1.
Theorem: State the Criterion for a Product of Cyclic Groups to be Cyclic.
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Worksheet Weekly Practice 4
Instructions: You may type up or handwrite your work, but it must be neat, professional, and organized and it must be saved as a PDF file and uploaded to the appropriate Gradescope assignment. Use a scanner or scanning app to convert handwritten work on paper to PDF. I encourage you to type your work using the provided template.
All tasks below must have a complete solution that represents a good-faith attempt at being right to receive engagement credits. If your submission is complete and turned in on time, you will receive full engagement credit for the assignment. All other submissions will receive zero engagement credit. Read the guidelines at Grading Specifications carefully.
To abide by the class academic honesty policy, your work should represent your own understanding in your own words. If you work with other students, you must clearly indicate who you worked with in your submission. The same is true for using tools like generative AI although I strongly discourage you from using such tools since you need to build your own understanding here to do well on exams.
True/False, Multiple Choice, & Fill-In.
For these problems a justification is not required for credit, but it may be useful for your own understanding to include one. True/False problems should be marked True if the statement is always true, and False otherwise. Multiple choice problems may have more than one correct answer if that is indicated in the problem statement; be sure to select all that apply. Fill-in problems require a short answer such as a number, word, or phrase.
2.
True/False: \(G/H\) is a group for any subgroup \(H\) of \(G\text{.}\)
Solution.
False, \(G/H\) is a group if and only if \(H\) is a normal subgroup of \(G\text{.}\)
3.
True/False: There are only two groups of order \(2p\) up to isomorphism.
Solution.
Short Response.
Your responses to these questions should be complete solutions with justifications, as per the Grading Specifications.
4.
The dihedral group \(D_n\) of order \(2n\) for \(n\geq 3\) has a cyclic subgroup of \(n\) rotations and a subgroup of order \(2\text{.}\) Explain why \(D_n\) cannot be isomorphic to the external direct product of two such groups.
Solution.
The cyclic subgroup \(C\) of \(n\) rotations is Abelian, since it is cyclic. Similarly, any subgroup \(F\) of order \(2\) is also Abelian. Therefore the external direct product \(C\times F\) is Abelian. But \(D_n\) is non-Abelian, so it cannot be isomorphic to this external direct product.
Problem Specs/Notes: This problem needs a clear description of a group property that is not shared between \(D_n\) and the product group that prevents the isomorphism for a Success.
5.
Find the largest order of a cyclic subgroup of \(\Z_6 \times \Z_{10}\times \Z_{15}\text{.}\) Give an example of such a subgroup. Is there a unique such subgroup?
Solution.
Let \(\langle (a,b,c)\rangle\) be a cyclic subgroup of this group \(G\text{.}\) Then
\begin{align*}
|(a,b,c)|\amp=\lcm(|a|,|b|,|c|)\\
\amp\leq \lcm (|\Z_6|,|\Z_{10}|,|\Z_{15}|)\\
\amp=\lcm(6,10,15)\\
\amp=30\text{.}
\end{align*}
So the largest order is at most \(30\text{.}\) To see that it is \(30\text{,}\) we need only choose elements \(a\text{,}\) \(b\text{,}\) \(c\) achieving this maximum. For example, we could take \((3,0,1)\text{,}\) which has order \(30\) since \(|3|=2\) in \(\Z_6\text{,}\) \(|0|=1\text{,}\) and \(|1|=15\) in \(\Z_{15}\text{.}\) On the other hand, the subgroup is not unique, since another element of order \(30\) is \((0,5,1)\) (as \(|5|=2\) in \(\Z_{10}\)), and the two subgroups generated by these elements intersect trivially.
Problem Specs/Notes: This problem needs a correctly derived order with examples provided for a Success.
6.
Construct the subgroup lattice for \(\Z_4\times \Z_2\text{,}\) justifying your work. It will be useful to use Lagrange’s Theorem, the Fundamental Theorem of Cyclic Groups, and the possible isomorphism classes of groups of size 2 and 4. Which subgroups are of the form \(H\times K \) where \(H\leq \Z_4 \) and \(K\leq \Z_2\text{?}\)
Hint.
