The Quotient by the Kernel Induces an Injective Homomorphism

Group Theory Problems and Solutions

Problem 4

Let $G$ and $G’$ be a group and let $\phi:G \to G’$ be a group homomorphism. 

Show that $\phi$ induces an injective homomorphism from $G/\ker{\phi} \to G’$.

 

LoadingAdd to solve later

Outline.

  1. Define $\tilde{\phi}([g])=\phi(g)$ and show that this is well-defined.
  2. Show that $\tilde{\phi}$ is a homomorphism.
  3. Show that $\tilde{\phi}$ is injective.

Proof.

Define the map $\tilde{\phi}: G/\ker{\phi} \to G’$ by sending $[g]$ to $\phi(g)$. Here $[g]$ is the element of $G/\ker{\phi}$ represented by $g\in G$. 


We need to show that this is well-defined.
Namely, we need to show that $\tilde{\phi}$ does not depend on the choice of representative. 

So suppose $[g]=[h]$ for $g, h \in G$. Then we have $x:=gh^{-1} \in \ker{\phi}$. Thus we have
\[e’=\phi(x)=\phi(gh^{-1})=\phi(g)\phi(h)^{-1},\]  where $e’\in G’$ is the identity element of $G’$. 
Here the third equality follows because $\phi$ is a homomorphism. 
Hence we obtain $\phi(g)=\phi(h)$, equivalently $\tilde{\phi}([g])=\tilde{\phi}([h])$. Thus $\tilde{\phi}$ is well-defined.


Now we show that $\tilde{\phi}: G/\ker{\phi} \to G’$ is a homomorphism. Let $[g], [h] \in G/\ker{\phi}$. Then we have
\[\tilde{\phi}([g][h])=\tilde{\phi}([gh])=\phi(gh)=\phi(g)\phi(h)=\tilde{\phi}([g])\tilde{\phi}([h]) \] and $\tilde{\phi}$ is a homomorphism.


Finally, we prove that $\tilde{\phi}$ is injective. Suppose that $\tilde{\phi}([g])=e’$. Then this means $\phi(g)=e’$, hence $g\in \ker{\phi}$.
Thus $[g]=[e]$, where $e$ is the identity element of $G$. 
Hence $\tilde{\phi}$ is injective and the proof is complete.


LoadingAdd to solve later

More from my site

  • A Group Homomorphism is Injective if and only if the Kernel is TrivialA Group Homomorphism is Injective if and only if the Kernel is Trivial Let $G$ and $H$ be groups and let $f:G \to K$ be a group homomorphism. Prove that the homomorphism $f$ is injective if and only if the kernel is trivial, that is, $\ker(f)=\{e\}$, where $e$ is the identity element of $G$.     Definitions/Hint. We recall several […]
  • A Group Homomorphism is Injective if and only if MonicA Group Homomorphism is Injective if and only if Monic Let $f:G\to G'$ be a group homomorphism. We say that $f$ is monic whenever we have $fg_1=fg_2$, where $g_1:K\to G$ and $g_2:K \to G$ are group homomorphisms for some group $K$, we have $g_1=g_2$. Then prove that a group homomorphism $f: G \to G'$ is injective if and only if it is […]
  • Group of $p$-Power Roots of 1 is Isomorphic to a Proper Quotient of ItselfGroup of $p$-Power Roots of 1 is Isomorphic to a Proper Quotient of Itself Let $p$ be a prime number. Let \[G=\{z\in \C \mid z^{p^n}=1\} \] be the group of $p$-power roots of $1$ in $\C$. Show that the map $\Psi:G\to G$ mapping $z$ to $z^p$ is a surjective homomorphism. Also deduce from this that $G$ is isomorphic to a proper quotient of $G$ […]
  • Injective Group Homomorphism that does not have Inverse HomomorphismInjective Group Homomorphism that does not have Inverse Homomorphism Let $A=B=\Z$ be the additive group of integers. Define a map $\phi: A\to B$ by sending $n$ to $2n$ for any integer $n\in A$. (a) Prove that $\phi$ is a group homomorphism. (b) Prove that $\phi$ is injective. (c) Prove that there does not exist a group homomorphism $\psi:B […]
  • Dihedral Group and Rotation of the PlaneDihedral Group and Rotation of the Plane Let $n$ be a positive integer. Let $D_{2n}$ be the dihedral group of order $2n$. Using the generators and the relations, the dihedral group $D_{2n}$ is given by \[D_{2n}=\langle r,s \mid r^n=s^2=1, sr=r^{-1}s\rangle.\] Put $\theta=2 \pi/n$. (a) Prove that the matrix […]
  • Group Homomorphism from $\Z/n\Z$ to $\Z/m\Z$ When $m$ Divides $n$Group Homomorphism from $\Z/n\Z$ to $\Z/m\Z$ When $m$ Divides $n$ Let $m$ and $n$ be positive integers such that $m \mid n$. (a) Prove that the map $\phi:\Zmod{n} \to \Zmod{m}$ sending $a+n\Z$ to $a+m\Z$ for any $a\in \Z$ is well-defined. (b) Prove that $\phi$ is a group homomorphism. (c) Prove that $\phi$ is surjective. (d) Determine […]
  • Normal Subgroups, Isomorphic Quotients, But Not IsomorphicNormal Subgroups, Isomorphic Quotients, But Not Isomorphic Let $G$ be a group. Suppose that $H_1, H_2, N_1, N_2$ are all normal subgroup of $G$, $H_1 \lhd N_2$, and $H_2 \lhd N_2$. Suppose also that $N_1/H_1$ is isomorphic to $N_2/H_2$. Then prove or disprove that $N_1$ is isomorphic to $N_2$.   Proof. We give a […]
  • Group Homomorphism, Preimage, and Product of GroupsGroup Homomorphism, Preimage, and Product of Groups Let $G, G'$ be groups and let $f:G \to G'$ be a group homomorphism. Put $N=\ker(f)$. Then show that we have \[f^{-1}(f(H))=HN.\]   Proof. $(\subset)$ Take an arbitrary element $g\in f^{-1}(f(H))$. Then we have $f(g)\in f(H)$. It follows that there exists $h\in H$ […]

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Group Theory
Normal Subgroups Problems and Solutions in Group Theory
A Condition that a Commutator Group is a Normal Subgroup

Let $H$ be a normal subgroup of a group $G$. Then show that $N:=[H, G]$ is a subgroup of $H$...

Close