Dr. Marques Sophie
Office 519
Algebra 1
Fall Semester 2014
marques@cims.nyu.edu
Problem Set 12
Exercise 1 :
Taking H “ Z{3Z and N “ Z{6Z
1. Describe all possible actions by automorphism of H on N.
2. How many different semi direct product groups G “ N ˆφ H can there be with
normal subgroup N “ Z{6Z and cross-sectioning subgroup H “ Z{3Z ? Explain.
Solution : Let H “ă a ą. Actions H ˆ N Ñ N correspond to arbitrary homomorphism
Φ : H Ñ AutpNq, which in turn are determined by where Φ sends the generator a
(Φpaq “ φa ) ; then Φpeq “ IdN , Φpaq “ φa and Φpa2 q “ Φpaq2 “ φa ˝ φa . However, since
a3 “ e, the automorphism φa assigned to a must also have the property pφa q3 “ idN , if
we want a well defined homomorphism Φ : H Ñ N.
Now N » Z{6Z ð AutpZ{6Z, `q » pU6 , ¨q where U6 “ tr1s, r5su “ tr1s, ´r1su are the
multiplicative units in Z{6Z, and we may regard Φ as a map from H “ Z{3Z “ă a ą
into U6 such that Φpaq “ rks satisfies rks3 “ rk3 s “ r1s. Now consider what happens
when we compute the third powers of elements in U6
k P U6 k3 in Z{6Z
1
1 OK
5 “ ´1 5
The only assignment of Φpaq with Φpaq3 “ r1s in U6 is Φpaq “ r1s. Thus Φpeq “ IdN ,
Φpaq “ φr 1s : r js Ñ r1s ¨ r js “ rjs is the identity map idN ; Φpa2 q “ Φpaq2 “ IdN . Thus Φ
is the trivial homomorphism, the vector HˆN Ñ N is trivial and the corresponding semi
direct product is the direct product Z{6Z ˆ Z{3Z (abelian). Note : since gcdp3, 6q ‰ 1
this is not isomorphic Z{18Z.
Remark : This does not classify all G with |G| “ 18. It only identifies those that are semi
direct products Z{6Z ˆ Z{3Z ; there is only one of this type, namely Z{6Z ˆ Z{3Z.
Exercise 2 :
1. Show that, up to isomorphism G “ pZ{15Z, `q is the only group of order |G| “ 15.
Hints : First show G is abelian by examining the pattern of p-Sylow subgroups.
Then show G is isomorphic to the direct product Z{5Z ˆ Z{3Z
2. If |G| “ 30 use the Sylow theorems to prove that the 3-Sylow and 5-Sylow
subgroup are both normal in G, and that G is a semi direct product G »
Z{15Z ¸ Z{2Z. Hints : Use 1.
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3. If |G| “ 30 “ 5 ¨ 3 ¨ 2, determine all the groups of this order. Hints : In 1. we show
that G must be a semi direct product Z{15Z ˆφ Z{2Z. You may assume this fact
as a given, even if you have not done it in 1. What are the possibilities for semi
direct products of this type ?
Solution :
1. Let |G| “ 15 “ 5 ¨ 3 and let H5 , ¨ ¨ ¨ , H5 , H3 , ¨ ¨ ¨ , H3 be the distinct Syp jq
piq
low subgroup in G. We have |H5 | “ 5 and |H3 | “ 3 and 7pp ´ Sylowq “
1, 6, 11, 16, 21, ¨ ¨ ¨ , 7p3 ´ Sylowq “ 1, 4, 7, 10, 13, 16, ¨ ¨ ¨ . But the 7pp ´ Sylowq
must also be a divisor of |G| “ 15. The only possibilities are : the 5-Sylow and
3-Sylow subgroups are unique, and hence both are normal in G.
But H5 XH3 “ teu by Lagrange, and then |H5 ¨H3 | “ |H3 |¨|H3 |{|H5 XH3 | “ 15 “ |G|,
so G “ H5 H3 and every g P G has a unique decomposition g “ ab, a P H5 , b P H3 .
