School of Mathematics and Statistics MT5824 Topics in Groups ...

CMRD 2010
School of Mathematics and Statistics
MT5824 Topics in Groups
Problem Sheet I: Revision and Re-Activation (Solutions)
1. Let H and K be subgroups of a group G. Define
HK = { hk | h ∈ H, k ∈ K }.
(a) Show that HK is a subgroup of G if and only if HK = KH.
(b) Show that if K is a normal subgroup of G, thenHK is a subgroup of G.
(c) Give an example of a group G and two subgroups H and K such that HK is not a
subgroup of G.
(d) Give an example of a group G and two subgroups H and K such that HK is a
subgroup of G but neither H nor K are normal subgroups of G.
Solution: (a) Suppose HK is a subgroup of G. Let h ∈ H and k ∈ K. Then
h = h1 ∈ HK and k =1k ∈ HK and since HK is closed under products, we
deduce kh ∈ HK. ThuswehaveKH ⊆ HK.
Also for the same elements h ∈ H and k ∈ K, we have (hk) −1 ∈ HK, so
(hk) −1 = xy where x ∈ H and y ∈ K. Then hk =(xy) −1 = y −1 x −1 ∈ KH (as
x −1 ∈ H and y −1 ∈ K). This shows that HK ⊆ KH.
Hence if HK is a subgroup of G, thenHK = KH.
Conversely suppose HK = KH. Let a, b ∈ HK; saya = h 1 k 1 and b = h 2 k 2
where h 1 ,h 2 ∈ H and k 1 ,k 2 ∈ K. Then k 1 h 2 ∈ KH = HK; sayk 1 h 2 = hk
where h ∈ H and k ∈ K. Now
since h 1 h ∈ H and kk 2 ∈ K. Also
ab = h 1 k 1 h 2 k 2 = h 1 hkk 2 ∈ HK
a −1 =(h 1 k 1 ) −1 = k −1
1 h−1 1 ∈ KH = HK.
Hence HK is closed under products and inverses, so it is a subgroup of G.
(b) Suppose K G. Let a, b ∈ HK;saya = h 1 k 1 and b = h 2 k 2 where h 1 ,h 2 ∈ H
and k 1 ,k 2 ∈ K. Then
ab = h 1 k 1 h 2 k 2 = h 1 h 2 (h −1
2 k 1h 2 )k 2 .
Now h 1 h 2 ∈ H. Furthermore h −1
2 k 1h 2 ∈ K as K G, so(h −1
2 k 1h 2 )k 2 ∈ K.
Hence
ab ∈ HK.
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Also
a −1 = k −1
1 h−1 1 = h −1
1 (h 1k −1
1 h−1 1 )
and here h 1 k1 −1 h−1 1 =(k1 −1 )h−1 1 is the conjugate of the element k1 −1 by the element
h −1
1 so belongs to K. Thus a −1 ∈ HK and we have shown that HK is
closed under products and inverses, so is a subgroup of G.
(c) Take G = S 3 , H = (1 2) and K = (2 3). Then H and K are subgroups
of G (each containing two elements) and
HK = {1, (1 2), (2 3), (1 3 2)}
a set of size 4. Therefore HK is not a subgroup of G (by Lagrange’s Theorem)
since 4 does not divide 6.
(d) Let G = D 8 generated by the permutation α of order 4 and β of order 2. Let
H = α 2 β and K = β. ThenH and K are subgroups of G of order 2 and
HK = {1,β,α 2 β,α 2 }
and since we can calculate that α 2 commutes with β it is easy to check that
HK is closed under products and inverses. On the other hand neither H nor K
are normal in G:
α −1 (α 2 β)α = αβα = β ∈ H
and
α −1 βα = α 2 β ∈ K.
2. Let M and N be normal subgroups of G. Show that M ∩N and MN are normal subgroups
of G.
Solution: The intersection of two subgroups is again a subgroup, so M ∩ N is
a subgroup of G. Let x ∈ M ∩ N and g ∈ G. Theng −1 xg ∈ M (as x ∈ M G)
and similarly g −1 xg ∈ N. Hence g −1 xg ∈ M ∩ N and so M ∩ N G.
Since N G, part (b) of Question 1 shows that MN is a subgroup of G. Let
x ∈ MN,sayx = mn where m ∈ M and n ∈ N. Ifg ∈ G, then
g −1 xg = g −1 mg · g −1 ng ∈ MN
as g −1 mg ∈ M and g −1 ng ∈ N. Hence MN G.
3. Let G be a group and H be a subgroup of G.
(a) If x and y are elements of G, showthatHx = Hy if and only if x ∈ Hy.
