 Solution to Sample Exam 1. Math 113 Summer 2014.

```Solution to Sample Exam 1. Math 113 Summer 2014.
These problems are practice for the first exam, on group theory. The actual exam will be
similar in size and difficulty level, and, like this one, scored out of 125 points.
1. (25 points)
(a) If G is a group, give the definition of a normal subgroup of G .
(b) If N is a normal subgroup of G , prove that there exists a group H and a homomorphism f : G → H such that ker f = N.
(c) Let Q be the group of quaternions, and H the subgroup {±1}. Prove that Q/H ∼
=
Z/2Z × Z/2Z. You may use without proof the fact that up to isomorphism there
are just two groups of order 4.
Solution:
(a) Let H ⊂ G . Then, H is a normal subgroup if 1) H is a subgroup: for every h, k ∈ H,
hk −1 ∈ H; 2) gHg −1 = {ghg −1 | h ∈ H} = H, for every g ∈ G .
(b) Let H = G /N, the quotient group defined in class, and f = πN : G → G /N, the
quotient homomorphism. Then,
g ∈ ker πN ⇔ eN = eG /N = πN (g ) = gN ⇔ g ∈ N.
Hence, ker πN = N.
(c) First, we observe that |Q/H| = |Q|/|H| = 8/2 = 4, so that Q/H has order 4. The
groups Z/4Z and Z/2Z × Z/2Z are groups of order four so that, by the given hint,
every group is isomorphic to exactly one of these. Note that
iH ∗ iH = i −2 H = (−1)H = H =⇒ o(iH) = 2.
Similarly, we have
o(jH) = o(kH) = 2.
Since
G /H = {1H, iH, jH, kH},
we see that there are no elements in G /H of order 4, so that G /H is not isomorphic
to Z/4Z. Hence, it must be isomorphic to Z/2Z × Z/2Z.
2. (20 points) True or False. You don’t need to explain your answers.
(a) If H is a subgroup of G such that gH = Hg for all g ∈ G , then H is contained in
the center of G .
(b) If f : G → H is a homomorphism between groups whose orders are coprime, the
ker f = G .
(c) The permutations (135)(1462)(65) and (234)(156) are conjugate in S6 .
(d) If C (g ) and C (h) are the conjugacy classes of elements g , h ∈ G , C (g ) = C (h) if
and only if g and h have the same order.
(e) If H and K are subgroups of a group G which have orders 5 and 6, respectively,
then H ∩ K = {eG }.
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(f) When a group G acts on a set X , the size of any orbit Ox divides |X |.
(g) If G is a group, and p is a prime dividing |G | such that there are n Sylow p-subgroups
of G , then there is a nontrivial homomorphism G → Sn .
(h) IF G is a group of order 10, generated by two elements, of orders 2 and 5, then
G∼
= D10 .
(i) For a finite group G , and a prime p dividing |G |, all p-subgroups of G are conjugate
to each other.
(j) If f : G → H is a homomorphism and H ∼
= G / ker f , then f is surjective.
Solution:
(a) False
(b) True
(c) False
(d) False
(e) True
(f) False
(g) True
(h) False
(i) False
(j) True
3. (30 points) Consider the subgroup H = {e, (12)(34), (13)(24), (14)(23)} ⊂ S4 .
(a) Prove that H ∼
= K4 = Z/2Z × Z/2Z.
(b) Prove that H can be written as a union of conjugacy classes in S4 .
(c) Use (b) to deduce that H is a normal subgroup of S4 .
(d) Find a group G and a homomorphism from S4 to G whose kernel is H (this gives
an alternate proof that H is normal) [note: do not just use G = S4 /H...]
(e) Determine to which familiar group S4 /H is isomorphic.
Solution:
(a) Since H has two elements, it must be isomorphic to K4 or Z/4Z; but since it has
no elements of order 4, it must be isomorphic to K4 .
(b) Write H = {e} ∪ {(12)(34), (13)(24), (14)(23)}. These are two conjugacy classes,
since they correspond to the cycle types (1,1,1,1) and (2,2), respectively.
(c) It is true in general that any subgroup which can be written as a union of conjugacy
classes is normal. Proof (when G is finite, which suffices for this problem): Suppose
N = C (e) ∪ C (g1 ) · · · ∪ C (gr ), and pick any element g in G . Then for any n ∈ N,
n lies in one of the classes C (gi ), and since gng −1 is also in C (gi ), this shows
that gng −1 ∈ N. So gNg −1 ⊆ N. The same argument, with g −1 replacing g ,
shows that g −1 Ng ⊆ N. and conjugating both sides by g gives N ⊆ gNg −1 . Thus
gNg −1 = N, so N is normal.