That means G is a direct product G » Z{5Z ˆ Z{3Z. We claim that G actually
contains an element x with opxq “ 15, and hence G » pZ{15Z, `q.
Since 5 and 3 are prime, we have H5 » pZ{5Z, `q and H3 » pZ{3Z, `q. Let
H5 “ă a ą with opaq “ 5 and H3 “ă b ą with opbq “ 3 and let x “ ab. We now
show that opxq “ 15. Obviously G, a direct product of abelian groups, is abelian,
so that for r “ 0, 1, 2 ¨ ¨ ¨
p1q
p5q
p1q
p5q
xr “ pabqr “ ar br “ e ô ar “ e in Z{5Z and br “ e in Z{3Z
ô r ” 0 pmod 5q and r ” 0 pmod 3q
The smallest integer r ą 0 satisfying both conditions is r “ 15, as claimed.
Alt. Proof : Let H “ă x ą. Then |H| “ opxq can only be 1, 3, 5, 15. Since a ‰ e,
b ‰ e, we cannot have |H| “ 1. If |H| “ 3 and H “ Z{3Z, we get a contradiction
because that would implies ab “ b j ô a “ b j´1 ô a P Z{3Z X Z{5Z “ teu ; but
a ‰ e. If |H| “ 3 and H ‰ Z{3Z, then H X Z{3Z “ teu, H ¨ Z{3Z is a subgroup
in G of order 9 “ |H| ¨ |Z{3Z| ; that is also impossible. Conclusion : we can’t have
|H| “ 3. Same argument also shows |H| ‰ 5. By default, |H| “ 15, opxq “ 15,
and G » pZ{15Z, `q.
2. Denote the distinct Sylow subgroups as H5 “ H5 , ¨, H5 ; H3 “ H3 , ¨ ¨ ¨ H3 ;
piq
p1q
ptq
H2 “ H2 ¨ ¨ ¨ H2 and let Sp “ Yi Hp for p “ 2, 3, 5. The divisors of |G| are
1, 2, 3, 5, 6, 10, 15, 30 and 7pp ´ Sylowq “ 1 ` pk for some k “ 0, 1, 2, ¨ ¨ ¨ for each
p. Since 7pp ´ Sylowq must divide |G|, the only possibilities are :
— p “ 5 7p5 ´ Sylowq “ 1 or 6 ;
— p “ 3 7p3 ´ Sylowq “ 1 or 10 ;
— 7p2 ´ Sylowq “ 1, 3, 5 or 15.
p jq
piq
By Lagrange Hp X Hp “ teu for all i, j if p ‰ q. Hence |S5 | “ 5 ` 4pr ´ 1q,
|S3 | “ 3 ` 2ps ´ 1q, etc. If neither is normal then 7p5 ´ Sylowq ě 6, 7p3 ´ Sylowq ě
10 and |S5 | ě 5 ` 4p5q “ 25, |S3 | ě 3 ` 2p9q “ 21. But S3 X S5 “ teu, so
§3 Y S5 | “ |S3 | ` |S5 | ´ 1 ě 25 ` 21 ´ 1 ą 0. This is impossible, so at least one of
the groups H3 and H5 are normal.
p1q
2
prq
p1q
p5q
Since at least one is normal, M “ H3 H5 is a subgroup in G and since H3 XH5 “ teu
by Lagrange, we have |M| “ 15. M must be abelian (and » Z{15Z). For p “ 3, 5
p1q
let xp “the set of p-Sylow subgroups and take Hp “ Hp as a base point in this
space. As in the 2nd Sylow theorem, there is a transitive action G ˆ Xp Ñ Xp
by conjugation g : H Ñ gHg´1 . Furthermore, |Xp | “ 7pp ´ Sylow subgroupsq “
|G|{|StabG pHp q| where StabG pHp q “ tg P G : gHp g´1 “ Hp u.
