sylow subgroups of S4 are H1,2 , H1,3 and H1,4 .
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Now I will explain how to prove the claims above. By an easy inspection, the elements
τi,j are all conjugate to each other under S4 ; to be more precise, the elements
τi,j 1 ≤ i < j ≤ 4
form a single conjugacy class (of size 6) inside S4 . Now let H ⊂ S4 be any 2-Sylow
subgroup. Since the subgroup hτ1,2 i generated by τ1,2 has order 2, it must be conjugate
to a subgroup of H, which in view of the previous observation implies that τi,j ∈ H for
some 1 ≤ i < j ≤ 4. We already saw that N ⊂ H, so H must contain the subgroup Hi,j
generated by N and τi,j . Furthermore, Hi,j must be bigger than N since τi,j 6∈ N . As
|N | = 4 and |H| = 8, Lagrange’s theorem implies that |Hi,j | = 8, so H = Hi,j . Thus we
showed that every 2-Sylow subgroup of S4 must be equal to Hi,j for some 1 ≤ i < j ≤ 4.
A technical but important point. At first glance, it may seem that the proof is
now complete. However, what we proved so far does not directly imply that the 2Sylow subgroups of S4 are exactly the subgroups Hi,j . To make the argument completely
rigorous, the following words must be said. First, we know that S4 has at least one 2Sylow subgroup, so by the previous paragraph, Hi,j is a 2-Sylow subgroup of S4 for at
least one pair 1 ≤ i < j ≤ 4. On the other hand, because N is normal and the elements
τi,j form a single conjugacy class, it follows that all the subgroups Hi,j are conjugate to
each other, and hence each of them is a 2-Sylow subgroup of S4 .
Remark. An alternative approach would be to first check directly that each of the subgroups Hi,j has order 8, and hence is a 2-Sylow subgroup of S4 . This is certainly possible
and not very difficult, but would require an additional argument. The method used above
circumvents this issue by using the fact that S4 has at least one 2-Sylow subgroup, even
if we don’t know a priori what that subgroup is.