Download ( )3 = ab3a 1 . ( )3 = ab a 1a ( )= 54 ( ) 23

Math 546
Problem Set 15
1. Let G be a finite group.
(a). Suppose that H is a subgroup of G and o(H) = 4. Suppose that K is a
subgroup of G and o(K) = 5. What is H ! K (and why)?
Solution: H ! K = {e} since H ! K is a subgroup of each of H and K and so
its order must divide each of 4 and 5. So, H ! K = 1 .
(b). Suppose that H is a subgroup of G and o(H) = 12. Suppose that K is a
subgroup of G and o(K) = 18. Also H ! K " 4 .
What is the value of H ! K (and why)?
Solution: Similar to part (a), H ! K = 6 since that is the only number that
satisfies the given conditions. (i.e., 6 ≥ 4 and 6 divides both 12 and 18.
(
2. Suppose that a and b are elements of G show that aba !1
(
Solution: aba !1
)
3
)
3
= ab 3a !1 .
= ab a !1a b a !1a ba !1 = abbba !1 = ab 3a !1 .
=!e
=!e
3. (a). What is the order of U(5000)?
Solution: U(5000) = ! (5000) = ! 5 4 2 3 = ! 5 4 ! 2 3 = 4 " 5 3 " 4 = 2000 .
(
) ( ) ( )
(b). Is U(15) cyclic?
(c). Is Z 8 cyclic?
Solution: Yes. Every Z n is cyclic – and is generated by any k in Z n that is
relatively prime to n.
(d). Is Z 4 isomorphic to the Klein 4-group?
Solution: No. Z 4 is cyclic but the Klein 4-group is not cyclic.
4. What is the remainder when 2100 is divided by 19?
Solution: By Fermat’s Theorem, 218 ! 1mod19 . Hence, 2 90 ! 1mod19 .
Thus, 2100 ! 210 mod19 . But simple division show that 210 ! 17 mod19 .
Hence the remainder is 17.
5. (a). How many generators are there for Z 30 ?
Solution: There are ! (30) = 8 generators.
(b). How many generators are there for (Z,!+) ?
Solution: Two – the numbers 1 and –1 generate Z.
6. Try various values of n in the Group Therapy program and make some
conjectures as to when U(n) is cyclic.
7. (a). Show that GL(2, R) is not an Abelian group.
" 0 !1%
(b). What is the order of the element $
' (GL(2, R) .
#1 0 &
Solution: The order is 4.
)!a b $
,
(c). Let H = * #
:
ad
'
bc
=
1,!a,!b,!c,!d
(R
- . Show that H is a subgroup
&
+" c d %
.
of GL(2, R) . H is called the Special Linear Group SL(2, R) .
Hint: for any two n ! n matrices A and B, det(AB) = det(A)det(B) .
8. Suppose that H is a subgroup of the group G and the order of G is 30.
You check all the non-identity elements of G and find that there are three nonidentity elements a, b, and c that you know are in H and three d, f, and g that you
are not certain of. None of the other non-identity elements belong to H. How
many of d, f, and g belong to H? Explain.
Solution: By LaGrange’s Theorem, either one or two of d, f, and g belong to H.
9. Let G be a group and a an element of G. The Centralizer of a is the set
C(a) = { x !G : ax = xa} .
(a). Show that C(a) is a subgroup of G.
(b). What is ! C(a) ?
a"G
Solution: ! C(a) = Z(G) , the center of G.
a"G
!1 1 $
(c). What is the Centralizer of #
& 'GL(2, R) ?
"1 0 %
b %
)"a
,
Solution: * $
: a,!b (R - Note this is a slightly different description
'
+#b a ! b &
.
than the one we did in class.
(d). Suppose that G is a group of order 20. Could the Centralizer of G contain
exactly three elements? Explain.
Solution: No, C(a) is a subgroup of G and so its order must divide 20.
(e). Suppose that a is an element of G with o(a) = 5 , and x !C(a 3 ) .
Show that x !C(a 2 )
Solution: Let x !C(a 3 ) . Then
x = xa 5 = xa 3a 2 = a 3 xa 2 ! a 2 x = a 2 a 3 xa 2 = a 5 xa 2 = xa 2 . There are many
other ways to argue this.
10. Is (Q,!+ ) a cyclic group?
Solution: No.
(!a a $
+
11. Show that K = ) #
: a 'R* , is a group under matrix multiplication.
&
*"a a %
You need not verify that matrix multiplication is associative.