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(d) (note: this isn’t really the type of thing we’d ask on an exam - just for fun!) We’ll
try to map S4 to a group of order 6, that way the kernel will have size 4. The group
of order 6 we’ll use is S3 , as follows. We can identify S4 with symmetries of the
tetrahedron, by some HW problem. Consider the set X of pairs of opposite edges;
since there are 6 edges, there are three such pairs, and elements of S4 permute
them. If we name the three pairs 1, 2, 3, this is nothing other than an action of
S4 on {1, 2, 3}, which is equivalent to a homomorphism f : S4 → S3 . Now finding
the kernel of this map is the same as finding those elements of S4 which fix all
three pairs of edges (they “act trivially”). By playing around with a model of a
tetrahedron, you should be able to convince yourself that the only elements which
do so are the four elements of H. Thus this action induces a map S4 → S3 whose
kernel is H.
(e) The map f constructed in (d) is actually surjective: since S3 can be generated by
the 2-cycles (12) and (23), it is enough to check that these two are in the image.
But the transposition of two pairs of opposite edges can be effected by a rotation of
the tetrahedron, so they are indeed in the image. Thus S3 im f ∼
= S4 / ker f = S4 /H,
using the first isomorphism theorem.
4. (25 points)
(a) Prove that if gcd(a, b) = 1, then Z/aZ × Z/bZ is cyclic.
(b) Prove that if G is a group of order pq, where p > q are primes and p 6≡ 1 mod q,
then G ∼
= Z/pZ × Z/qZ.
(c) Deduce that every group of order 15 is cyclic.
Solution:
(a) Suppose that gcd(a, b) = 1. Then, we claim that (1, 1) ∈ Z/aZ × Z/bZ is a
generator: note that lcm(a, b) = ab, using the Fundamental Theorem of Arithmetic
(for example); we claim that o((1, 1)) = lcm(a, b). Denote k = lcm(a, b). Then,
k(1, 1) = (k1, k1) = (k, k) = (0, 0) =⇒ o((1, 1)) ≤ k.
If k 0 (1, 1) = (0, 0), k 0 6= 0, then we must have that k 0 > 0 is divisible by both a
and b; hence, we have k ≤ k 0 . Therefore, k is the least positive integer such that
k((1, 1)) = (0, 0), so that o((1, 1)) = k, by definition of the order of an element.
Since |Z/aZ × Z/bZ| = ab, then Z/aZ × Z/bZ = h(1, 1)i is a cyclic group.
(b) Using Sylow’s Theorems, we see that there exists a Sylow p-subgroup H ⊂ G , and
a Sylow q-subgroup K ⊂ G . The number of Sylow p-subgroups divides q and is
equivalent to 1 modulo p; hence, as p > q, there is only one Sylow p-subgroup.
Similarly, the number of Sylow q-subgroups divides p and is equivalent to 1 modulo
q. The given condition p 6≡ 1 mod q implies that there is only one such Sylow
q-subgroup. Let hai = H, hbi = K (because H and K have prime order, so must
be cyclic). Then, using that both H and K are normal (because there exist no
conjugates of them, by what we’ve seen) we have that
ab = b 0 a, as aK = Ka, and ab = ba0 , as Hb = bH.
Hence, b 0 a = ba0 =⇒ b −1 b 0 = a0 a−1 . Denote this element x(= b −1 b 0 = a0 a−1 ).
Then, x ∈ H ∩ K ⊂ H. However, the only subgroups of H are H and {e} (because
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H has prime order): if H ∩ K = H, then H ⊂ K is a subgroup of order p > |K | = q,
which is absurd. Hence, we must have H ∩ K = {e}. Hence, x = e and b = b 0 ,
a = a0 . So, ab = ba. Define a function, where 0 ≤ i < p, 0 ≤ j < q,
f : G → Z/pZ × Z/qZ ; ai b j 7→ (i, j).
Here we have use that G /H ∼
= Z/qZ, so that G = ∪b j H. Then, f is a homomorphism
f (ai b j ak b l ) = f (ai+k b j+l ) = (i + k, j + l) = (i, j) + (k, l) = f (ai b j ) + f (ak b l ).
