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Section 15.4 Normal subgroups and quotient groups: Discovery guide

Subsection Recognizing when two cosets are actually the same coset

Discovery 15.4.

Prove that if \(H\) contains \(\inv{y} x\text{,}\) then the coset \(x H\) contains \(y\text{.}\)

Hint.

All elements of \(x H\) can be expressed in the form \(x h\) for some element \(h\) of \(H\text{.}\)

Subsection Normal subgroups and the quotient group

Discovery 15.5.

Using the pattern of conjugacy classes in \(D_6\) from Discovery 6 of Discovery 14.4, check which of the following subgroups of \(D_6\) satisfy Definition 15.3.2.

(a)

\(\gen{r^3}\text{.}\)

(b)

\(\gen{r}\text{.}\)

(c)

\(\gen{s}\text{.}\)

Discovery 15.7.

Suppose \(A\) and \(B\) are subgroups in a group \(G\text{,}\) with \(A\) entirely contained in \(B\text{.}\) Prove that if \(A\) is normal in \(G\text{,}\) then \(A\) is also normal in \(B\text{.}\)

Careful: It's not as simple as just applying the definition of normal, because conjugacy classes in \(B\) will be different from conjugacy classes in \(G\text{.}\)

Hint.

Apply Condition 2 of Fact 15.3.5 twice, first to the assumption that \(A\) is normal in \(G\) (using conjugates \(\inv{g} a g\) for \(a\) in \(A\) and \(g\) in \(G\)), and second to the desired conclusion that \(A\) is normal in \(B\) (using conjugates \(\inv{b} a b\) for \(a\) in \(A\) and \(b\) in \(B\)).

Discovery 15.9.

In this activity we will explore what happens if we “divide out” by the whole group.

(a)

Verify that \(G\) is always a normal subgroup of itself.

(b)

How many cosets of \(G\) are there? So how many elements does the group \(G/G\) have?

(c)

What “form” does the group \(G/G\) take? That is, to what group is it isomorphic?

Discovery 15.10.

In this activity we will explore what happens if we “divide out” by the trivial subgroup.

(a)

Verify that \(\{e\}\) is always a normal subgroup of the group in which it resides.

(b)

How many cosets of \(\{e\}\) are there? So how many elements does the group \(G/\{e\}\) have?

(c)

What “form” does the group \(G/\{e\}\) take? That is, to what group is it isomorphic?

Discovery 15.11.

When \(H\) is a normal subgroup of \(G\text{,}\) which coset of \(H\) acts as the identity element in the quotient group \(G/H\text{?}\)

Hint.

Re-read the identity group axiom to remind yourself of the defining properties of the identity element in a group, and then consider equality (✶) in Fact 15.3.3.

Discovery 15.12.

(a)

Verify that every subgroup of an Abelian group is normal.

(b)

Using equality (✶) in Fact 15.3.3, verify that if \(G\) is Abelian and \(H\) is a subgroup (normal by Task a), then

\begin{equation*} x H \cdot y H \qquad \text{and} \qquad y H \cdot x H \end{equation*}

always evaluate to the same result. (This verifies that \(G/H\) is always Abelian when \(G\) is Abelian.)

Discovery 15.13.

Recall that the index of a subgroup is the number of cosets. Suppose \(H\) is a subgroup of index two. Without peeking in the textbook, use Condition 5 of Fact 15.3.5 to prove that \(H\) must be normal.

Hint.

The left cosets of \(H\) partition \(G\text{.}\) And so do the right cosets. Compare these two partitions.

Discovery 15.14.

Use the result of Discovery 15.13 to verify that each of the following subgroups is normal in the stated parent group.

(a)

\(A_n\) in \(S_n\text{.}\)

(b)

\(\gen{r}\) in \(D_n\text{.}\)

(c)

\(\SO_n(\R)\) in \(\Or_n(\R)\text{.}\)

Discovery 15.15.

In this activity we will practise analyzing a quotient group in a specific example. Consider \(G = D_n\text{,}\) wherein we have algebra rules

\begin{align*} r^n \amp = e, \amp s^2 \amp = e, \amp s r^k = r^{n - k} s \text{.} \end{align*}

Assume that \(n\) is even, and let \(H = \gen{r^2}\text{.}\)

(a)

We know \(\grporder{G} = 2 n\) and \(\grporder{r^2} = n/2\text{.}\) So how many left cosets of \(H\) in \(G\) must there be?

(b)

Choose a representative and write out the elements of each of the cosets of \(H\text{.}\) Then verify that each left coset is the same as the corresponding right coset. (This will verify that \(H\) is normal.)

