Download 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, then HK 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. Thus we have KH ⊆ 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, then HK = KH.
Conversely suppose HK = KH. Let a, b ∈ HK; say a = h1 k1 and b = h2 k2
where h1 , h2 ∈ H and k1 , k2 ∈ K. Then k1 h2 ∈ KH = HK; say k1 h2 = hk
where h ∈ H and k ∈ K. Now
ab = h1 k1 h2 k2 = h1 hkk2 ∈ HK
since h1 h ∈ H and kk2 ∈ K. Also
a−1 = (h1 k1 )−1 = k1−1 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; say a = h1 k1 and b = h2 k2 where h1 , h2 ∈ H
and k1 , k2 ∈ K. Then
ab = h1 k1 h2 k2 = h1 h2 (h−1
2 k1 h2 )k2 .
−1
Now h1 h2 ∈ H. Furthermore h−1
2 k1 h2 ∈ K as K � G, so (h2 k1 h2 )k2 ∈ K.
Hence
ab ∈ HK.
1
Also
−1
−1 −1
a−1 = k1−1 h−1
1 = h1 (h1 k1 h1 )
−1
−1 h1
is the conjugate of the element k1−1 by the eleand here h1 k1−1 h−1
1 = (k1 )
−1 ∈ HK and we have shown that HK is
ment h−1
1 so belongs to K. Thus a
closed under products and inverses, so is a subgroup of G.
(c) Take G = S3 , 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 = D8 generated by the permutation α of order 4 and β of order 2. Let
H = �α2 β� and K = �β�. Then H and K are subgroups of G of order 2 and
HK = {1, β, α2 β, α2 }
and since we can calculate that α2 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 (α2 β)α = αβα = β �∈ H
and
α−1 βα = α2 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. Then g −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 M N is a subgroup of G. Let
x ∈ M N , say x = mn where m ∈ M and n ∈ N . If g ∈ G, then
g −1 xg = g −1 mg · g −1 ng ∈ M N
3.
as g −1 mg ∈ M and g −1 ng ∈ N . Hence M N � G.
Let G be a group and H be a subgroup of G.
(a) If x and y are elements of G, show that Hx = 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, then x = 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� , first show that Hu = Hu� and deduce
u = u� , and then show that t = t� .
(c) Deduce that T U = { 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, so Hg = Hu for
some u ∈ U (since U is a transversal). Then g ∈ Hu, so g = hu for some
h ∈ H. Now consider the coset Kh. As T is a transversal to K in H, we have
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, we see u(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 T U ; that is, T U 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, then H is a normal subgroup of G.
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Solution: (a) Suppose H � G. Let h ∈ H. Then x−1 hx ∈ H, say x−1 hx =
k ∈ H. Then hx = xk ∈ xH and we deduce Hx ⊆ xH.
Equally xhx−1 = hx
deduce xH ⊆ Hx.
−1
∈ H, say xhx−1 = l ∈ H. Then xh = lx ∈ Hx and we
Hence if H � G, then Hx = xH for all x ∈ G.
Conversely suppose Hx = xH for all x ∈ G. Let h ∈ H and x ∈ G. Then
hx ∈ Hx = xH, so hx = xk where k ∈ H. Then x−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. If x ∈ H, then xH = H = Hx. If x �∈ 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):
xH = G \ H = Hg = Hx.
6.
Hence, by (a), H � G.
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 = A4 . Then 6 divides |A4 | = 12. Suppose that H is a
subgroup of G of order 6. Then |G : H| = 2 so H � G.
Consider the quotient group G/H. This has order 2, so (Hx)2 = H1 for all Hx
in G/H. Therefore Hx2 = H for all x ∈ G; that is, x2 ∈ H.
Hence H contains all squares of elements in A4 . Since (α β γ)2 = (α γ β) 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 A4 and this contradicts |H| = 6.
Thus A4 has no subgroup of order 6.
7.
Let G = �x� be a cyclic group.
(a) If H is a non-identity subgroup of G, show that H contains an element of the form xk
with k > 0.
