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Section 16.2 Basics: Discovery guide

Subsection Examples

Discovery 16.1.

For each of the following maps, verify that it is a homomorphism, and then either compute or try to describe the elements in the image and the elements in the kernel.

(a)

\(\funcdef{\varphi}{\Z}{\Z_n}\) defined by taking \(\varphi(k)\) to be the residue of \(k\) modulo \(n\text{.}\)

(b)

\(\funcdef{\varphi}{\Z_3}{\Z_6}\) defined by setting

\begin{align*} \varphi(0) \amp = 0, \amp \varphi(1) \amp = 2, \amp \varphi(2) \amp = 4 \text{.} \end{align*}
(c)

\(\funcdef{\varphi}{\Z_6}{D_6}\) defined by \(\varphi(k) = r^k\text{.}\)

(d)

\(\funcdef{\det}{\GL_n(\R)}{\multunits{\R}}\text{.}\)

(e)

\(\funcdef{\sgn}{S_n}{\{\pm 1\}}\text{,}\) where \(\sgn(p)\) is the sign of the permutation \(p\text{,}\) either \(1\) for an even permutation or \(-1\) for an odd permutation.

(Note: the group \(\{\pm 1\}\) is the multiplicative version of the cyclic group \(\Z_2\text{.}\))

(f)

\(\funcdef{\trace}{\matrixring_n(\R)}{\R}\text{,}\) where \(\matrixring_n(\R)\) is the additive group of \(n \times n\) matrices (both invertible and singular), and the trace of a square matrix is defined to be the sum of its diagonal entries.

(g)

\(\funcdef{\iota}{\Z}{\Q}\) defined by \(\iota(m) = {m \over 1}\text{.}\)

(h)

\(\funcdef{\iota}{\multunits{\Q}}{\multunits{\R}}\) defined by taking \(\iota(a/b)\) to be the decimal number that results from performing long division of \(b\) into \(a\text{.}\)

(Note: You may skip the operation-preserving check for this example.)

(i)

\(\funcdef{\varphi}{\R^2}{\R}\) defined by taking \(\varphi(\vec{v})\) to be the (oriented) distance from the origin to the orthogonal projection of \(\vec{v}\) onto the line \(y = x\text{,}\) where positive distances are to the upper right and negative distances are to the lower left.

(Note: You should perform the check that this map is operation-preserving geometrically. You still remember how to draw a geometric diagram of vector addition in \(\R^2\text{,}\) right?)

(j)

\(H\) is a normal subgroup of group \(G\text{,}\) and \(\funcdef{\rho}{G}{G/H}\) is the “projection” map defined by

\begin{equation*} \rho(x) = x H \text{.} \end{equation*}

That is, \(\rho\) sends each element in \(G\) to the coset in \(G/H\) that contains it.

Subsection Some properties preserved by homomorphisms

Just as with isomorphisms, because a group homomorphism \(\funcdef{\varphi}{G}{G'}\) is operation-preserving it transfers many group-algebra related properties from \(G\) to \(G'\text{.}\)

Once again, distinguish between the two different identity elements by writing \(e\) for the identity element in \(G\) and \(e'\) for the identity element in \(G'\text{.}\)

Discovery 16.2. Homomorphism preserves identity.

Unfortunately the reasoning of Discovery 7.5 does not hold for homomorphisms because in that activity we relied on the bijective property. Here is some slightly more complicated reasoning that works in both the case of a homomorphism and of an isomorphism.

(a)

Verify that

\begin{gather} \varphi(e) \cdot \varphi(e) = \varphi(e)\text{.}\tag{✶} \end{gather}
(b)

Now multiply both sides of (✶) by \(\inv{\bigl(\varphi(e)\bigr)}\) and simplify to arrive at the desired conclusion.

Warning: Do not change \(\inv{\bigl(\varphi(e)\bigr)}\) into \(\varphi(\inv{e})\text{,}\) as we have not yet verified that homomorphisms preserve inverses.

Discovery 16.3. Homomorphism preserves (positive) powers.

Verify that the reasoning of Discovery 7.6 holds for homomorphisms and not just for isomorphisms.

Remark 16.2.1.

As we remarked with isomorphisms, by taking Discovery 16.2 into account, and recalling the convention that \(g^0 = e\) in every group, we can consider the result of Discovery 16.3 to hold for all non-negative integers \(k\text{.}\)

Discovery 16.4. Homomorphism does not preserve order.

