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Abstract Algebra I MATH 4107 Summer 2026

Dr. Hunter Lehmann

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Section Group Homomorphisms

Worksheet Part 1: Homomorphisms and the First Isomorphism Theorem

The goal of these problems is to explore group homomorphisms and their properties and practice using the First Isomorphism Theorem.

1.

Let \(\R^*\) be the group of nonzero real numbers under multiplication and \(r\) be a positive integer. Show that the map that takes \(x\) to \(x^r\) is a homomorphism from \(\R^*\) to \(\R^*\) and determine the kernel.
Solution.
Let \(\phi:\R^*\to \R^*\) be the given map. Then for any \(x,y\in \R^*\text{,}\) we have
\begin{equation*} \phi(xy)=(xy)^r=x^ry^r=\phi(x)\phi(y)\text{,} \end{equation*}
so \(\phi\) is a homomorphism. The kernel of \(\phi\) is the set of all \(x\in \R^*\) such that \(\phi(x)=1\text{,}\) which is equivalent to \(x^r=1\text{.}\) This means \(x=1\) if \(r\) is odd, and \(x\in\{\pm 1\}\) if \(r\) is even.

2.

In this problem you will determine all possible homomorphisms \(\phi\) from \(\Z_{12}\) to \(\Z_{30}\text{.}\)
(a)
Use the properties of homomorphisms to explain why such a \(\phi\) is totally determined by \(\phi(1)\text{.}\)
Solution.
We have \(\phi(n)=\phi(\underbrace{1+\cdots+1}_{n\text{ times}})=\underbrace{\phi(1)+\cdots+\phi(1)}_{n\text{ times}}=n\phi(1)\text{.}\) So knowing \(\phi(1)\) allows us to compute \(\phi(n)\) for any \(n\in \Z_{12}\text{.}\)
(b)
Using Lagrange’s Theorem and property 3 of Properties of Homomorphisms: Elements, what can we conclude about \(|\phi(1)|\text{?}\)
Solution.
Since \(\phi(1)\in \Z_{30}\text{,}\) we from Lagrange’s Theorem that \(|\phi(1)|\) divides 30. We also have by property 3 of Properties of Homomorphisms: Elements that \(|\phi(1)|\) divides \(|1|=12\text{.}\) Therefore, \(|\phi(1)|\) divides the greatest common divisor of 12 and 30, which is 6.
(c)
Explicitly list all of the homomorphisms from \(\Z_{12}\) to \(\Z_{30}\text{.}\)
Solution.
The homomorphisms from \(\Z_{12}\) to \(\Z_{30}\) are determined by where they send the generator 1. Since \(|\phi(1)|\) must divide 6, the possible values for \(\phi(1)\) are the elements of \(\Z_{30}\) of order dividing 6, i.e. the elements of the unique subgroup of order 6 in \(\Z_{30}\text{.}\) These are \(0, 5, 10, 15, 20, 25\text{.}\) So there are 6 homomorphisms from \(\Z_{12}\) to \(\Z_{30}\text{,}\) given by \(\phi_k(n)=nk\) for \(k=0,5,10,15,20,25\text{.}\)

3.

Show that the map \(\pi_G: G\times H \to G\) given by \(\pi_G(g,h)=g\) is a homomorphism (called the projection of \(G\times H\) onto \(G\) ). What is the kernel of this map?
Solution.
To show that \(\pi_G\) is a homomorphism, let \((g_1,h_1),(g_2,h_2)\in G\times H\text{.}\) Then
\begin{equation*} \pi_G((g_1,h_1)(g_2,h_2))=\pi_G((g_1g_2,h_1h_2))=g_1g_2=\pi_G(g_1,h_1)\pi_G(g_2,h_2)\text{.} \end{equation*}
So \(\pi_G\) is a homomorphism. The kernel of \(\pi_G\) is the set of all \((g,h)\in G\times H\) such that \(\pi_G(g,h)=e\text{,}\) which is equivalent to \(g=e\text{.}\) Therefore, the kernel of \(\pi_G\) is \(\{e\}\times H\text{.}\)

4.

Prove that \((G\times H)/(\{e\}\times H)\isom G\text{.}\)
Solution.
The map \(\pi_G\) is a homomorphism and surjects onto \(G\) since \(\pi_G(g,e)=g\) for any \(g\in G\text{.}\) The kernel of \(\pi_G\) is \(\{e\}\times H\text{,}\) so by the First Isomorphism Theorem we have that \((G\times H)/(\{e\}\times H)\isom G\text{.}\)

5.

Show that the map \(\phi: \Z\times \Z \to \Z\) given by \(\phi(a,b)=a-b\) is a homomorphism. What is the kernel of this map?
Solution.
To show that \(\phi\) is a homomorphism, let \((a_1,b_1),(a_2,b_2)\in \Z\times \Z\text{.}\) Then
\begin{equation*} \phi((a_1,b_1)+(a_2,b_2))=\phi((a_1+a_2,b_1+b_2))=(a_1+a_2)-(b_1+b_2)=(a_1-b_1)+(a_2-b_2)=\phi(a_1,b_1)+\phi(a_2,b_2). \end{equation*}
So \(\phi\) is a homomorphism. The kernel of \(\phi\) is the set of all \((a,b)\in \Z\times \Z\) such that \(\phi(a,b)=0\text{,}\) which is equivalent to \(a=b\text{.}\) Therefore, the kernel of \(\phi\) is \(\{(a,a) \mid a\in \Z\}=\subgroup{(1,1)}\text{.}\)

6.

Prove that \((\Z \times \Z)/\subgroup{(1,1)}\isom \Z\text{.}\)
Solution.
The map \(\phi\) is a homomorphism and surjects onto \(\Z\) since \(\phi(a,0)=a\) for any \(a\in \Z\text{.}\) The kernel of \(\phi\) is \(\subgroup{(1,1)}\text{,}\) so by the First Isomorphism Theorem we have that \((\Z \times \Z)/\subgroup{(1,1)}\isom \Z\text{.}\)

7.

Let \(G\) be an Abelian group, \(G_n=\{g \mid g^n=e\}\text{,}\) and \(G^n=\{g^n \mid g \in G\}\text{.}\) (\(G_n\) is the subgroup of all elements with order dividing \(n\) and \(G^n\) is the subgroup of all \(n\)-th powers.)
Show that \(G/G_n \isom G^n\text{.}\) Why is it important that \(G\) is Abelian in this problem?
Solution.
The map \(\phi: G \to G\) given by \(\phi(g)=g^n\) is a homomorphism since
\begin{equation*} \phi(gh)=(gh)^n=g^nh^n=\phi(g)\phi(h)\text{.} \end{equation*}
The kernel of \(\phi\) is the set of all \(g\in G\) such that \(\phi(g)=e\text{,}\) which is equivalent to \(g^n=e\text{.}\) Therefore, the kernel of \(\phi\) is \(G_n\text{.}\) The image of \(\phi\) is the set of all elements of the form \(g^n\text{,}\) which is exactly \(G^n\text{.}\) So by the First Isomorphism Theorem we have that \(G/G_n \isom G^n\text{.}\) It is important that \(G\) is Abelian in this problem because otherwise the map given by taking an element to its \(n\)th power would not be a homomorphism.
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