A Group Homomorphism is Injective if and only if Monic

Group Theory Problems and Solutions in Mathematics

Problem 243

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 monic.

 
LoadingAdd to solve later

Proof.

$(\implies)$ Injective implies monic

Suppose that $f: G \to G’$ is an injective group homomorphism.
We show that $f$ is monic.

So suppose that we have $fg_1=fg_2$, where $g_1:K\to G$ and $g_2:K \to G$ are group homomorphisms for some group $K$.
Then for any $x\in K$, we have
\[f(g_1(x))=f(g_2(x)).\] Since $f$ is injective, it follows that
\[g_1(x)=g_2(x)\] for any $x\in K$, and thus we obtain $g_1=g_2$. Thus $f$ is monic.

$(\impliedby)$ Monic implies injective

For the opposite implication, we prove the contrapositive statement. Namely, we prove that if $f$ is not injective, then $f$ is not monic.

Suppose that $f$ is not injective. Then the kernel $\ker(f)$ is a non-trivial subgroup of $G$.
We define the group homomorphism $g_1: \ker(f)\to G$ to be the identity map on $\ker(f)$. That is $g_1(x)=x$ for all $x\in \ker(f)$.
Also we define the group homomorphism $g_2:\ker(f)\to G$ by the formula $g_2(x)=e$ for all $x\in \ker(f)$, where $e$ is the identity element of $G$.
Since $\ker(f)$ is a nontrivial group, these two homomorphisms are distinct: $g_1\neq g_2$.

However, we have
\[fg_1=fg_2.\] In fact, we have for $x\in \ker(f)$
\[fg_1(x)=f(x)=e’,\] where $e’$ is the identity element of $G’$,
and
\[fg_2(x)=f(e)=e’.\] Thus, by definition, the homomorphism $f$ is not monic as required to complete the proof.


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 […]
  • Subgroup of Finite Index Contains a Normal Subgroup of Finite IndexSubgroup of Finite Index Contains a Normal Subgroup of Finite Index Let $G$ be a group and let $H$ be a subgroup of finite index. Then show that there exists a normal subgroup $N$ of $G$ such that $N$ is of finite index in $G$ and $N\subset H$.   Proof. The group $G$ acts on the set of left cosets $G/H$ by left multiplication. Hence […]
  • Pullback Group of Two Group Homomorphisms into a GroupPullback Group of Two Group Homomorphisms into a Group Let $G_1, G_1$, and $H$ be groups. Let $f_1: G_1 \to H$ and $f_2: G_2 \to H$ be group homomorphisms. Define the subset $M$ of $G_1 \times G_2$ to be \[M=\{(a_1, a_2) \in G_1\times G_2 \mid f_1(a_1)=f_2(a_2)\}.\] Prove that $M$ is a subgroup of $G_1 \times G_2$.   […]
  • Surjective Group Homomorphism to $\Z$ and Direct Product of Abelian GroupsSurjective Group Homomorphism to $\Z$ and Direct Product of Abelian Groups Let $G$ be an abelian group and let $f: G\to \Z$ be a surjective group homomorphism. Prove that we have an isomorphism of groups: \[G \cong \ker(f)\times \Z.\]   Proof. Since $f:G\to \Z$ is surjective, there exists an element $a\in G$ such […]
  • 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 $h\in H$ […]
  • The Quotient by the Kernel Induces an Injective HomomorphismThe Quotient by the Kernel Induces an Injective Homomorphism 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 $G/\ker{\phi} \to G'$.   Outline. Define $\tilde{\phi}([g])=\phi(g)$ and show that this is well-defined. Show […]
  • 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 $G$ […]
  • 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 […]

You may also like...

Leave a Reply

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

More in Group Theory
Abelian Group problems and solutions
No Finite Abelian Group is Divisible

A nontrivial abelian group $A$ is called divisible if for each element $a\in A$ and each nonzero integer $k$, there...

Close