Each Hp lies in the abelian group M and hence mHp m´1 “ Hp , @m P M.
M Ď StabG pHp q and |StabG pHp q| ě |M| “ 15, so that |Xp | ď 30{15 “ 2. Since
|Xp | “ 1or6, if p “ 5, |Xp | “ 1or10 if p “ 3 we must actually have |X5 | “ |X3 | “ 1.
Thus H5 , H3 are both normal in G.
Once we know H5 , H3 E G, the subgroup M “ H3 H5 is also normal p» Z{15Zq.
piq
Finally, H2 X M “ teu because gcdp2, 15q “ 1 so if H2 is any 2-Sylow subgroup
we have a semi direct product G “ M ¸ H2 » Z{15Z ¸ Z{2Z.
3. We need to classify the possible actions of Z{2Z by automorphisms on Z{15Z “
M. These correspond to homomorphisms Φ : pZ{2Z, `q Ñ AutpZ{15Z, `q “
pU15 , ¨q. Here U15 “ trks : gcdpk, nq “ 1u “ tr1s, r2s, r4s, r7s, r8s, r11s, r13s, r14su
is an abelian group of order |U15 | “ 8. Let’s write Z{2Z “ te, au ; to define a
homomorphism Φ : Z{2Z Ñ U15 we only have to specify the image Φpaq of the
generator a ; but the assignment Φpaq “ b P U15 must yield a subgroup ΦpZ{2Zq
whose order is either 1 ( if Φpaq “ a “ r1s) or 2 in U15 . This means we must have
b2 “ e. A simple calculation allows us to list the orders of the elements in U15
rbs
order
1
1‹
2
4
4
2‹
4
7
8
2‹
2‹
11
13
4
´1 “ 14 2‹
with those such that b2 “ e marked with ‹. The only possible homomorphism
Φi : Z{2Z Ñ U15 are :
— b1 “ r1s : Φ1 mapping e Ñ r1s, a Ñ r1s (Z{2Z “ te, au) ;
— b2 “ r4s : Φ2 mapping e Ñ r1s, a Ñ r4s (with Φpa2 q “ r1s as expected) ;
— b3 “ r8s : Φ3 mapping e Ñ r1s, a Ñ r8s ;
— b4 “ r11s : Φ4 mapping e Ñ r1s, a Ñ r11s ;
— b5 “ r14s : Φ5 mapping e Ñ r1s, a Ñ r14s “ ´r1s.
Given Φi paq “ rbi s P U15 , the corresponding automorphism in AutpZ{15Z, `q is the
map φia : rks Ñ rbi srks “ rbi ks for all rks P Z{15Z, i “ 1, 2, 3, 4, 5.
Thus the homomorphisms Φi : Z{2Z Ñ AutpZ{15Z, `q are determined by assigning
Φi paq “ φia as follows (We have φie “ IdN in every case).
1. i “ 1, φia “ φr1s “ IdZ{15Z . Then φ1g “ IdZ{15Z . Then φ1g “ IdZ{15Z for all
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g P Z{2Z, the action pg, nq Ñ φ g pnq, n P Z{15Z, is trivial, and N ¸ H is
actually the direct product Z{15Z ˆ Z{2Z » Z{30Z.
2. i “ 2 φ2a “ φr4s , multiplication law (modeled in cartesian product set N ˆ H) is
pn, hqpn1 , h1 q “ pn ` φ2h ph1 q, h ` h1 q (φ2e “ IdN ; φ2a pn1 q “ r4s ¨ n1 , for n1 P Z{15Z)
when the group operations in N “ Z{15Z and H “ Z{2Z are written as p`q.
3. i “ 3 φ3a “ φr8s : rks Ñ r8s ¨ rks and now
pn, hqpn1 , h1 q “ pn ` φ3h pn1 q, h ` h1 q
(φ3e “ IdN , φ3a pn1 q “ r8s ¨ n1 for n1 P Z{15Z.)
