# A Group Homomorphism that Factors though Another Group

## Problem 490

Let $G, H, K$ be groups. Let $f:G\to K$ be a group homomorphism and let $\pi:G\to H$ be a surjective group homomorphism such that the kernel of $\pi$ is included in the kernel of $f$: $\ker(\pi) \subset \ker(f)$.

Define a map $\bar{f}:H\to K$ as follows.
For each $h\in H$, there exists $g\in G$ such that $\pi(g)=h$ since $\pi:G\to H$ is surjective.
Define $\bar{f}:H\to K$ by $\bar{f}(h)=f(g)$.

(a) Prove that the map $\bar{f}:H\to K$ is well-defined.

(b) Prove that $\bar{f}:H\to K$ is a group homomorphism.

## Proof.

### (a) Prove that the map $\bar{f}:H\to K$ is well-defined.

Let $h\in H$. Suppose that there are two elements $g, g’\in G$ such that $\pi(g)=h, \pi(g’)=h$.
Then we have
\begin{align*}
\pi(gg’^{-1})=\pi(g)\pi(g’)^{-1}=hh^{-1}=1
\end{align*}
since $\pi$ is a homomorphism.
Thus,
$gg’^{-1}\in \ker(\pi) \subset \ker(f).$ It yields that $f(gg’^{-1})=1$.
It follows that
\begin{align*}
1=f(gg’^{-1})=f(g)f(g’)^{-1},
\end{align*}
and hence we have
$f(g)=f(g’).$

Therefore, the definition of $\bar{f}$ does not depend on the choice of elements $g\in G$ such that $\pi(g)=h$, hence it is well-defined.

### (b) Prove that $\bar{f}:H\to K$ is a group homomorphism.

Our goal is to show that for any elements $h, h’\in H$, we have
$\bar{f}(hh’)=\bar{f}(h)\bar{f}(h’).$

Let $g, g’$ be elements in $G$ such that
$\pi(g)=h \text{ and } \pi(g’)=h’.$ Then by definition of $\bar{f}$, we have
$\bar{f}(h)=f(g) \text{ and } \bar{f}(h’)=f(g’) \tag{*}.$

Since $\pi$ is a homomorphism, we have
\begin{align*}
hh’=\pi(g)\pi(g’)=\pi(gg’).
\end{align*}
By definition of $\bar{f}$, we have
$\bar{f}(hh’)=f(gg’).$

Since $f$ is a homomorphism, we obtain
\begin{align*}
\bar{f}(hh’)&=f(gg’)\\
&=f(g)f(g’)\\
&\stackrel{(*)}{=} \bar{f}(h)\bar{f}(h’).
\end{align*}
This proves that $\bar{f}$ is a group homomorphism.

### More from my site

• 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 […]
• A 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 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 $h\in H$ […]
• Subgroup 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 […]
• Group 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 $G$ […]
• Surjective 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 Homomorphisms From Group of Order 21 to Group of Order 49 Let $G$ be a finite group of order $21$ and let $K$ be a finite group of order $49$. Suppose that $G$ does not have a normal subgroup of order $3$. Then determine all group homomorphisms from $G$ to $K$.   Proof. Let $e$ be the identity element of the group […]
• Abelian Normal subgroup, Quotient Group, and Automorphism Group Let $G$ be a finite group and let $N$ be a normal abelian subgroup of $G$. Let $\Aut(N)$ be the group of automorphisms of $G$. Suppose that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime. Then prove that $N$ is contained in the center of […]

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

##### If a Finite Group Acts on a Set Freely and Transitively, then the Numbers of Elements are the Same

Let $G$ be a finite group and let $S$ be a non-empty set. Suppose that $G$ acts on $S$ freely...

Close