Group Homomorphism, Preimage, and Product of Groups

Group Theory Problems and Solutions in Mathematics

Problem 208

Let $G, G’$ be groups and let $f:G \to G’$ be a group homomorphism.
Put $N=\ker(f)$. Then show that we have
\[f^{-1}(f(H))=HN.\]

Add to solve later

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 $h\in H$ such that $f(g)=f(h)$.
Since $f$ is a group homomorphism, we obtain
\[f(h^{-1}g)=e’,\] where $e’$ is the identity element of the group $G’$.

This implies that $h^{-1}g\in \ker(f)=N$, hence we have
\begin{align*}
g\in hN\subset HN.
\end{align*}
Therefore we have $f^{-1}(f(H)) \subset HN$.

$(\supset)$ On the other hand, let $g\in HN$ be an arbitrary element.
Then we can write $g=hn$ with $h \in H$ and $n\in N$.
We have
\begin{align*}
f(g)&=f(hn)=f(h)f(n)\\
&=f(h)e’=f(h)\in f(H)
\end{align*}
since $f$ is a group homomorphism and $f(n)=e’$.
Thus we have
\[g\in f^{-1}(f(H))\] and $f^{-1}(f(H)) \supset HN$.

Therefore, putting the two continents together gives
\[f^{-1}(f(H))=HN\] as required.

Add to solve later

More from my site

$Surjective Group Homomorphism to $\Z$ and Direct Product of Abelian Groups$ 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 […]
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 […]
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 […]
The Preimage of a Normal Subgroup Under a Group Homomorphism is Normal Let $G$ and $G'$ be groups and let $f:G \to G'$ be a group homomorphism. If $H'$ is a normal subgroup of the group $G'$, then show that $H=f^{-1}(H')$ is a normal subgroup of the group $G$. Proof. We prove that $H$ is normal in $G$. (The fact that $H$ is a subgroup […]
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 […]
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 $G$ […]
A Group Homomorphism that Factors though Another Group 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 […]