4. φ4a “ φr11s : rks Ñ r11s ¨ rks, and now
pn, hqpn1 , h1 q “ pn ` φ4h pn1 q, h ` h1 q
(φ4e “ IdN ; φ4a pn1 q “ r11s ¨ n1 , for n1 P Z{15Z).
5. φ5a “ φr14s : rks Ñ r14s ¨ rks “ ´r1s ¨ rks “ ´rks, for rks P Z{15Z. This is just the
inversions map Jpxq “ x´1 , written in additive notation ; it is an automorphism
of any abelian group, such as Z{15Z. Now
pn, hqpn1 , h1 q “ pn ` φ5h pn1 q, h ` h1 q
pφ5e “ IdN ; φ5a “ J)
The last semi direct product is just the dihedral group D15 . To see this identify N “
Z{15Z » Ñ “ tpn, 0q : n P Z{15Zu and H “ Z{2Z » H̃ “ t0, hq : h P Z{2Zu in G.
Then pn, 0qp0, hq “ pn, hq, p0, hq´1 “ p0, ´hq, pn, 0q´1 “ p´n, 0q, e “ p0, 0q and
p0, hqpn, 0qp0, hq´1 “ pφ5h pnq, hqp0, ´hq “ pφ5h pnq, 0q
Now let p “ pr1s15 , r0sq and satisfy the commutation relation
ρ15 “ e, σ2 “ e, σρσ´1 “ σρσ “ ρ´1 “ ρ14
used to define D15 . In fact, by p2q above,
σρσ´1 “ pφ5σ pr1s15 q, r0s2 q “ pJpr1s15 , r0sq “ p´r1s15 , r0sq “ ρ´1
as required. From this, it is easy to see G » D15 in case 5. The other semi direct products are all crow.
Exercise 3 :
Discuss the number and nature of the 3-Sylow and 5-Sylow subgroups of a group whose
order is |G| “ 225 “ 32 ¨ 52 .
Solution :
The divisor of G are 1, 3, 9, 15 “ 3 ¨ 5, 25 “ 52 , 45 “ 32 ¨ 5, 75 “ 3 ¨ 52 , 225. There
exist p-Sylow subgroups H3 , H5 such that |H3 | “ 9, |H5 | “ 25. By Cauchy’s theorem
both are abelian because they have order p2 for primes p ą 1. The only possibilities are
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H3 » tZ{9Z or Z{3Z ˆ Z{3Zu and H5 » tZ{25 or Z{5Z ˆ Z{5Zu.
Let’s look at H5 . The 7p of 5-Sylow subgroupsq must be of form 1 ` 5k “ 1, 6?11, 16, ¨ ¨ ¨
and it must also be divisor of |G| “ 32 ¨ 52 . The only divisors of |G| in the arithmetic
progression t1 ` 5k : k P Nu is 1. That implies H5 E G since all p-Sylow subgroups are
conjugate in G.
As for 3- Sylow subgroups, 7p3 ´ Sylowq “ 1 ` 3k for some k “ 0, 1, 2, ¨ ¨ ¨ . In this progression we find two divisors of |G|, namely 1 and 25 “ 1 ` 3p8q. Thus 7p3 ´ Sylowq “ 1
or 25.
If 7p3 Sylowq “ 1, then H3 E G, H5 E G, |H3 H5 | “ |G|, implies H3 H5 “ G, H3 X H5 “ teu
(by Lagrange, since |H3 | “ 9 and |H5 | “ 25 have no common divisor and |H3 X H5 | must
divide both |H3 | and |H5 |. Thus G “ H3 ˆ H5 (direct product) in this case. The possibilities in this case are : Z{9Z ˆ Z{25Z, Z{3Z ˆ Z{3Z ˆ Z{25Z, Z{9Z ˆ Z{5Z ˆ Z{5Z,
Z{3Z ˆ Z{3Z ˆ Z{5Z ˆ Z{5Z. (all distinct i.e. no two are isomorphic).
