Mathematics 243, section 1 – Algebraic Structures Selected Solutions – Problem Set 8 December 2, 2003

General Comment:

It seems from the work I saw on this problem set that some very basic material (such as what “set builder notation” means, what it means for a mapping to be 1-1 or onto, what an inverse function is, what composition of mappings means, and how to write these things down in a correct fashion) has become hazy again in some people’s minds. If you are in that category, begin your review for Exam 3 by looking back at Chapter 1 and learning that material once and for all now. It is part of the basic language of mathematics that you must have reliably at your fingertips now and for the rest of your careers as mathematics majors. Even though I did not mention this explicitly on the Review Sheet for Exam 3, it is a certainty that there will be questions dealing with these ideas!

4.1/14. The problem asks for all of the distinct cyclic subgroups of the alternating group A_{4} (all even permutations of {1, 2, 3, 4}). Many people just wrote down the cycle decompositions of the 12 elements of A_{4}. Recall that the cyclic subgroup generated by an element a in a group G is the set

⟨a⟩ = {a^{k}|k ∈ Z}

This means that you must take all of the elements a in A_{4}, determine their powers, and find the elements in ⟨a⟩. There are exactly eight distinct cyclic subgroups:

\begin{aligned}⟨()⟩ & = \{()\} \\ ⟨(12)(34)⟩ & = \{(), (12)(34)\} \\ ⟨(13)(24)⟩ & = \{(), (13)(24)\} \\ ⟨(14)(23)⟩ & = \{(), (14)(23)\} \\ ⟨(123)⟩ & = \{(), (123), (132)\} = ⟨(132)⟩ \\ ⟨(124)⟩ & = \{(), (124), (142)\} = ⟨(142)⟩ \\ ⟨(134)⟩ & = \{(), (134), (143)\} = ⟨(143)⟩ \\ ⟨(234)⟩ & = \{(), (234), (243)\} = ⟨(243)⟩ \end{aligned}

4.1/16. The full group operation table for the “octic” group can be found inside the back cover of the textbook (in the “endpapers”).

4.2/1,4. Most people followed the format of the proof of Cayley’s Theorem in the text for these, and expressed the permutations as the one-to-one and onto mappings f_{a} : G → G (or m_{a} using the notation I used in class). That’s fine, but you should also be aware that you can also get subgroups of S_{n} (with elements written in cycle decomposition form) isomorphic to each of these groups. This way of writing the groups comes by labeling the elements of the group using the integers {1, 2, 3, …, n} in some fixed fashion, and seeing what the rows of the group operation table look like as permutations of {1, 2, 3, …, n}. For instance, for the Klein 4-group, letting

\begin{matrix}e ↔ 1 \\ a ↔ 2 \\ b ↔ 3 \\ ab ↔ 4 \end{matrix}

the group table becomes:

1 2 3 4
1 1 2 3 4
2 2 1 4 3
3 3 4 1 2
4 4 3 2 1

So we also get an isomorphism

ϕ : G → {(), (12)(34), (13)(24), (14)(23)} ⊂ S_{4}

by letting ϕ(e) = (), ϕ(a) = (12)(34), etc. The subgroup of S_{6} corresponding to the group in problem 4 in this way is

{(), (12)(34)(56), (13)(26)(45), (164)(235), (15)(24)(36), (146)(253)}

written in the same order as the permutation matrices (i.e. ϕ(I_{3}) = (), ϕ(P_{1}) = (12)(34)(56), etc. defines the isomorphism).

4.1/5. a) Let h_{a} : G → G be the mapping defined by h_{a}(x) = xa. The problem says to show that h_{a} is a permutation of G. This means that we must show that h_{a} is a 1-1 and onto mapping. To show that h_{a} is 1-1, consider the equation h_{a}(x) = h_{a}(y) for x, y ∈ G. Then xa = ya, so multiplying on the right by a^{−1} ∈ G we get

(xa)a^{−1} = (ya)a^{−1} ⇒ x(aa^{−1}) = y(aa^{−1}) ⇒ x = y.

Hence h_{a} is 1-1. To show h_{a} is onto, let y ∈ G be arbitrary and consider x = ya^{−1}. Then

h_{a}(x) = h_{a}(ya^{−1}) = (ya^{−1})a = y(aa^{−1}) = y

Since y was arbitrary, we see that every element of G is in the image of h_{a}, which is what we wanted to show.

b) Consider H = {h_{a}|a ∈ G}. This means the set of all the mappings h_{a} where a runs over all the elements of G (“set builder notation!”). We must show that this set is a group under the operation of composition of mappings. That is, we need to show:

i. H is closed under composition ii. composition is associative as a binary operation on H iii. there is an identity element for composition in H iv. every element of H has an inverse in H (under composition).

i. We must show h_{a} ∘ h_{b} is one of the elements of H, that is that h_{a} ∘ h_{b} = h_{c} for some c ∈ G. If x is any element of G,

(h_{a} ∘ h_{b})(x) = h_{a}(h_{b}(x)) = h_{a}(xb) = (xb)a = x(ba) = h_{ba}(x)

This means that h_{a} ∘ h_{b} = h_{ba} (i.e. the c above is the product c = ba ∈ G). Hence H is closed under composition.

ii. Composition is always associative:

((h_{a} ∘ h_{b}) ∘ h_{c})(x) = h_{a}(h_{b}(h_{c}(x))) = (h_{a} ∘ (h_{b} ∘ h_{c}))(x)

(Alternatively, you could work this out and use associativity of multiplication in G to see that x((ab)c) = x(a(bc)) for all x ∈ G.)

iii. Since G is a group, it has an identity element e. The mapping h_{e}(x) = xe = x for all x ∈ G. Hence h_{e} is the identity mapping from G to itself, which is the identity element for composition: h_{a} ∘ h_{e} = h_{a} = h_{e} ∘ h_{a} for all a.

iv. The inverse mapping of h_{a} is the mapping h_{a^{−1}} since for all x ∈ G,

(h_{a} ∘ h_{a^{−1}})(x) = x(a^{−1}a) = xe = h_{e}(x)

Similarly, h_{a^{−1}} ∘ h_{a} = h_{e}.

c) The mapping ϕ : G → H defined by ϕ(a) = h_{a} is not necessarily an isomorphism of groups because

ϕ(ab) = h_{ab} = h_{b} ∘ h_{a} = ϕ(b) ∘ ϕ(a)

by what we said in b i. above. For a group isomorphism we would need ϕ(ab) = ϕ(a)∘ϕ(b).