There are a total of 8 subgroups.
Solution.
The lattice is shown below. All of the non-trivial proper subgroups have order 2 or 4 by Lagrange’s Theorem. Computing powers of the group elements, there are five cyclic subgroups: three of order 2, generated by \((2,0)\text{,}\) \((0,1)\text{,}\) and \((2,1)\text{,}\) and two of order 4, generated by \((1,0)\) and \((1,1)\rangle\text{.}\) The remaining subgroup is the unique non-cyclic subgroup of order 4, which is isomorphic to \(\Z_2\times \Z_2\text{,}\) and contains all three elements of order 2.
The subgroups of the form \(H\times K\) are
\begin{align*}
\langle (0,1),(1,0)\rangle \amp \cong \Z_4\times \Z_2\\
\langle (1,0)\rangle \amp \cong \Z_4\times \{e\}\\
\langle (0,1)\rangle \amp \cong \{e\}\times \Z_2 \\
\langle (2,0)\rangle \amp \cong \subgroup{(2,0)}\times \{e\} \\
\subgroup{(2,0),(0,1)} \amp \isom\subgroup{(2,0)}\times\subgroup{(0,1)}\\
\{e\} \amp \isom \{e\}\times \{e\}\text{.}
\end{align*}
Problem Specs/Notes: This problem needs a correctly drawn lattice with clear justifications given for why it is complete for a Success.
7.
For each of the normal subgroups \(\langle (2,1)\rangle\) and \(\langle (0,1)\rangle\) of \(G=\Z_4\times \Z_2\text{,}\) carry out the following steps. Below I am writing \(N\) for the normal subgroup.
(a)
Partition \(G\) by the left (or right) cosets of \(N\text{.}\)
Solution.
Part 1: \(N=\subgroup{(2,1)}\). The cosets are
\begin{gather*}
(0,0)+N=\{(0,0),(2,1)\}, (1,0)+N=\{(1,0),(3,1)\},\\
(2,0)+N=\{(2,0),(0,1)\}, (3,0)+N=\{(3,0),(1,1)\}\text{.}
\end{gather*}
Part 2: \(N=\subgroup{(0,1)}\). The cosets are
\begin{gather*}
(0,0)+N=\{(0,0),(0,1)\}, (1,0)+N=\{(1,0),(1,1)\},\\
(2,0)+N=\{(2,0),(2,1)\}, (3,0)+N=\{(3,0),(3,1)\}\text{.}
\end{gather*}
(b)
Create a Cayley table of the quotient group \(G/N\text{.}\)
Solution.
Both quotient groups have the same Cayley table, which is shown below. The elements of the quotient group are the cosets of \(N\) in \(G\text{,}\) and the operation is coset addition.
| \((G/N,+)\) | \((0,0)+N\) | \((1,0)+N\) | \((2,0)+N\) | \((3,0)+N\) |
|---|---|---|---|---|
| \((0,0)+N\) | \((0,0)+N\) | \((1,0)+N\) | \((2,0)+N\) | \((3,0)+N\) |
| \((1,0)+N\) | \((1,0)+N\) | \((2,0)+N\) | \((3,0)+N\) | \((0,0)+N\) |
| \((2,0)+N\) | \((2,0)+N\) | \((3,0)+N\) | \((0,0)+N\) | \((1,0)+N\) |
| \((3,0)+N\) | \((3,0)+N\) | \((0,0)+N\) | \((1,0)+N\) | \((2,0)+N\) |
(c)
Create a Cayley graph of \(G/N\text{.}\)
Solution.
Since both quotient groups have the same Cayley table, they also have the same Cayley graphs. A graph is shown below, with the cosets of \(N\) as vertices and edges labeled by the generator \((1,0)+N\text{.}\)
(d)
Determine the isomorphism type of both \(N\) and \(G/N\text{.}\)
Solution.
Both subgroups are isomorphic to \(\Z_2\) because they have size 2, and both quotient groups are isomorphic to \(\Z_4\) because they have size 4 and are cyclic.