[But it is not a subgroup of GL(2, R) .]
What is the identity of K?
Solution: Under matrix multiplication, any two matrices of the form in K,
have a product of the form required to belong to K. More precisely,
! a a $ !b b $ ! 2ab 2ab $
# a a & #b b & = # 2ab 2ab & for any real numbers a and b. Thus K is closed
"
%"
% "
%
under matrix multiplication. We already know that matrix multiplication is
associative, so there is no need to check that.
! 12 12 $
It is simple to check that E = # 1 1 & is an identity element for K. And for
" 2 2%
!a a $
! 14a 14a$
any A = #
,
the
matrix
serves as an inverse for A.
'K
A'
=
&
#1
1 &
"a a %
" 4a 4a%
12. Suppose that the order of the element g in the group G is 12. What is the order
of x = g 3 ? Verify your answer.
Solution: g k = e ! k " 12 , we know that none of x,!x 2 ,!x 3 can be e.
However, x 4 = g12 = e and so x has order 4.
13. Suppose that a and b are elements of a group G, and o(a) = 12,!o(b) = 22 .
Suppose that H =!< a > ! < b >!" {e} .
(a). What is the order of H?
Solution: The order of H is 2.
(b). Show that a 6 = b11 .
Hint: A non-trivial cyclic group has exactly one element of order 2.
Solution: Let x be the non-identity element of H. Then x has order 2.
However, the only element of order 2 in < a > is a 6 and the only element of
order 2 in < b > is b11 . Thus a 6 = x = b11 .
14. Suppose that x is an element of the group G and x 8 = e . If x 2 ! e and the
order of x is not 8, then what is the order of x?
Solution: 4
15. What is the order of the element
1
[1 + i ] in the non-zero complex numbers
2
under multiplication?
Solution: Note that x 2 = i,!x 4 = !1,!x 8 = 1 . The order is 8.
16. If a, b and c are elements of the group G, then what is the inverse of abc?
!1
Solution: ( abc ) = c !1b !1a !1
17. (a). Suppose that H is a subgroup of U(56) . Given that 9 is in H,
show that 25 is also in H.
Solution: In U(56) , 9 2 = 25 .
(b). What is the order of 5 in U(17)? – check with the group therapy software.
(c). What is the order of 7 in U(15)? - check with the group therapy software.
18. The index of a subgroup H of the finite group G is the number of left cosets of
H in G and is denoted by [G : H ] . So, by LaGrange’s Theorem,
[G : H ] = o(G) o(H ) .
Suppose that G is a finite group and H is a subgroup of G and K is a subgroup
of H. Show that [G : K ] = [G : H ] ![H : K ] .
G
G
H
Solution: [G : K ] =
,![G : H ] =
,![H : K ] =
. Now just substitute.
K
H
K
19. What is the index of the subgroup {1, –1} in the Quaternions Q8 ?
Solution: 4.
20. Show that if every element of G has order 2, then G is Abelian.
Solution: See your notes.
21. Show that for any prime p > 2, there are exactly two elements of U(p) such
that satisfy x 2 = 1 (i.e., there is exactly one element of order 2).
Hint: If p is a prime and p divides the product nm of integers n and m, then p
divides n or p divides m.
Solution: Note that x 2 = 1 in U(p) is equivalent to
x 2 ! 1mod p " p!divides!x 2 # 1 = (x + 1)(x # 1) .
This in turn is equivalent to p divides x ! 1 or x + 1 .
So x ! 1mod p or x ! "1mod p .
22. Let H be a subgroup of the finite group G and g an element of G with
o(g) = n . Let m be the smallest positive integer such that g m !H . Show that
m divides n.
Hint: You know that n = mq + r where q and r are integers and 0 ! r < m .
Solution: Suppose that n = mq + r .
( )
q
( )
"q
Then e = g n = g mq + r = g m g r ! g r = g m #H .
Hence by the choice of m, r cannot be both positive and less than m. So r = 0.
23. Show that if x,!y !G and xy !Z(G) , then xy = yx .
Note: Recall that Z(G) denotes the center of G.
Hint: Note that xy = yx ! xyx "1 y "1 = e .
Solution: See your notes.
24. No group can have exactly two elements of order 2.
Hint: Consider the two cases where ab = ba and ab ≠ba.
Solution: Suppose that a and b are elements of order 2 in G. Then if ab = ba
then, ab is a third element of order 2, while otherwise, aba is a third element
of order 2.