(a)

Unfortunately the reasoning of Discovery 7.7 does not hold for homomorphisms. Use one of the example homomorphisms in Discovery 16.1 to demonstrate a specific example of an element \(x\) in \(G\) so that \(x\) and \(\varphi(x)\) have different orders.

(b)

Use Discovery 7.6 to verify that if \(x\) has order \(n\text{,}\) then

\begin{equation*} \bigl(\varphi(x)\bigr)^n = e' \end{equation*}

will hold. So what can be said about the relationship between the order of \(x\) and the order of \(\varphi(x)\text{?}\)

Discovery 16.5. Homomorphism preserves inverses.

(a)

Verify both of the equalities

\begin{align*} \varphi(x) \cdot \varphi(\inv{x}) \amp = e' \text{,} \amp \varphi(\inv{x}) \cdot \varphi(x) \amp = e' \text{.} \end{align*}

Warnings: Do not assume that \(G'\) is Abelian, so the second equality does not follow from the first. Also, do not change \(\varphi(\inv{x})\) into \(\inv{\bigl(\varphi(x)\bigr)}\text{,}\) as that would be assuming our desired final conclusion.

(b)

Explain why the equalities established in Task a imply that we must have

\begin{equation*} \inv{\bigl(\varphi(x)\bigr)} = \varphi(\inv{x}) \text{.} \end{equation*}
Hint.

Apply the conclusion of Discovery 2.8 to the group element \(\varphi(x)\) in \(G'\text{.}\)

Subsection Image and kernel

Discovery 16.7. Image is a subgroup.

This activity will guide you through using The Subgroup Test (Version 1) to verify that the image of a homomorphism \(\funcdef{\varphi}{G}{G'}\) is always a subgroup of \(G'\text{.}\) (Clearly \(\im \varphi = \varphi(G)\) is always non-empty because \(G\) is non-empty.)

(a)

Closed under multiplication. Suppose \(y_1, y_2\) are elements of \(\im \varphi\text{.}\) By definition, this means that there are corresponding elements \(x_1,x_2\) in \(G\) so that

\begin{align*} y_1 \amp = \varphi(x_1), \amp y_2 \amp = \varphi(x_2) \text{.} \end{align*}

Your task is to now demonstrate that the product \(y_1 y_2\) is also in \(\im \varphi\) by filling in the blank in the equality

\begin{equation*} y_1 y_2 = \varphi(\fillinmath{XXXXX}) \end{equation*}

with some element of \(G\text{.}\)

(b)

Closed under taking inverses. Suppose \(y\) is an element of \(\im \varphi\text{.}\) By definition, this means that there is a corresponding element \(x\) in \(G\) so that

\begin{equation*} y = \varphi(x) \text{.} \end{equation*}

Your task is to now demonstrate that the inverse \(\inv{y}\) is also in \(\im \varphi\) by filling in the blank in the equality

\begin{equation*} \inv{y} = \varphi(\fillinmath{XXXXX}) \end{equation*}

with some element of \(G\text{.}\)

Discovery 16.8. Kernel is a subgroup.

Use your preferred version of the Subgroup Test to prove that the kernel of a homomorphism \(\funcdef{\varphi}{G}{G'}\) is always a subgroup of \(G\text{.}\)

Discovery 16.9.

In the case that \(\funcdef{\varphi}{G}{G'}\) is an isomorphism, what must \(\im \varphi\) be? What must \(\ker \varphi\) be?

Discovery 16.10.

Suppose \(\funcdef{\varphi}{G}{G'}\) is a homomorphism so that \(\im \varphi\) is all of \(G'\text{.}\)

(a)

Prove that if \(\varphi\) is actually an isomorphism, then \(\ker \varphi\) contains only \(e\text{.}\)

(b)

Prove the converse: If \(\ker \varphi\) contains only \(e\text{,}\) then \(\varphi\) must be an isomorphism.

Hint.

We are already assuming that \(\varphi\) is a homomorphism, so you only need to check that it is a bijective correspondence. For this, suppose you have elements \(x_1,x_2\) in \(G\) so that

\begin{equation*} \varphi(x_1) = \varphi(x_2) \text{.} \end{equation*}

Try to turn this into

\begin{equation*} \varphi(\fillinmath{XXXXX}) = e' \text{,} \end{equation*}

so that you can appeal to the assumption about \(\ker \varphi\text{.}\)