Download Solutions. - University of Bristol

Introduction to Group Theory
Solutions 3
Unit website: http://www.maths.bris.ac.uk/~majcr/IGT.html
Questions 1,2,3 to be handed in for tutorials.
1. Let G = (Z, +) be the group of integers under addition. Let H = 2Z
be the subgroup consisting of all the even integers, and K = 3Z the
subgroup consisting of all the integers divisible by 3.
Prove that H ∪ K is not a subgroup of G.
Solution: 2 ∈ H and 3 ∈ K, so 2, 3 ∈ H ∪K, but 2+3 = 5 6∈ H ∪K.
So H ∪ K is not closed under addition and so is not a subgroup. [The
other conditions needed for a subgroup are satisfied: it contains the
identity element 0 and is closed under taking inverses.]
2. Let G be a multiplicatively-written group, and let H be a non-empty
subset of G such that, for all x, y ∈ H, x−1 y is also an element of H.
Prove that H is a subgroup of G.
Solution: We need to check H is closed under multiplication, contains the identity element and is closed under taking inverses.
Since H is non-empty, there is some h ∈ H. Taking x = h and y = h,
we have that h−1 h = e ∈ H. So H contains the identity.
Let x ∈ H. Taking y = e (which we now know is an element of H),
we have x−1 e = x−1 ∈ H. So H is closed under taking inverses.
Let g, h ∈ H. We now know that also g −1 ∈ H, so taking x = g
and y = h we have that (g −1 )−1 h = gh ∈ H. So H is closed under
multiplication.
3.
(i) Let G be the dihedral group D6 of order 6. Find two different
subgroups of G with order 2.
(ii) Let G be the dihedral group D12 of order 12. Find two different
subgroups of G with order 6.
Solution:
(i) Let b and b0 be two different reflections (e.g., with our standard
notation for elements of dihedral groups, take b and ab). Then
{e, b} and {e, b0 } are both subgroups. [They’re closed, since
b2 = e when b is a reflection. They certainly contain the identity,
and they are closed under taking inverses since e−1 = e and
b−1 = b. In fact, since there are three different reflections, there
are three subgroups of order 2.]
(ii) {e, a, a2 , a3 , a4 , a5 }, the group of rotational symmetries, is a subgroup of D12 of order 6. Another is {e, a2 , a4 , b, a2 b, a4 b}. [Imagine an equilateral triangle inscribed in a regular hexagon, with
its vertices at every other vertex of the hexagon. This is the
subgroup consisting of those symmetries of the hexagon that
are also symmetries of the triangle.]
4. (This generalizes question 1.) Let H and K be subgroups of a group
G. Prove that H ∪ K is a subgroup of G if and only if either H ≤ K
or K ≤ H.
Solution: If H ≤ K, then H ∪ K = K is a subgroup. Similarly,
if K ≤ H, then H ∪ K = H is a subgroup. If neither H nor K
is a subgroup of the other, then there is some element h ∈ H \ K
and some k ∈ K \ H. Consider hk. This can not be in H, or else
k = h−1 (hk) would be in H, since H is a subgroup and therefore closed
under taking inverses and multiplication. Similarly it can not be in K,
or else h = (hk)k −1 would be in K, since K is a subgroup.
5. Let H be a subgroup of a group G, and x ∈ G. Define
xHx−1 = xhx−1 : h ∈ H .
Prove that xHx−1 is also a subgroup of G, and that (if H is finite)
|H| = |xHx−1 |.
Solution: Let xhx−1 and xh0 x−1 be two elements of xHx−1 , where
h, h0 ∈ H. Then
(xhx−1 )(xh0 x−1 ) = xheh0 x−1 = x(hh0 )x−1 ,
which is in xHx−1 since hh0 ∈ H (H is a subgroup, and therefore is
closed under multiplication).
e ∈ H, since H is a subgroup. So xex−1 ∈ xHx−1 . But xex−1 =
xx−1 = e. So xHx−1 contains the identity element.
Let xhx−1 ∈ xHx−1 , where h ∈ H. Then
(xhx−1 )−1 = (x−1 )−1 h−1 x−1 = xh−1 x−1 ,
which is in xHx−1 since H is closed under taking inverses and so h−1 ∈
H. So xHx−1 is closed under taking inverses.
This shows that H is a subgroup. Alternatively, we could use question 2.
To show that |H| = |xHx−1 | we can find a bijection between the
two subgroups. Define α : H → xHx−1 by α(h) = xhx−1 , and β :
xHx−1 → H by β(k) = x−1 kx. Note that if k = xhx−1 then β(k) = h,
so β is indeed a function from xHx−1 and β (α(h)) = h for h ∈ H.
Also α (β(k)) = k for k ∈ xHx−1 , so α and β are inverse functions and
therefore bijections.
c
University
of Bristol 2016. This material is copyright of the University unless explicitly stated otherwise. It is provided exclusively
for educational purposes at the University and is to be downloaded or
copied for your private study only.