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Abstract Algebra I MATH 4107 Summer 2026

Dr. Hunter Lehmann

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Worksheet Weekly Practice 6

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.

1.

True/False: A group of size \(15\) can act on a set of size \(7\) with a single orbit.
Solution.
False. By the Orbit-Stabilizer Theorem, a group acting with a single orbit on a set of size 7 would have size divisible by 7, which is not the case for a group of size 15.

3.

Fill-In: If \(G\) acts on a set \(X\) and \(x\in X\text{,}\) the set
\begin{equation*} \{g \in G \mid g\cdot x = x\} \end{equation*}
is the .
Solution.
The stabilizer of \(x\) is the subgroup \(\{g \in G \mid g\cdot x = x\}\text{.}\)

Short Response.

Your responses to these questions should be complete solutions with justifications, as per the Grading Specifications.

4.

Let \(G\) act on the set \(X=\{H \mid H \leq G\}\) by conjugation: \(g\cdot H=gHg^{-1}\text{.}\) What is the stabilizer of each subgroup \(H\text{?}\) What are the fixed points of this action?
Solution.
The stabilizer of each subgroup \(H\) is the normalizer \(N_G(H)\) since
\begin{equation*} gHg^{-1} = H \Leftrightarrow g\in N_G(H)\text{.} \end{equation*}
The fixed points of this action are the subgroups \(H\) that are normal in \(G\text{,}\) i.e., those for which \(gHg^{-1} = H\) for all \(g \in G\text{.}\)

5.

Consider the map \(S_3\times \R^3\to \R^3\) by \(\sigma\cdot (x_1,x_2,x_3)=(x_{\sigma(1)},x_{\sigma(2)}, x_{\sigma(3)})\text{,}\) i.e. \(S_3\) acting on \(\R^3\) by permuting the coordinates. Determine whether this is a (left) group action of \(S_3\) on \(\R^3\text{.}\)
Solution.
This is not a (left) group action, because products of permutations do not act as expected. For example, let \(\sigma=(1\;2)\) and \(\tau=(2\;3)\text{.}\) Then
\begin{equation*} \sigma\cdot(\tau\cdot(a,b,c)) = (1\;2)\cdot((2\;3)\cdot(a,b,c)) = (1\;2)\cdot(a,c,b) = (c,a,b) \end{equation*}
but
\begin{equation*} (\sigma \tau)\cdot (a,b,c) = (1\;2\;3)\cdot(a,b,c)= (b,c,a) \end{equation*}
and these are not equal. Thus the group action property
\begin{equation*} \sigma\cdot(\tau\cdot x) = (\sigma\tau)\cdot x \end{equation*}
does not hold for all \(\sigma,\tau\in S_3\) and \(x\in \R^3\text{.}\)
Problem Specs/Notes: This problem is tricky and trying a concrete example will be useful. Solutions earning a Success will need a careful analysis of the group action properties.

6.

Show that there is no simple group of order \(45\text{.}\) Recall that a simple group is one with no non-trivial proper normal subgroups, i.e. one which cannot be meaningfully quotiented.
Solution.
We will show that a group of order \(45\) has a normal Sylow \(3\)-subgroup. By Sylow Theorem III the number of Sylow \(3\)-subgroups, \(n_3\text{,}\) divides \(45/3^2=5\) and is congruent to \(1\bmod 3\text{.}\) The only divisor of 5 that is congruent to 1 mod 3 is 1, so there is a unique Sylow \(3\)-subgroup, which must be normal. Thus any group of order 45 has a non-trivial proper normal subgroup and cannot be simple.
Actually, since we also have \(n_5\) divides \(45/5=9\) and is congruent to \(1\bmod 5\text{,}\) we also have \(n_5=1\) and so there is also a unique Sylow \(5\)-subgroup, which is also normal. So any group of order 45 has two non-trivial proper normal subgroups. It then also follows quickly that elements of the Sylow \(3\)-subgroup and the Sylow \(5\)-subgroup commute, so the group is actually isomorphic to the direct product of these two subgroups and is therefore isomorphic to \(\Z_3\times \Z_3\times \Z_5\) or \(\Z_9\times \Z_5\text{.}\)
Problem Specs/Notes: To earn a Success you need to carefully apply the Sylow Theorems to show that such a group cannot exist.
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