Moreover, f is injective: if ai b j ∈ ker f then i = j = 0. Hence, |im f | = |G | = pq
so that im f = Z/pZ × Z/qZ. Hence, G is isomorphic to Z/pZ × Z/qZ.
(c) As 15 = 3.5, and 5 6≡ 1 mod 3, we are in the situation considered in (b). Thus,
G∼
= Z/3Z × Z/5Z
part (a)
∼
=
Z/15Z.
5. (25 points)
(a) Determine all automorphisms of Z/8Z.
(b) Let G be any group. Prove that the following defines an action of G on Aut G :
(g , f ) 7→ g · f , where g · f is the function on G defined by g · f (h) = f (g −1 hg ).
(c) In the case when G = Z/8Z, prove that the homomorphism G → Perm(Aut G )
induced by this action is trivial.
Solution:
(a) Let f be an automorphism of Z/8Z. It must be the case that f (1) is a generator
of Z/8Z. Hence, we must have f (1) = 1, 3, 5, 7, as these are the generators of
Z/8Z. Moreover, if g and f are two homomorphisms such that f (1) = g (1), then
f (i) = g (i), for any i ∈ Z/8Z.
(b) Let f ∈ AutG . Then, (e · f )(g ) = f (e −1 ge) = f (g ), for any g ∈ G . Hence,
e · f = f , for any f ∈ AutG . If f ∈ AutG and g , h ∈ G , then
(g · (h · f ))(k) = (h · f )(g −1 kg ) = f (h−1 g −1 kgh) = f ((gh)−1 k(gh)) = (gh · f )(k),
for any k ∈ G . Hence, g · (h · f ) = (gh) · f .
(c) Since Z/8Z is abelian, for any g , h ∈ Z/8Z, we have −g + h + g = h. Hence,
(g · f )(h) = f (−g + h + g ) = f (h) =⇒ g · f = f ,
for any g ∈ Z/8Z, f ∈ AutZ/8Z. Therefore, the homomorphism
Z/8Z → Perm(AutZ/8Z) ; g 7→ ag ,
has the property that ag (f ) = g · f = f . Hence, ag is the identity, for each g ∈ G ,
so that the above homomorphism is trivial.
6. (25 points)
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(a) Determine, up to isomorphism, all abelian groups of order 24.
(b) Let G = Z/6Z × Z/4Z. Find a subgroup H of G such that the quotient group G /H
is isomorphic to K4 = Z/2Z × Z/2Z.
(c) Prove that there is no subgroup H of G such that G /H is isomorphic to Z/8Z.
(d) If a group G (no longer necessarily abelian), has 3 Sylow 2-subgroups, prove that
G admits a nontrivial homomorphism to S3 via its action on the set Syl2 of Sylow
2-subgroups, and use this to conclude that G is not simple.
Solution:
(a) Using the prime factorization 24 = 23 · 3, We find that they are:
Z/8Z × Z/3Z
Z/4Z × Z/2Z × Z/3Z
(Z/2Z)3 × Z/3Z
(b) We use the notation hxi to mean “cyclic subgroup generated by x. Let H = h3i×h2i.
This has the desired quotient.
(c) Suppose for contradiction that there were such a subgroup H. Then we would have
an element of order 8 in the quotient group G /H, which would be a coset a + H
(written additively because this is an abelian group), for some a ∈ G . This coset is
the image π(a) = a + H of a under the canonical homomorphism. By HW2 problem
1, then, 8 would divide the order of a, so o(a) would be at least 8, which is not
possible in G .
(d) We have seen that G acts on Syl2 by conjugation, simply because conjugating any
subgroup produces another subgroup with the same order. SYL2 asserts that this
action is nontrivial (in fact, that there is exactly one orbit, but we won’t actually use
that here). So we can interpret this action as a nontrivial homomorphism f : G → S3
(we have written S3 in place of Perm(Syl2 ), but the groups are isomorphic since
both Syl2 and {1, 2, 3} are sets with 3 elements). Since the action is nontrivial,
ker f 6= G . Moreover, ker f is normal, as we have seen in class, so it remains only to
check that ker f 6= {e}, and then we will have produced a proper nontrivial normal
subgroup, showing that G is not simple. We know 1 < |im f | ≤ 6 (since f is
nontrivial and S3 has order 6) and also that | ker f | = |G |/|im f | = 24/|im f |, by
the first isomorphism theorem. Since |im f | is at most 6, this means | ker f | is at
least 4, so we’re done.
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``` Course 3413 — Group Representations Sample Paper II Dr Timothy Murphy December 2010 