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Section 10.2 Discovery guide

Note 10.2.1.

Make sure you recall what you read about demonstrating that two groups are isomorphic or not isomorphic in the Pre-read section.

Subsection External products

In the first few activities, assume \(G\) and \(H\) are groups. To distinguish between them, write \(e_G\) for the identity element in \(G\) and \(e_H\) for the identity element in \(H\text{.}\)

Discovery 10.1.

What ordered pair \((g,h)\) acts as the identity element in \(G \times H\text{?}\)

Discovery 10.2.

What is the inverse of an ordered pair \((g,h)\) in \(G \times H\text{?}\) Be sure you can justify your answer.

Discovery 10.3.

If \(G\) and \(H\) are both finite, express the number of elements in \(G \times H\) as a formula in the number of elements in \(G\) and the number of elements in \(H\text{.}\) Be sure you can justify your answer.

Discovery 10.4.

Consider

\begin{equation*} \R^2 = \R \times \R \end{equation*}

as in Example 10.1.2. Determine two “obvious” examples of a subgroup of \(\R^2\) that is isomorphic to \(\R\text{.}\) (Make sure you describe the elements of your example subgroups as ordered pairs.)

Discovery 10.5.

Building off the patterns in Discovery 10.4, demonstrate that a product group \(G \times H\) always contains a subgroup that is isomorphic to \(G\) and a subgroup that is isomorphic to \(H\text{.}\)

Discovery 10.6.

Verify that the collection of elements of the form

\begin{equation*} (g,g) \end{equation*}

(i.e. with repeated components) is always a subgroup of \(G \times G\text{,}\) and that this subgroup is isomorphic to \(G\text{.}\)

(This is called the diagonal embedding of \(G\) into \(G \times G\text{.}\))

Discovery 10.7.

Verify that \(H \times G\) is always isomorphic to \(G \times H\) by providing an explicit isomorphism between them.

Discovery 10.8.

Verify that if \(G\) and \(H\) are both Abelian, then so is \(G \times H\text{.}\)

Subsection Product of cyclic groups

We have seen that a product of Abelian groups is Abelian. Will the same result hold if we replace “abelian” by “cyclic”? Let's find out.

Discovery 10.9.

Every finite cyclic group is isomorphic to \(\Z_n\text{,}\) where \(n\) is the order of the generator. So let's concentrate on different instances of that group.

(a)

Determine whether \(\Z_2 \times \Z_3\) is cyclic.

Hint.

To show that a group is cyclic, you need to demonstrate a generator element for the group, so that every other element is somehow equal to a power (or, in this case of an additive group, an integer multiple) of that generator.

(b)

Determine whether \(\Z_2 \times \Z_2\) is cyclic.

(c)

Determine whether \(\Z_2 \times \Z_4\) is cyclic.

(d)

Determine whether \(\Z_5 \times \Z_6\) is cyclic.

(e)

Determine whether \(\Z_4 \times \Z_6\) is cyclic.

Discovery 10.10.

Make a conjecture about the pattern of Discovery 10.9: the product \(\Z_m \times \Z_n\) will be cyclic precisely when .

Subsection Internal products

Discovery 10.11.

Let \(H\) represent the subgroup of \(\GL_3(\R)\) consisting of the invertible \(3 \times 3\) matrices of the form

\begin{equation*} \begin{bmatrix} a \amp b \amp 0 \\ c \amp d \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix} \text{,} \end{equation*}

and let \(K\) represent the subgroup of \(\GL_3(\R)\) consisting of the invertible \(3 \times 3\) matrices of the form

\begin{equation*} \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp x \end{bmatrix} \text{.} \end{equation*}

(See Discovery 9.3 and Discovery 9.4.)

(a)

What is the form of matrices in the product set \(H K\text{?}\)

(b)

Based on Task a, demonstrate that \(H K\) is a subgroup of \(\GL_3(\R)\text{.}\)

(c)

Devise an isomorphism between the product group \(H \times K\) and the product set \(H K\text{.}\)

In the Pre-read section, you were warned that an internal product \(H K\) of subgroups \(H\) and \(K\) is not always itself a subgroup. The previous activity provided an example where the product \(H K\) is a subgroup, and moreover is isomorphic to the external product of the two subgroups. The next activity demonstrates that the latter condition doesn't always hold either.

Discovery 10.12.

Recall that

\begin{equation*} D_3 = \{ e, r, r^2, s, r s, r^2 s \} \text{,} \end{equation*}

with

\begin{align} r^3 \amp = e, \amp s^2 \amp = e, \amp s r = r^2 s\text{,}\tag{✶} \end{align}

is the group of rotational symmetries of a thin triangular plate.

(a)

Our presentation of \(D_3\) sure looks like a product of the two subgroups

\begin{align*} H \amp = \gen{r}, \amp K = \amp \gen{s} \text{.} \end{align*}

Verify that

\begin{equation*} D_3 = H K \text{.} \end{equation*}
(b)

Demonstrate that \(D_3\) (hence, \(H K\)) is not isomorphic to the external product group \(H \times K\text{.}\)

Hint.

