Math 312
Assignment #1 Answers
Due on 23 September 2015
• Solutions should be correct, complete, clear, and neatly presented to get full marks.
• Questions 1 -7 are mostly short answer questions. Others require full answers.
(1) Describe the group of symmetries of a non-square rectangle and a non-square rhombus. Are
the groups isomorphic?
The symmetries of the non-square rectangle are rotations by 0 and π and two reflections
about lines passing through the midpoints of two parallel sides.
The symmetries of the non-square rhombus are rotations by 0 and π and two reflections
about the two diagonals.
The two groups are isomorphic to each other and to the Klein 4-group.
(2) Describe the symmetries of an infinite strip of equally spaced H’s
······ H H H H ······
Is the group abelian? Justify!
[3]
[4]
Right translations Tn by a nonnegative integer n (i.e., each H moves n places to the right);
left translations T−n by a positive integer n; reflection about the x-axis (the line that goes
through the middle of every H); reflections about each vertical line that goes through the
middle of an H and the vertical lines that are midway between two adjacent H’s.
It is nonabelian. A nontrivial translation and a vertical reflection don’t commute.
[2]
(3) Write down an element in S9 of largest order and an element of A9 of largest order.
In Sn , a product of a 5-cycle and a disjoint 4-cycle with order 20.
in An , a product of a 5-cycle and a disjoint 3-cycle with order 15.
(4) Write down all elements of the group GL2 (Z2 ) and write down the order of each element.
Guess what familiar group is this isomorphic to.
Answer: I, ( 10 11 ), ( 11
isomorphic to S3 .
0 ),
1
( 01
1 ),
0
( 11
1 ),
0
( 01
1 ).
1
[4]
The orders are 1, 2, 2, 2, 3, 3. The group is
(5) Find elements a, b, c in a group such that |a| = |b| = 2 and |ab| = 3.
0 1
Take a = (12) and b = (13) in S3 or a = 10 −1
−1 , b = ( 1 0 ) in GL2 (R).
(6) Write down the centralizer of a reflection f in Dn and the centralizer of of a rotation r by
the smallest possible degree. The answers may depend on the parity of n.
[2]
[3]
The centralizer of r is the subgroup of all rotations.
If n is odd, the centralizer of f is {id, f }.
If n is even, the centralizer of f is {id, f, rn/2 , f rn/2 }. The last one is a reflection about the
line perpendicular to the axis of reflection of f .
(7) Divide the following collection of groups into subsets of isomorphic groups.
(2Z, +), (3Z, +), (Q, +), Z2 , Z6 , S2 , S3 , A3 , D3 , (R, +), (R∗ , ·), (Q∗ , ·), (C∗ , ·), (R+ , ·),
< 123)(45) > (subgroup of S5 ), A =< π > (subgroup of (R∗ , ·)), B =< π > (subgroup of (R, +))
Answer: (a) 2Z, 3Z, A, B,
(d) Z6 , < 123 >< 45 >,
(b) Z2 , S2 ,
(e) R, R+
(c) S3 , D3 ,
Each of the other groups is isomorphic only to itself. This question will not be marked.
1
2
[3]
(8) Prove that An is generated by the set of 3-cycles.
Hint: (1 2)( 3 4)= (1 2)(2 3)(2 3)(3 4).
A permutation τ in An is a product of an even number of transpositions which can then be
bracketed into pairs. A pair of identical transposition can be canceled. A pair (ij)(jk) with
exactly one common symbol equals a 3-cycle (ijk) . For a pair (ij)(k`) with distinct symbols,
insert (jk)(jk) (which is the identity) and you get (ijk)(jk`), a product of 3-cycles.
(9) If H is a subgroup of Sn and K = H ∩ An . Prove that either K = H or |K| = |H|/2. Hint:
The proof of Theorem 5.7 is useful.
[3]
If every permutation in H is even, then K = H. If there an odd permutation σ in H, then
the mapping α 7→ σα is a bijection from the set of even permutation in H and the set of odd
permutations in H for the same reason given in the proof of Theorem 5.7. So each subset
includes half of the elements in H.
(10) (a) Give an example of elements a, b, c in a group such that ab = bc but a 6= c..
(b) Let G be a group with the property that ab = bc implies that a = c. Prove that G is
abelian.
[3]
(a) Take a = (12), b = (23), c = (13) in S3 .
(b) For x, y ∈ g, take a = xy, b = x, c = yx. Then ab = xyx = bc and by unusual cancelation
assumption a = b, i.e., xy = yx proving commutativity.
(11) Let G be a finite abelian group. Prove that the map g 7→ g 2 is an automorphism of G if and
only if G has no element of order 2. Give counterexamples for infinite abelian G and for a
finite nonabelian G.
[6]
Let ϕ(g) = g 2 . First assume that ϕ is an automorphism. If a has order 2, then ϕ(a) = e =
ϕ(e) contradicting the injectivity of ϕ. So there is no elements of order 2.
To prove the converse assume that G has no elements of order 2.
ϕ(ab) = abab = aabb = ϕ(a)ϕ(b) by commutativity.
To prove injectivity, assume that a2 = b2 , then by commutativity (a−1 b)= e. This implies
that a−1 b = e since the group has no elements of order 2. Thus a = b proving injectivity.
Surjectivity follows since G is finite.
Counterexample 1 for infinite abelian group. G = Z
Counterexample 2. Ask me! (no marks given or taken away)
(12) Let G be an infinite group. Prove that it has infinitely many subgroups.
[3]
Hint: Consider two cases according to whether there an element of infinite order or not.
If there is an element a of infinite order, then < aj >; (j ≥ 0) are infinitely many distinct
groups
If every element has finite order and if there is a finite number of cyclic groups, then their
union U is a finite set so there exists an element g ∈ G \ U and the group < g > is a new
cyclic subgroup, a contradiction.