Choose k to be the smallest positive integer such that xk ∈ H. Show that every
element in H has the form xkq for some q ∈ Z and hence that H = �xk �. [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 = �xk � where k is the smallest positive integer such that xk ∈ H,
and suppose that |H| = m.
Show that k divides n. [Hint: Why does xn ∈ H?]
Show that |xk | = n/k and deduce that m = n/k.
Conclude that, if G is a cyclic group of finite order n, then G 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 = �xk � where k is the smallest positive integer such that xk ∈ H.
Show that {1, x, x2 , . . . , xk−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 xn ∈ Hxr where
0 � r < k.]
Conclude that, if G is a cyclic group of infinite order, then G has a unique subgroup
of index k for each positive integer k and that every non-trivial subgroup of G is
equal to one of these subgroups.
Solution: (a) Since H �= 1, it must contain an element of the form xl with
l �= 0. It is also contains all inverses, so x−l = (xl )−1 ∈ H. One of l or −l is
positive, so we deduce xk ∈ H for some k > 0.
Now choose k to be the smallest positive integer satisfying xk ∈ H. Let h be any
element in H. Then, since x generates G, we have h = xn for some n ∈ Z. Use
the Division Algorithm to write n = kq +r where 0 � k < r. Then xn , xk ∈ H, so
xr = xn−kq = (xn )(xk )−q ∈ H. The minimal choice of k now forces r = 0. Hence
n = kq and we have h = xn = xkq = (xk )q . This shows that an arbitrary element
in H necessarily lies in �xk �. On the other hand, since xk ∈ H we certainly have
�xk � � H. Hence
H = �xk �
for k the smallest positive integer with xk ∈ H.
The above shows that every non-trivial subgroup of G is cyclic (and also specifies
a generator). The remaining case is that 1 = �1�, so we conclude every subgroup
of a cyclic group is cyclic.
(b) Suppose now that G is cyclic of order n, so |x| = n. Let H be a non-identity
subgroup of G, so H = �xk � where k is the smallest positive integer such that
xk ∈ H. Define m = |H|, so m = |xk |.
Attempt to divide n by k, so n = kq + r where 0 � r < k, by the Division
Algorithm. Now xn = 1 ∈ H, so
xr = xn−kq = (xn )(xk )−q ∈ H.
By the minimality of k, we deduce r = 0. Thus k divides n.
If we now calculate successive powers of xk :
xk , x2k , . . . ,
then the first time we obtain a multiple of n in the exponent is the element
(xk )n/k = xn = 1 (since k | n). Thus |xk | = n/k and we deduce m = |xk | = n/k.
The above shows that if H is a (non-trivial) subgroup of G of order m, then
H = �xn/m � (since k = n/m in the above). On the other hand, if m divides n,
then xn/m is an element of order m (by the same argument as used above) so
�xn/m � is a subgroup of order m. Hence G has a unique subgroup of order m for
each positive divisor m of n, namely the subgroup �xn/m �.
(c) Now suppose G = �x� is of infinite order. Let H = �xk � as in (a). If g ∈ G,
then g = xn for some n. Use the Division Algorithm to write n = kq + r where
0 � r < k. Then g(xr )−1 = xn−r = xkq ∈ H = �xk �. Hence Hg = Hxr , so
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every coset of H has the form Hxr where r ∈ {0, 1, . . . , k − 1}. Note that if
Hxi = Hxj where 0 � i � j � k − 1, then xj−i ∈ H. Then 0 � j − i < k
and the minimality of k forces i = j. Hence H has precisely k cosets, namely
H, Hx, . . . , Hxk−1 , and we see that {1, x, . . . , xk−1 } is a transversal to H in G.
Therefore |G : H| = k.
If H is a subgroup of G of index k, then H �= 1, so H = �x� � for some smallest
integer � by part (a) and from the previous paragraph we have � = |G : H| = k.
Hence a subgroup of index k has the form �xk �, while conversely �xk � always
does have index k. Thus G has a unique subgroup of index k for each k, namely
the subgroup �xk �. Furthermore, part (a) shows that every non-trivial subgroup
of G equals one of these subgroups.
8.