(b) Suppose T is a subset of G containing precisely one element from each (right) coset
of H in G (such a set T is called a (right) transversal to H in G and has the property
that |T | = |G : H|). Deduce that { Ht | t ∈ T } is the set of all (right) cosets of H
in G with distinct elements of T defining distinct cosets.
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Solution: (a) If Hx = Hy, thenx =1x ∈ Hx = Hy. Conversely suppose
x ∈ Hy. Then x = hy for some h ∈ H and so xy −1 = h ∈ H and we deduce
Hx = Hy.
(b) If Hx is any coset of H in G, then by assumption there is some element t ∈ T
with t ∈ Hx and by part (a) we have Hx = Ht. Thus{ Ht | t ∈ T } is the set of
all cosets of H in G. Suppose Ht = Hu for some t, u ∈ T . Then both t and u
belong to Ht (since t =1t ∈ Ht and u =1u ∈ Hu = Ht) and our assumption
on T forces t = u. Thus a transversal to H in G parametrises the cosets of H in
a one-one manner.
4. Let G be a (not necessarily finite) group with two subgroups H and K such that K H
G. The purpose of this question is to establish the index formula
|G : K| = |G : H|·|H : K|.
Let T be a transversal to K in H and U be a transversal to H in G.
(a) By considering the coset Hg or otherwise, show that if g is an element of G, then
Kg = Ktu for some t ∈ T and some u ∈ U.
(b) If t, t ∈ T and u, u ∈ U with Ktu = Kt u ,firstshowthatHu = Hu and deduce
u = u ,andthenshowthatt = t .
(c) Deduce that TU = { tu | t ∈ T, u ∈ U } is a transversal to K in G and that
|G : K| = |G : H|·|H : K|.
(d) Show that this formula follows immediately from Lagrange’s Theorem if G is a finite
group.
Solution: (a) Let g ∈ G. Then Hg is a coset of H in G, soHg = Hu for
some u ∈ U (since U is a transversal). Then g ∈ Hu, sog = hu for some
h ∈ H. Now consider the coset Kh. As T is a transversal to K in H, wehave
Kh = Kt for some t ∈ T . Therefore h ∈ Kt so h = kt for some k ∈ K. Then
g = hu = ktu ∈ Ktu and we deduce Kg = Ktu.
(b) Suppose Ktu = Kt u . Then tu(t u ) −1 ∈ K, that is tu(u ) −1 (t ) −1 = k for
some k ∈ K. Since t, t ,k ∈ H, weseeu(u ) −1 = t −1 kt ∈ H and we deduce
Hu = Hu . As U is a transversal to H in G, we must therefore have u = u .
Therefore t(t ) −1 = k and we have Kt = Kt . The fact that T is a transversal
to K in H forces t = t .
(c) Part (a) tells us that each coset of K in G contains an element tu for some
t ∈ T and some u ∈ U. Furthermore part (b) tells us that there is precisely one
choice of t and one choice of u achieving this. Thus each coset of K in G contains
exactly one element from TU; that is, TU is a transversal to K in G. Note that
the number of cosets of K in G then equals |T |·|U|; that is,
|G : K| = |U|·|T | = |G : H|·|H : K|.
(d) If G is finite, Lagrange’s Theorem tells us |G : K| = |G|/|K|, and similarly
for the other indices. Thus
|G : H|·|H : K| =(|G|/|H|) · (|H|/|K|) =|G|/|K| = |G : K|.
5. Let G be a group and H be a subgroup of G.
(a) Show that H is a normal subgroup of G if and only if Hx = xH for all x ∈ G.
(b) Show that if |G : H| =2,thenH is a normal subgroup of G.
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Solution: (a) Suppose H G. Let h ∈ H. Then x −1 hx ∈ H, sayx −1 hx =
k ∈ H. Thenhx = xk ∈ xH and we deduce Hx ⊆ xH.
Equally xhx −1 = h x−1 ∈ H, sayxhx −1 = l ∈ H. Then xh = lx ∈ Hx and we
deduce xH ⊆ Hx.
Hence if H G, thenHx = xH for all x ∈ G.
Conversely suppose Hx = xH for all x ∈ G. Let h ∈ H and x ∈ G. Then
hx ∈ Hx = xH, sohx = xk where k ∈ H. Thenx −1 hx = k ∈ H and we deduce
H G.
(b) Suppose |G : H| = 2. Then G has two right cosets which are necessarily H =
H1 and Hg where g ∈ H. Note then that Hg consists of all the elements not
in H:
Hg = G \ H.
Now consider any left coset xH. Ifx ∈ H, thenxH = H = Hx. Ifx ∈ H, then
xH must be another left coset of H and this must equal the remaining elements
of G (since the left cosets also partition G into the same number of left cosets as
there are right cosets):
Hence, by (a), H G.
xH = G \ H = Hg = Hx.