Both \(H\) and \(K\) are cyclic. Consider what you learned in Discovery 10.9.

Discovery 10.13.

Recall that

\begin{equation*} D_6 = \{ e, r, r^2, r^3, r^4, r^5, s, r s, r^2 s, r^3 s, r^4 s, r^5 s \} \text{,} \end{equation*}

with

\begin{align*} r^6 \amp = e, \amp s^2 \amp = e, \amp s r = r^5 s \text{,} \end{align*}

is the group of rotational symmetries of a thin hexagonal plate. Similar to Discovery 10.12, it will still fail to be the case that this dihedral group is isomorphic to

\begin{equation*} \gen{r} \times \gen{s} \text{.} \end{equation*}

However, we might have more success if we “decompose” \(D_6\) into a product in a different way.

(a)

Demonstrate that

\begin{equation*} K = \{ e, r^2, r^4, s, r^2 s, r^4 s \} \end{equation*}

is a subgroup of \(D_6\text{.}\)

(b)

Demonstrate that \(D_6 = H K\text{,}\) where \(H = \gen{r^3}\text{.}\)

(c)

Convince yourself that \(D_6\) is isomorphic to the external product group \(H \times K\text{,}\) by

  • devising a bijective correspondence (your calculations in Task b should essentially take care of this), and

  • computing a few examples to check that the correspondence appears to be operation-preserving (but here, on the \(D_6\) side don't merely restrict yourself to the types of product computations you performed in Task b).

Discovery 10.14.

Interpret the result of Discovery 10.13 geometrically using Figure 10.2.2. (Be careful about the axis of rotation you use for \(s\text{.}\))

A thin triangular plate inscribed in a thin hexagonal plate.
Figure 10.2.2. A thin triangular plate inscribed in a thin hexagonal plate.

Discovery 10.15.

In Discovery 9.10 and Discovery 9.11 we discovered that

\begin{gather} \Or_2(\R) = H K \tag{✶✶} \end{gather}

for

\begin{align*} H \amp = \gen{A_0}, \amp K = \SO_2(\R) \text{,} \end{align*}

where \(A_0\) is any fixed choice of reflection matrix. (Recall that reflections always have order \(2\text{.}\)) In Chapter 9 of the textbook we read that the same decomposition works for \(\Or_3(\R)\text{.}\)

For \(n = 2\text{,}\) let's take \(A_0\) to be reflection in the \(y\)-axis:

\begin{equation*} A_0 = \left[\begin{array}{rr} -1 \amp 0 \\ 0 \amp 1 \end{array}\right] \text{.} \end{equation*}
(a)

A general \(2 \times 2\) rotation matrix takes the form

\begin{equation*} R_\theta = \left[\begin{array}{rr} \cos \theta \amp -\sin \theta \\ \sin \theta \amp \cos \theta \end{array}\right] \text{.} \end{equation*}

Compute \(R_\theta A_0\text{.}\)

(b)

As \(\Or_2(\R)\) is a group, the result of \(R_\theta A_0\) is also in that group. And since this result is a reflection (it has determinant \(-1\)), the decomposition of \(\Or_2(\R)\) in (✶✶) says that there should be another rotation matrix \(R_{\theta'}\) so that

\begin{equation*} R_\theta A_0 = A_0 R_{\theta'} \text{.} \end{equation*}

Determine that second rotation matrix \(R_{\theta'}\text{.}\)

(c)

Explain why Task a and Task b together demonstrate that the “obvious” correspondence between

\begin{equation*} \Or_2(\R) = \gen{A_0} \cdot \SO_2(\R) \end{equation*}

and the external product group

\begin{equation*} \gen{A_0} \times \SO_2(\R) \end{equation*}

defined by

\begin{equation*} A_0^m R_\theta \leftrightarrow \bigl(A_0^m, R_\theta\bigr) \end{equation*}

(where \(m\) is \(0\) for a rotation and \(1\) for a reflection) is not operation-preserving.

Hint.
You've computed \(R_\theta A_0 \) in Task 10.15.a. Use Task 10.15.b to determine to which element of the external group product this computation result corresponds. Then transfer
\begin{equation*} R_\theta = A_0^0 R_\theta \end{equation*}
and
\begin{equation*} A_0 = A_0^1 R_0 \end{equation*}
over to the external product group separately, perform multiply these two elements together (in the correct order!) in that product group, and see if you get the same result.
(d)

In contrast to the patterns for \(n = 2\) so far in this activity, things do work out for \(n = 3\) is we use the reflection matrix \(A_0 = - I_3\text{,}\) where \(I_3\) is the \(3 \times 3\) identity matrix (though you don't need to verify this).

See if you can figure out what property of our \(2 \times 2\) choice of \(A_0\) caused

\begin{equation*} \Or_2(\R) \isoto \gen{A_0} \times \SO_2(\R) \end{equation*}

to fail, and compare with our \(3 \times 3\) choice of \(A_0\) above, for which

\begin{equation*} \Or_3(\R) \isoto \gen{A_0} \times \SO_3(\R) \end{equation*}

actually works.

Hint.

Order of matrix multiplication (usually) matters!