If p73 Sylowq “ 25, H3 is not normal but we still how H3 XH5 “ teu, H5 EG, H3 H5 “ G,
so G is some non commutative semi direct product H5 ¸ H3 . Additional analysis would
be needed to classify the possible actions of H3 on H5 by automorphisms in each of the
cases H3 “ Z{9Z or Z{3Z ˆ Z{3Z and H5 “ Z{25Z or Z{5Z ˆ Z{5Z. We won’t do
that here.
Exercise 4 :
If a group has order |G| “ 231 “ 11 ¨ 7 ¨ 3, prove that the 11-Sylow subgroup must lie
in the center of G.
Solution :
If s “ 7p11 ´ Sylowq, we have s “ 1, 12, 23, ¨ ¨ ¨ and s also divides |G|. The only possibility is s “ 1, so H11 is normal in G. Since |H11 | “ 11 is prime, H11 is abelian and
» pZ{11Z, `q.
An element x P H11 certainly commutes with all elements in H11 . Recall that the various p-Sylow subgroups p “ 3, 7, 11 generate all of G, so if x commutes with @g in
an arbitrary 7-Sylow subgroup, and with @g in an arbitrary 3-Sylow subgroups, then x
commutes with all products of such elements, and hence x commutes with all g P G.
Thus we will show every element of H11 is in ZpGq.
Let N “ H11 and Hp any p-Sylow subgroup, p “ 3 to 7. Since N E G we get a homomorphism Φp : Hp Ñ AutpNq “ AutpZ{11Z, `q » pU11 , ¨q “ tr1s, r2s, ¨, r10su (11 is
prime, so all rks ‰ r0s are units) ; Just take Φp pyqpnq “ yny´1 ie Φp pyq “ α y |N where
α y pgq “ ygy´1 , @g P G.
Note that |U11 | “ 10 and that rangepΦp q, being a subgroup of U11 must have order dividing U11 | “ 10. However |rangepΦp q| “ |Hp |{|KerpΦp q| and |Hp | “ p is a prime, so the
only possibilities for |rangepΦp q| are 1 and p. Since p “ 3 or 7, neither of which divide
10, we must in fact have |rangepΦp q| “ 1, kerpΦp q “ Hp ; the action Hp ˆ N Ñ N is
trivial (which means every y P Hp commutes with every n P N). That proves N Ď ZpGq.
Exercise 5 :
5
If a group has order |G| “ 385 show that its 11-Sylow subgroup is normal in G and its
7-Sylow subgroup lies in the center of G.
Solution :
p1q
p1q
p1q
p5q
Let Hp “ Hp , ¨ ¨ ¨ Hp and Hq “ Hq ¨ ¨ ¨ Hq be the distinct p-Sylow and q-Sylow subgroups are all conjugate to each other ; likewise for the q-Sylow subgroups are all conjupjq
piq
gate to each other ; likewise for the q-Sylow subgroups, we have Hq X Hq “ teu for
piq
piq
pjq
i ‰ j ; but all we can say about the Hp is |Hp X Hp | “ 1 or p if i ‰ j.
p jq
piq
piq
The conjugates Hq overlap only in the single point e, so | Y5j“1 Hq | “ 1 ` 5 ¨ |Hq zteu| “
p jq
spq ´ 1q ` 1, or | Y5j“1 Hq | “ r7pq ´ Sylowqpq ´ 1qs ` 1. (We cannot make this argument
for the Sylow subgroups).