(e)
Consider the sublattice of the subgroup lattice of \(G\) consisting of all subgroups between \(G\) and \(N\text{.}\) What do you notice about the structure of this lattice?
Solution.
In both cases, the sublattice is a chain of length 2, with \(G\) at the top, \(N\) at the bottom, and a single subgroup of order 4 in between. This is the same structure as the subgroup lattice of \(\Z_4\) (except that the groups are twice as large)
Problem Specs/Notes: This problem needs all five responses for both normal subgroups for a Success.
8.
Let
\begin{equation*}
H=\left\{ \begin{bmatrix}
a \amp b \\ 0 \amp d
\end{bmatrix} \mid a,b,d\in \R, ad\neq 0\right\}\text{.}
\end{equation*}
Is \(H\) a normal subgroup of \(\GL(2,\R)\text{?}\)
Hint.
The characterization that \(H\) is normal if and only if \(ghg^{-1}\in H\) for all \(g\in G, h\in H\) is helpful on this problem.
Solution.
First, note that the condition on \(H\) is that \(H\) is an invertible \(2\times2\) matrix with lower left entry 0. So \(H\) is a subgroup of \(\GL(2,\R)\) by the one-step subgroup test, since \(I_2 \in H\) and if
\begin{equation*}
A=\begin{bmatrix}
a \amp b \\ 0 \amp d
\end{bmatrix}, B=\begin{bmatrix}
a' \amp b' \\ 0 \amp d'
\end{bmatrix} \in H
\end{equation*}
then
\begin{equation*}
AB^{-1}=\begin{bmatrix}
a \amp b \\ 0 \amp d
\end{bmatrix} \frac{1}{a'd'}\begin{bmatrix}
d' \amp -b' \\ 0 \amp a'
\end{bmatrix} = \frac{1}{a'd'}\begin{bmatrix}
ad' \amp a'b-b'a \\ 0 \amp a'd
\end{bmatrix}
\end{equation*}
and this matrix has the required form since it is the product of two invertible matrices and its lower left entry is 0. Now to show that \(H\) is normal, we need to show that \(XAX^{-1}\) is in \(H\) for any \(X\in \GL(2,\R)\) and \(A\in H\text{.}\) This product will be invertible for any such product, so we only need to check that it has a 0 in its lower left entry. Let \(X=\begin{bmatrix} x \amp y \\ z \amp w\end{bmatrix}\in \GL(2,\R)\text{,}\) we have \(X^{-1}=\frac{1}{xw-yz}\begin{bmatrix} w \amp -y \\ -z \amp x\end{bmatrix}\text{.}\) Then
\begin{align*}
XAX^{-1}\amp=\frac{1}{xw-yz}\begin{bmatrix} x \amp y \\ z \amp w\end{bmatrix}\begin{bmatrix}
a \amp b \\ 0 \amp d
\end{bmatrix}\begin{bmatrix} w \amp -y \\ -z \amp x\end{bmatrix}\\
\amp=\frac{1}{xw-yz}\begin{bmatrix}
ax \amp xb+yd \\ az \amp bz+wd
\end{bmatrix}\begin{bmatrix} w \amp -y \\ -z \amp x\end{bmatrix}\\
\amp=\frac{1}{xw-yz}\begin{bmatrix}
axw-xbz-yzd \amp -axy+x^2b+xyd \\ azw-bz^2-wzd \amp -azy+bzx+wxd
\end{bmatrix}\text{.}
\end{align*}
The lower left entry \(azw-bz^2-wzd\) is not always 0, so this subgroup is not normal. For example, take \(A=\begin{bmatrix}
1 \amp 1 \\ 0 \amp1
\end{bmatrix}\) and \(X=\begin{bmatrix}
1 \amp 0 \\ 1 \amp 1
\end{bmatrix}\text{.}\) Then \(XAX^{-1}=\begin{bmatrix}
0 \amp 1 \\ -1 \amp 2
\end{bmatrix} \not \in H\text{.}\)
Problem Specs/Notes: This problem needs careful computation of the matrix multiplications for a Success.