Let V4 denote the Klein 4-group: that is V4 = {1, a, b, c} where a = (1 2)(3 4), b = (1 3)(2 4)
and c = (1 4)(2 3) (permutations of four points). Find three distinct subgroups H1 , H2
and H3 of V4 of order 2. Show that Hi ∩ Hj = 1 for all i �= j and V4 = Hi Hj for all
i and j.
[Note that I am using the more conventional notation V4 for the Klein 4-group, rather
than the less frequently used K4 from MT4003. Here V stands for Viergruppe.]
Solution: Let H1 = �a�, H2 = �b� and H3 = �c�. Since a, b and c are
permutations of order 2, we have |Hi | = 2 for all i. Since these subgroups are
distinct, clearly Hi ∩ Hj is a proper subgroup of Hi for all i and j, so Hi ∩ Hj = 1
by Lagrange’s Theorem.
Since V4 is an abelian group, Hi Hj is a subgroup of V4 (by Question 1(a)) while
it contains at least three elements (namely those in Hi ∪ Hj ). Hence |Hi Hj | = 4
by Lagrange’s Theorem and we deduce V4 = Hi Hj .
[Alternatively this last step can be done by simply calculating the elements in
the product Hi Hj directly.]
9.
The dihedral group D2n of order 2n is generated by the two permutations
α = (1 2 3 . . . n),
β = (2 n)(3 n−1) · · · .
(a) Show that α generates a normal subgroup of D2n of index 2.
(b) Show that every element of D2n can be written in the form αi βj where i ∈ {0, 1, . . . , n−
1} and j ∈ {0, 1}.
(c) Show that every element in D2n which does not lie in �α� has order 2.
Solution: (a) Since α is an n-cycle, |α| = n and therefore |�α�| = n. Hence
|D2n : �α�| = 2 so that �α� � D2n .
(b) Let A = �α�. Since β �∈ A, we see that the two cosets of A are A and Aβ.
Hence D2n = A ∪ Aβ, so an element of D2n either has the form αi (0 � i � n − 1)
when it lies in A or the form αi β when it lies in Aβ.
(c) Let x ∈ D2n \ A. Then x = αi β. Now
x2 = αi βαi β = αi (βαβ)i
(here we are using the fact that β 2 = 1 so (βαβ)i = βαβ 2 αβ 2 . . . αβ = βαi β).
Since βα = α−1 β we now have
x2 = αi (α−1 β 2 )i = αi α−i = 1.
6
10.
Hence x has order 2. (It is not the identity since x �∈ A.)
The quaternion group Q8 of order 8 consists of eight elements
1, −1, i, −i, j, −j, k, −k
with multiplication given by
i2 = j 2 = k2 = −1,
ij = −ji = k,
jk = −kj = i,
ki = −ik = j.
(a) Show that Q8 is generated by i and j.
(b) Show that �i� is a normal subgroup of Q8 of index 2.
(c) Show that every element of Q8 can be written as im j n where m ∈ {0, 1, 2, 3} and
n ∈ {0, 1}.
(d) Show that every element in Q8 which does not lie in �i� has order 4.
(e) Show that Q8 has a unique element of order 2.
Solution: (a) Consider the subgroup �i, j� generated by i and j. It contains −1 = i2 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� = Q8 .
(b) i2 = −1 and i3 = −1 · i = −i. Then i4 = (−1)2 = 1. Hence i is an element
of order 4, so |�i�| = 4. Therefore |Q8 : �i�| = 2 and we deduce �i� � Q8 .
(c) Let I = �i�. Since j �∈ I, the two cosets of I are I and Ij. Hence Q8 = I ∪ Ij
and an element of Q8 either has the form im if it lies in I or the form im j if it
lies in Ij (and here 0 � m � 3).
(d) Let x ∈ Q8 \ I. Then x = im j (where 0 � m � 3). The
x2 = im jim j = im jim j −1 j 2 = im (jij −1 )m (−1).
Now jij −1 = (−k)(−j) = kj = −i. Hence
x2 = im (−i)m (−1) = (i · (−i))m (−1) = 1m · (−1) = −1.
Then x4 = (x2 )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, while I contains the identity and
two elements of order 4 (namely i and −i). Hence Q8 has a unique element of
order 2 namely the element −1.
7