If either Hp E G of Hq E G we are done, so let’s assume neither is normal. If NpHp q=
normalizer of Hp “ tx P G : xHp x´1 “ Hp u then this is the stabilizer of the acp1q
prq
tion of G by conjugation on the set Xp “ tHp , ¨ ¨ ¨ , Hp u of distinct p-Sylow subgroups. Hence, 7pp ´ Sylowq “ |G|{|NpHp q| “ p2 q{|NpHp q|. But Hp certainly normalizes itself, so |NpHp q| ě |Hp | “ p2 . That forces 7pp ´ Sylowq “ p2 q{p2 “ q, for if
NpHp q “ G, Hp would be normal G. Now let’s take a different view of the conjugation
action (transitive action) G ˆ Xp Ñ Xp on the distinct p-distinct p-Sylow subgroups
pqq
p1q
tHp , ¨ ¨ ¨ , Hp u. The ”action kernel” is K “ kerpΦq “ tg P G : φ g “ IdXp u “ tg P G :
piq
con jugation αp leaves invariant each subgroup Hp , 1 ď i ď qu. K is normal subgroup
p1q
in G. It cannot equals G because that would imply Hp “ Hp is normal G ; if K “ teu
we have found a proper normal subgroup in G, and we are done. The remaining possibility is that K “ teu which means Φ : G Ñ PerpXp q is one-to-one. Then ΦpGq is a
subgroup of PerpΦp q and by Lagrange |ΦpGq| “ |G| “ p2 q must divide |PerpXp q| “ q!
(q “ 7pp ´ Sylow subgroupsq ; as we showed in the preceding paragraph) ; we are assuming p ‰ q, so we are reduced to study of the study of the case p ă q. Still assuming
Hp , Hq non-normal, and p ă q, we have s “ 7pq ´ Sylowq “ |G|{|StabG pHq q| where
StabG pHq q “ NG pHq q “ tg P G : gHq g´1 “ Hq u. (This is the stabilizer of the base point
p1q
Hq “ Hq in the action G ˆ Xq Ñ Xq , Xq “ pall q ´ Sylow subgroupsq.
Exercise 6 :
If a group has order p2 q where p ą 1 and q ą 1 are primes, prove that G must contain
a proper normal subgroup. Solution :
Obviously Hq normalizes itself, so Hq Ď StabG pHq q and since |StabG pHq q| must divide
|G| “ p2 q we can only have S “ 7pq ´ Sylowq “ |Xq | equal to p or p2 . We shall consider
now the transitive action G ˆ Xq Ñ Xq .
Case 1 : p ă q and S “ |Xq | “ p. Then |G| “ p2 q cannot divide the factorial p! where
p “ |Xq |. Thus |G| cannot divide |PerpXq q| “ p! and the action cannot be one-to-one.
Thus K “ kerpΦq must be ‰ teu ; KerpΦq “ tg P G : gHq g´1 “ Hq u cannot equal G
either because then Hq would be normal in G. Hence kerpΦq is a proper, normal subgroup
in this case.
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Case 2 : p ă q and s “ |Xq | “ p2 . We now examine the sets
p jq
Sp “ Yri“1 Hp , Sq “ Ysj“1 Hq , r “ 7pp ´ Sylowq “ q; s “ 7pq ´ Sylowq “ p2
piq
piq
pjq
we must have Sp XSq “ teu. (In fact, if x ‰ e is in Sp XSq we would have Hp XHq ‰ teu
which is impossible since |Hp | “ p2 and |Hq | “ q.)
Because of their essential disjointness we get |Sp Y Sq | “ |Sp | ` |Sq | ´ 1 (left side counts
e twice).
But in our present situation we have |Sp | ą |Hp | “ p2 (because Hp is not normal G and
7pp ´ Sylowq ą 1) ; and as we computed earlier we also have
|Sq | “ 7pq ´ Sylowqpq ´ 1q ` 1
“ p2 pq ´ 1q ` 1 under our Case 2 hypotheses)
“ p2 q ´ p2 ` 1
|Sp Y Sq | ą p2 ` pp2 q ´ p2 ` 1q ´ 1
|Sp Y Sq | ą p2 q “ |G|
This impossible, so Case 2 cannot occur. Thus in every possible case we find a proper,
normal subgroup in G.
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