# A Group Homomorphism is Injective if and only if Monic ## 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. Add 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. Add to solve later

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