Math 546
Practice Exam 2B
1. (i). Let a, b, c be elements of the group G, then (abc) !1 = __________
Solution: (abc)!1 = c !1b !1a !1 .
(ii). What is the index of <(1,2,3)(4,5)> in S7 ?
7!
Solution:
= 7 ! 5! = 840
6
(iii). List all the generators of (Z12 , + ) .
Solution: 1, 5, 7, 11.
(iv). The center of a group G is the set Z(G) of all x such that __________
Solution: xa = ax fro every a in G.
(v). The remainder when 2 64 is divided by 21 is __________
Solution: 2! (21) " 1mod 21 # 212 " 1mod 21 # 2 60 " 1mod 21 # 2 64 " 16 mod 21 .
So the remainder is 16.
"1 2 3 4 5 6 7 8
2. Let ! = $
#2 6 9 7 3 8 5 1
9%
'
4&
(i). Express ! as a product of disjoint cycles.
Solution: ! = (1,!2,!6,!8 ) ( 3,!9,!4,!7,!5 ) .
(ii). The order of ! is ______ Solution: 20
(iii). Express ! "1 as a product of transpositions.
Solution: ! "1 = ( 8,!1) ( 8,!2 ) ( 8,!6 ) ( 5,!3) ( 5,!9 ) ( 5,!4 ) ( 5,!7 )
(iv). ! 103 = _________________________________________
Solution: ! 103 = ! 3 = (1,!8,!6,!2 ) ( 3,!7,!9,!5,!4 )
(v). Find a permutation ! such that ! " = (1, 2, 3, 4, 5, 6, 7, 8, 9) .
Solution: ! = ( 2,!5 ) ( 3,!9,!8 ) ( 4,!7,!6 )
3. Does there exist an element of S10 .
(a). Having order 21? Justify your answer.
Solution: Yes, for example, ! = (1,!2,!3) ( 4,!5,!6,!7,!8,!9,!10 ) .
(b). Having order 22? Justify your answer.
Solution: No, 22 does not divide 10! – which is the order of S10 .
(c). Having order 8? Justify your answer.
Solution: Yes, for example, (1,!2,!3,!4,!5,!6,!7,!8 ) .
4. Prove: If x and y are elements of a group G and n and m are relatively prime
integers such that x n = y n and xm = y m , then x = y.
Solution: See Your Notes.
5. (a). Suppose that H is a subgroup of a group G of order 36 and that you have
determined that the elements {x, y,z, w,t} are all elements of H. The only
element of G that you have not yet checked for inclusion into H is the
element d. Does d necessarily belong to H? Explain your answer.
Solution: Yes 5 does not divide 36,and so H cannot be the entire group, it must
contain at least one more element.
(b). Show that Sn is not Abelian for any n ≥ 3.
Solution: (1,!2 ) ( 2,!3) = (1,!2,!3) and ( 2,!3) (1,!2 ) = (1,!3,!2 ) and
(1,!3,!2 ) ! (1,!2,!3) .
6. Let a denote an element of the group G. Show that the centralizer
Ca = {x:ax = xa} is a subgroup of G.
Solution: See Your Notes.
7. Define the relation ~ on G by x ~ y ! there exists a g "G such that y = gxg #1 .
(a). Show that ~ is a transitive relation.
(It’s actually an equivalence relation, but you need not verify that.)
Solution: Suppose that x ~ y and y ~ z . We claim that x ~ z .
Since x ~ y , there is some g in G such that y = gxg !1 and since y ~ z , there is
some h in G such that z = hyh !1 .
(
)
Thus, z = hyh !1 = h gxg !1 h !1 = (hg)x(g !1h !1 ) = (hg)x(hg)!1 .
Hence, x ~ z .
(b). Suppose that x is an element of Z(G), then determine [x ] .
Solution: Suppose that x ~ y . Then there is some g in G such that y = gxg !1 and
so since x must commute with each element of G, y = gxg !1 = xgg !1 = xe = x .
Thus, [ x ] = { x} .
!1 2 $
!1
8. (a). For the element A = #
& 'GL(2, R) , what is A ?
1
3
"
%
" 3 !2 %
Solution: A !1 = $
'.
# !1 1 &
(b). Given that U(19)is cyclic, how many generators does U(19) have?
Solution: The order of U(19) is 18. So since it is cyclic it must have
! (18) = 6 generators.
(c). Suppose that A and B are subgroups of G and that A = 18,!! B = 30 .
What are the possible orders of A ! B ?
Solution: By Lagrange’s Theorem the order of A ! B must divide the order of A
and also the order of B. So the order must be 1, 2, 3, or 6,
!1 2 $
9. What is the centralizer of #
& 'GL(2, R) ?
"1 3 %
2b $
' !1 2 $* . ! a
1
Solution: C ) #
=
:
a,!b
-R
/
2.
&
#
&
( "1 3 %,+ 0 " b a + 2b %
3