6. Give an example of a finite group G and a divisor m of |G| such that G has no subgroup
of order m.
Solution: Let G = A 4 . Then 6 divides |A 4 | = 12. Suppose that H is a
subgroup of G of order 6. Then |G : H| =2soH G.
Consider the quotient group G/H. This has order 2, so (Hx) 2 = H1 for all Hx
in G/H. Therefore Hx 2 = H for all x ∈ G; that is, x 2 ∈ H.
Hence H contains all squares of elements in A 4 . Since (αβγ) 2 =(αγβ) for all
choices of α, β and γ, we deduce that H contains all 3-cycles. But there are
(4 × 3 × 2)/3 = 8 such 3-cycles in A 4 and this contradicts |H| = 6.
Thus A 4 has no subgroup of order 6.
7. Let G = x be a cyclic group.
(a) If H is a non-identity subgroup of G, showthatH contains an element of the form x k
with k>0.
Choose k to be the smallest positive integer such that x k ∈ H. Show that every
element in H has the form x kq for some q ∈ Z and hence that H = x k . [Hint: Use
the Division Algorithm.]
Deduce that every subgroup of a cyclic group is also cyclic.
(b) Suppose now that G is cyclic of order n. Let H be the subgroup considered in
part (a), so that H = x k where k is the smallest positive integer such that x k ∈ H,
and suppose that |H| = m.
Show that k divides n. [Hint: Why does x n ∈ H?]
Show that |x k | = n/k and deduce that m = n/k.
Conclude that, if G is a cyclic group of finite order n, thenG has a unique subgroup
of order m for each positive divisor m of n.
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(c) Suppose now that G is cyclic of infinite order. Let H be the subgroup considered in
part (a), so that H = x k where k is the smallest positive integer such that x k ∈ H.
Show that {1,x,x 2 ,...,x k−1 } is a transversal to H in G. Deduce that |G : H| = k.
[Hint: Use the Division Algorithm to show that if n ∈ Z, then x n ∈ Hx r where
0 r0.
Now choose k to be the smallest positive integer satisfying x k ∈ H. Let h be any
element in H. Then,sincex generates G, wehaveh = x n for some n ∈ Z. Use
the Division Algorithm to write n = kq+r where 0 k

every coset of H has the form Hx r where r ∈{0, 1,...,k − 1}. Note that if
Hx i = Hx j where 0 i j k − 1, then x j−i ∈ H. Then 0 j − i

Hence x has order 2. (It is not the identity since x ∈ A.)
10. The quaternion group Q 8 of order 8 consists of eight elements
1, −1,i,−i, j, −j, k, −k
with multiplication given by
i 2 = j 2 = k 2 = −1, ij = −ji = k, jk = −kj = i, ki = −ik = j.
(a) Show that Q 8 is generated by i and j.
(b) Show that i is a normal subgroup of Q 8 of index 2.
(c) Show that every element of Q 8 can be written as i m j n where m ∈{0, 1, 2, 3} and
n ∈{0, 1}.
(d) Show that every element in Q 8 which does not lie in i has order 4.
(e) Show that Q 8 has a unique element of order 2.
Solution: (a) Consider the subgroup i, j generated by i and j. It contains
−1 =i 2 and therefore it contains −i = −1 · i and similarly −j. Finally
k = ij lies in this subgroup and also so does −k = −1 · k. Hence i, j = Q 8 .
(b) i 2 = −1 and i 3 = −1 · i = −i. Then i 4 =(−1) 2 = 1. Hence i is an element
of order 4, so |i| = 4. Therefore |Q 8 : i| = 2 and we deduce i Q 8 .
(c) Let I = i. Sincej ∈ I, the two cosets of I are I and Ij. Hence Q 8 = I ∪ Ij
and an element of Q 8 either has the form i m if it lies in I or the form i m j if it
lies in Ij (and here 0 m 3).
(d) Let x ∈ Q 8 \ I. Thenx = i m j (where 0 m 3). The
x 2 = i m ji m j = i m ji m j −1 j 2 = i m (jij −1 ) m (−1).
Now jij −1 =(−k)(−j) =kj = −i. Hence
x 2 = i m (−i) m (−1) = (i · (−i)) m (−1) = 1 m · (−1) = −1.
Then x 4 =(x 2 ) 2 =(−1) 2 = 1. Hence o(x) divides 4, but is not 1 or 2, so o(x) =4
for all x ∈ I.
(e) There is no element of order 2 outside I, whileI contains the identity and
two elements of order 4 (namely i and −i). Hence Q 8 has a unique element of
order 2 namely the element −1.
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