(c)

Recall that we know the forms of all possible groups of the order you calculated in Task a. (Refer to your work from Discovery 13.6.) Determine which form \(D_n/\gen{r^2}\) must take by determining whether it is cyclic or not.

(d)

Based on Task c, is \(D_n/\gen{r^2}\) Abelian? Is \(D_n\) Abelian?

Discovery 15.16.

In this activity, we will verify that equality (✶) in Fact 15.3.3 is independent of the choice of coset representatives \(x\) and \(y\text{.}\)

Assume that \(H\) is a subgroup of a group \(G\text{.}\)

(a)

Prove that if \(y_1 H = y_2 H\text{,}\) then also \((x y_1) H = (x y_2) H\text{.}\)

Hint.

Apply Fact 15.3.1 to both the assumption and the desired conclusion. Don't forget to reverse order when you invert a product!

(b)

Assume that \(H\) is normal. Prove that if \(x_1 H = x_2 H\text{,}\) then also \((x_1 y) H = (x_2 y) H\text{.}\)

Hint.

Apply Fact 15.3.1 to both the assumption and the desired conclusion. Don't forget to reverse order when you invert a product!

Subsection Commutators

Discovery 15.17.

In this activity we will compute the commutator elements in \(D_n\) for \(n\) even.

Recall that every element of \(D_n\) is of one of the following two forms,

\begin{equation*} r^m, \qquad r^m s \text{,} \end{equation*}

for some exponent \(m\) between \(0\) and \(n - 1\text{,}\) inclusive. Also recall the algebra rules in \(D_n\text{:}\)

\begin{align*} r^n \amp = e, \amp s^2 \amp = e, \amp s r^k = r^{n - k} s \text{.} \end{align*}
(a)

Simplify each of the four basic commutator element types below into one of the two forms of \(D_n\) elements above.

(i)

\(\commutator{(r^m)}{(r^p)}\)

(ii)

\(\commutator{(r^m s)}{(r^p)}\)

(iii)

\(\commutator{(r^m)}{(r^p s)}\)

(iv)

\(\commutator{(r^m s)}{(r^p s)}\)

(b)

From your computations, describe which of the \(2 n\) elements of \(D_n\) are commutator elements.

Discovery 15.18.

Verify that the identity \(e\) is always a commutator element of the group in which it resides.

Discovery 15.19.

Verify that an inverse of a commutator element is also a commutator element. (Careful: When you invert a product you must reverse the order of multiplication!)

Discovery 15.20.

We know \(\commutatorgrp{G}\) is always a subgroup because we have defined it in terms of a generating set (the commutator elements). In this activity we will verify that it is normal.

(a)

By inserting some \(g \inv{g}\) factors (which is just \(e\text{,}\) after all) and rearranging brackets, verify that if we conjugate a commutator element,

\begin{equation*} \inv{g} (\commutator{x}{y}) g \text{,} \end{equation*}

the result is also a commutator element.

(b)

Use the same sort of trick as in Task a to verify that a conjugate of a product of any (finite) number of commutator elements is also a product of commutator elements.

Discovery 15.21.

In this activity we will verify that \(G/\commutatorgrp{G}\) is always Abelian. To do this, we must verify that coset products

\begin{equation*} x \commutatorgrp{G} \cdot y \commutatorgrp{G} \qquad \text{and} \qquad y \commutatorgrp{G} \cdot x \commutatorgrp{G} \end{equation*}

always have the same result.

Use equality (✶) in Fact 15.3.3 and Fact 15.3.1 to do so.

(Remember: When you invert a product you must reverse the order of multiplication!)

Discovery 15.22.

In this activity we will verify that if \(K\) is a normal subgroup of \(G\) so that \(G/K\) is Abelian, then \(K\) must contain all of \(\commutatorgrp{G}\text{.}\) As \(K\) is assumed to be a subgroup, it is closed under products and inverses, so it suffices to show that \(K\) contains each of the generating elements of \(\commutatorgrp{G}\) (i.e. the commutator elements).

So consider commutator \(\commutator{x}{y}\text{.}\) The assumption that \(G/K\) is Abelian means that coset products

\begin{equation*} \inv{x} K \cdot \inv{y} K \qquad \text{and} \qquad \inv{y} K \cdot \inv{x} K \end{equation*}

have the same result. (There is a reason those inverses are in there, hang tight ….)

Now apply equality (✶) in Fact 15.3.3 and Fact 15.3.1 to arrive at the conclusion that \(\commutator{x}{y}\) must be in \(K\text{.}\)

(Remember: When you invert a product you must reverse the order of multiplication!)