A Group Homomorphism that Factors though Another Group

Group Theory Problems and Solutions in Mathematics

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.

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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.

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