Math 546
Practice Exam 3
1. Let H be a normal subgroup of G, then
(a), the set G H is a group under the operation _____________________.
Solution: (aH )(bH ) = abH
(b). What is the identity element of G H ?
Solution: H
(c). Describe the inverse of a typical element of G H ?
Solution: (aH )!1 = a !1 H
2. Let G be a group and suppose that H = {x : x 3 = e} is a subgroup of G.
Show that H ! G .
Solution: Let x be any element of H. It is enough to show that for any g !G ,
gxg !1 "H . So we must show that (gxg !1 )3 = e . However,
(gxg !1 )3 = gxg !1gxg !1gxg !1 = gxxxg !2 = gx 3 g !3 = geg !1 = gg !1 = e .
3. Let G be a cyclic group and H a subgroup of G. Show that G H is cyclic.
Solution: Since G is cyclic, there is some element g in G that generates G.
Now consider any element aH in G H . Since a !G , a = g k for some integer k,
k
and so ( aH ) = g k H = ( gH ) . Hence every element of G H is some power of gH .
Thus, G H is a cyclic group generated by gH .
4. (a). Let H be a normal subgroup of G and let ! : G " G H be the homomorphism
defined by ! (a) = aH . What is the kernel of ! ?
Solution: H.
(b). Let M be the set of all 2 ! 2 real matrices under the operation of addition.
Let R denote the set of real numbers also under addition.
( " a b %+
Let ! : M " R be defined by ! * $
' = a + d . (i.e, ! (A) = trace(A) )
) # c d &-,
Show that ! is a homomorphism and determine its kernel.
Solution: Suppose that A and B are elements of M with say,
!a b $
!x y $
A=#
,!!B = #
&
& . Then
"c d %
" z w%
( " a b % " x y %+
( " a + x b + y %+
! (A + B) = ! * $
+$
=
!
'
'
*) $ c + z d + w '-, = a + x + d + w ,
) # c d & # z w &-,
#
&
And ! (A) + ! (B) = (a + d) + (x + w) = a + x + d + w , and so ! is a homomorphism.
!a b $
Now, A = #
& is in the kernel of ! if ! (A) = 0 " a + d = 0 " d = #a .
"c d %
*#a b &
So, ker ! = + %
: a,b, c )R . .
(
, $ c "a '
/
5. Suppose that every subgroup of the group G is a normal subgroup.
Show that for any a and b in G there exists an integer k such that ab = ba k .
Hint: Consider the subgroup H =!< a > .
Solution: Let a,!b !G . Since H =!< a > is a normal subgroup of G, it follows
that b !1ab "H . But that means that b !1ab = a k for some integer k. Now
multiplying on the left by b gives ab = ba k .
6. (a). Show that H = {i,!(2,!3)} is not a normal subgroup of S3 .
(b). What condition on the index [ G : H ] will guarantee that H ! G ?
Solution:
(a). (1, 3)(2, 3)(1, 3)!1 = (1, 3)(2, 3)(1, 3) = (1,!2) "H .
(b). If [ G : H ] = 2, then H is a normal subgroup.
7. Let G and H be groups and let f :G ! H be a homomorphism that is 1-1. Show
that ker( f ) = {eG } .
Solution: Suppose that f :G ! H is 1-1. Let x !ker(" ) . Then ! (x) = e = ! (eG ) .
Thus since f is 1-1, x = eG .
8. Suppose that a,!b !G and that !a ,! !b are the corresponding inner automorphisms.
(a). Show that !a !b = !ab .
(b). What is the value of !a (a) ?
Solution: The ! x ' s are functions so to show that !a !b = !ab we must show that
have the same functional values. But for any g in G,
!a !b (g) = !a ( !b (g)) = !a bgb "1 = a bgb "1 a "1 = (ab)g(b "1a "1 ) = (ab)g(ab)"1 = !ab (g).
(
) (
)
9. Cayley’s Theorem guarantees that any group G is isomorphic to a
subgroup of SG .
Moreover, the proof of Cayley’s Theorem provides an explicit isomorphism from
G to SG . For the group G = U(15) = {1,!2,!4,!7,!8,!11,!13,!14} , what element of
SG corresponds to 4 of U(15) under this isomorphism?
Solution: ! 4 = (1,!4)(2,!8)(7,!13)(11,!14) .
[Recall, that the isomorphism from G to SG given in the proof of Cayley’s
Theorem, is ! (a) = " a where ! a (x) = ax .]
10. Suppose that in the puzzle below jumps are allowed in either a vertical or
horizontal direction from any square over an occupied square into an unoccupied
square. The peg jumped over is removed. Suppose that a solution exists for which
exactly one peg remains. What squares could possibly be the final location of the
single peg?
Solution: Label the holes with the non-zero elements of the Klein 4-group so that
very vertical and horizontal line of three contains each of a, b and c exactly once.
Then proceed exactly as in the case of the pyramid puzzle. Note that no matter
how these holes are labeled, the value of the position must be the same as that of
the central hole.
If the central hole here is labeled a, then the value of this position is a, and so the
places where a single peg could occur would be at any hole labeled a.
Extra Credit: Show that every permutation in S4 can be written as a product using just
the transpositions (1, 2), (2, 3) and (3, 4).
Big Hint: What is the product (1,!2)(2,!3)(3,!4)(2,!3)(1,!2)