1.
List the elements of the group \(\Z_2 \times \Z_3\text{.}\)
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Section Direct Products
Worksheet Part 1: Computing in Direct Products
The goal of these problems is to practice computing in direct products and lay the foundation for classifying more finite groups.
2.
Theorem 84. Order of an Element in a Direct Product.
The order of an element in a direct product of a finite number of finite groups is the least common multiple of the orders of the components of the element.
\begin{equation*}
|(g_1,g_2,\ldots,g_n)|=\lcm(|g_1|,|g_2|,\ldots,|g_n|)
\end{equation*}
3.
Find the order of the element \((18,3) \in \Z_{36}\times \Z_9\)
4.
In this problem, you will determine the number of cyclic subgroups of order 6 in \(G=\Z_{36}\times \Z_9\text{.}\)
(a)
Count the number of elements \((a,b)\) of order 6 in \(G\text{.}\) You may want to proceed in cases on \(|a|\text{.}\)
(b)
How many elements of order 6 does a cyclic subgroup of order 6 have?
(c)
How many cyclic subgroups of order 6 are in \(G\text{?}\)
(d)
Is \(G\) cyclic?
Worksheet Part 2a: \(U(n)\) as a Product of Cyclic Groups
The goal of these problems is to build the tools needed to classify the unit groups \(U(n)\) as products of cyclic groups \(\Z_{k}\text{.}\)
Suppose \(k\mid n\) and define \(U_k(n)=\{x \in U(n) \mid x\pmod{k} =1 \}\text{.}\) For the purposes of examples, you will work in \(U(42)=\{1,5,11,13,17,19,23,25,29,31,37,41 \}\)
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Remark 85. Isomorphism Class of the Group of Units mod a Prime Power.
It is true that \(U(2)\cong \{0\}\text{,}\) \(U(4)\cong \Z_2\text{,}\) \(U(2^n)\cong \Z_2 \times \Z_{2^{n-2}}\text{,}\) and \(U(p^n)\cong \Z_{p^{n-1}(p-1)}\) for odd primes \(p\text{.}\)
5.
Use the above isomorphisms to determine what direct product of cyclic groups \(U(42)\) is isomorphic to.
Worksheet Part 2b: Groups of Order at Most 8
The goal of these problems is to classify all groups of order at most 8.
1.
There is only one group of order 1. What is it?
2.
Lagrange’s theorem says the the order of any non-identity element in a group of order \(p\) is \(p\text{.}\) What does this say about groups of order 2, 3, 5, and 7?
3.
Lagrange’s theorem says the the order of any non-identity element in a group of order 4 is 2 or 4. Show that either \(G=\subgroup{a}\) or \(G=\{e,a,b,ab\}\isom K_4\text{.}\)
4.
What does Groups of Order 2p are Cyclic or Dihedral tell you about the possible isomorphism classes of groups of order 6?
We are now left with the groups of order 8. By Lagrange’s theorem, the order of any non-identity element is 2, 4, or 8. It is easiest to deal with the extreme cases first.
5.
Suppose there is an element of order 8. What can you conclude about the group?
6.
Now suppose that all non-identity elements have order 2. Use the fact that such a group is Abelian to help you show that the group is isomorphic to \(\Z_2 \times \Z_2 \times \Z_2\text{.}\) It will be useful to choose three distinct elements \(a,b,c\) such that \(c\neq ab\text{.}\)
Now we are left with the case where there is an element of order 4, but no element of order 8. There will be three possibilities for the isomorphism class of the group, depending on how many elements of order 4 there are and how they commute. In all of these cases, let \(a\in G\) have order 4, and let \(b \in G\) be an element in \(G\setminus \subgroup{a}\text{.}\) Then we have \(G=\subgroup{A}\cup \subgroup{A}b=\{e,a,a^2,a^3,b,ab,a^2b,a^3b\}\text{.}\)
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Next, suppose that we can find \(b\) with order 2. Show that either \(ba=ab\) or \(ba=a^3b\) by showing that if \(ba=ab^2\) then \(a=ea=b^2a\) is the identity.
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For the last case, we will need to know about a new group of order 8, the quaternion group \(Q_8=\{1,-1,i,-i,j,-j,k,-k\}\text{,}\) where \(i^2=j^2=k^2=-1\text{,}\) and \(ij=k\text{,}\) \(jk=i\text{,}\) and \(ki=j\text{.}\) This group is non-Abelian, and can represent 3D rotations, or the cross-product of vectors in 3D space, and can be used to extend the complex numbers to the quaternions. This group is generated by \(i\) and \(j\text{,}\) and has the presentation
\begin{equation*}
Q_8=\langle i,j \mid i^4=j^4=e, i^2=j^2, ji=i^3j \rangle\text{.}
\end{equation*}
Below are a Cayley diagram and subgroup lattice for \(Q_8\text{.}\)
11.
Show that if \(b\) must be chosen to have order 4 (i.e. all elements in \(G\setminus \subgroup{a}\) have order 4), then \(b^2=a^2\text{,}\) \(ba\neq ab\) (by computing \((a^3b)^2\)), and \(ba\neq a^2b\) (by showing this implies \(a=b^2\) and deriving a contradiction). Conclude that \(G\isom Q_8\) by using the description above.
