UA-AUG-AUMAT229 Groups acting on themselves: Discovery guide

Section 17.3 Groups acting on themselves: Discovery guide

Subsection Some of the ways a group can act on itself

Discovery 17.14.

Check whether each of the following will create an action of a group on itself. That is, suppose we have a group $G,$ we take the set $X$ to be $G$ itself, and then consider the following descriptions of group elements $g$ as function $g : G \to G .$ Use the same sort of reasoning as in Discovery 17.3. (You may wish to return to Section 17.2 and read the lead-in paragraph above Discovery 17.3.)

For those that define group actions, determine what the orbits and stabilizers will be. Where possible, try to describe these orbits and stabilizers using group-theory terminology previously encountered in this course.

🔗

(a)

By left multiplication. Given group element $g$ in $G$ and set element $x$ in $X$ (so $x$ is also in $G$ in this case), set

g (x) = g x .

🔗

(b)

By right multiplication. Given group element $g$ in $G$ and set element $x$ in $X$ (so $x$ is also in $G$ in this case), set

g (x) = x g .

🔗

(c)

By right-inverse multiplication. Given group element $g$ in $G$ and set element $x$ in $X$ (so $x$ is also in $G$ in this case), set

g (x) = x g^{- 1} .

🔗

(d)

By left conjugation. Given group element $g$ in $G$ and set element $x$ in $X$ (so $x$ is also in $G$ in this case), set

g (x) = g x g^{- 1} .

🔗

(e)

By right conjugation. Given group element $g$ in $G$ and set element $x$ in $X$ (so $x$ is also in $G$ in this case), set

g (x) = g^{- 1} x g .

🔗

Subsection The centre of a group that has prime-power order

🔗

Discovery 17.15.

In Task 17.14.e you should have determined that right conjugation does define an action of a group on itself. Let's consider $G = D_{4}$ acting on itself by right conjugation.

One of the consequences of the Orbit-Stabilizer Theorem for a group action involving a finite group is that the size of each orbit must divide the order of the group.

🔗

(a)

Reinterpret the above statement about the Orbit-Stabilizer Theorem using the group-theoretic description for right-conjugation-orbits you came up with in Task 17.14.e.

🔗

(b)

Based only on Task a and the fact that $| D_{4} | = 8$ (so, without actually calculating the orbits or using your prior knowledge of the pattern of these orbits when interpreted using the theory of conjugacy), list the possibilities for the sizes of these orbits.

🔗

(c)

We know that orbits partition the set $X,$ so the size of $X$ is the sum of the sizes of the orbits. But in this case $X$ is our group $D_{4},$ so the order of $D_{4}$ is equal to a sum involving only those numbers you listed in Task b (possibly with repeats or with some not used).

The number $1$ should have been included in your list in Task b. Given what was stated just above about the order of $D_{4}$ as a sum of sizes of orbits, is it possible that one, and only one, orbit actually has size $1 ?$

🔗

(d)

Without actually calculating or using your prior knowledge of the pattern of the centre $Z (D_{4}),$ combine Task 17.15.c with Fact 14.2.3 into a statement about the centre $Z (D_{4}) .$

🔗

Discovery 17.16.

If you look back over Discovery 17.15, you should notice that we didn't really need to know anything about the group $D_{4}$ itself, other than the fact that $| D_{4} | = 8 = 2^{3} .$

Convince yourself that the overall logic of Discovery 17.15 works if $D_{4}$ is replaced by a finite group $G$ of prime-power order. Then re-state your conclusion from Task 17.15.d as a statement about $Z (G)$ in that case.

🔗

Subsection Groups of prime-squared order

🔗

In this subsection we will use the result of Discovery 17.16 to investigate the possibilities for the structure of a group of prime-squared order. So for all of the following activities, assume

p

is a prime number and

G

is a group with

| G | = p^{2} .

🔗

Discovery 17.17.

Using the assumption $| G | = p^{2}$ and Lagrange's Theorem, list the possibilities for the order of a group element in $G .$

🔗

Discovery 17.18.

Suppose $G$ contains an element that actually has the largest of the possible orders you listed in Discovery 17.17. What does this imply about the type of group $G$ must be?

🔗

Now that the case of Discovery 17.18 is figured out, let's further assume

G

has no elements that have the largest of the possible orders you listed in Discovery 17.17.

🔗

Let

x

represent some fixed choice of non-identity element in the centre

Z (G) .

(Remember, we are assuming

| G | = p^{2};

see Discovery 17.16.) Let

y

represent some fixed choice of element in

G

that is not in

⟨ x ⟩ .

Let

H

represent

⟨ x ⟩

and

K

represent

⟨ y ⟩ .

Our goal is to verify the conditions we learned for recognizing

G

as an internal product of

H

and

K .

🔗

Discovery 17.19.

🔗

(a)

Based on Discovery 17.17 and our new assumption on $G$ above, what are the orders of $x$ and $y ?$ So then what are the orders of the subgroups $H = ⟨ x ⟩$ and $K = ⟨ y ⟩ ?$

🔗

(b)

Use Task a to justify the statement that $⟨ y^{b} ⟩ = K$ for every integer $b,$ $0 < b < | y | .$

🔗

Discovery 17.20.

🔗

(a)

Recall that we have assumed that $y$ is not in $H = ⟨ x ⟩ .$ This means that $y$ cannot be equal to a power of $x .$ But can a power of $y$ be equal to a power of $x ?$

Hint.

Use Task 17.19.b.

🔗

(b)

What does Task a say about the intersection of $H$ and $K ?$

🔗

Discovery 17.21.

Verify that every element of $H$ commutes with every element of $K .$

Hint.

Look back at how we defined $x$ in the first place.

🔗

Discovery 17.22.

Finally, we will verify that $G = H K .$ But we will do so indirectly, by verifying that $H K$ has the same number of elements as $G .$

🔗

(a)

Suppose that $a, b, c, d$ are integers with $0 \leq a, b, c, d < p$ and at least one of $a \neq c$ and/or $b \neq d .$ Verify that it is not possible for $x^{a} y^{b}$ to equal $x^{c} y^{d} .$

Hint.

By contradiction, assume $x^{a} y^{b} = x^{c} y^{d} .$ Manipulate algebraically and compare with your answer to Task 17.20.a.

🔗

(b)

Based on Task a, how many elements are there in the product set $H K ?$ And how many elements are there in $G$ again?

🔗

Now let's summarize both cases, where either

G

contains an element of order

p^{2}

or it doesn't.

🔗

Discovery 17.23.

Fill in the question marks in the subscripts in the statement below.

If $p$ is a prime number and $| G | = p^{2},$ then $G$ is isomorphic to either $Z_{?}$ or $Z_{?} \times Z_{?} .$

Hint.

Task 17.19.a.