Let $G$ be a group. Define a map $f:G \to G$ by sending each element $g \in G$ to its inverse $g^{-1} \in G$.
Show that $G$ is an abelian group if and only if the map $f: G\to G$ is a group homomorphism.

$(\implies)$ If $G$ is an abelian group, then $f$ is a homomorphism.

Suppose that $G$ is an abelian group. Then we have for any $g, h \in G$
\begin{align*}
f(gh)&=(gh)^{-1}=h^{-1}g^{-1}\\
&=g^{-1}h^{-1} \text{ since } G \text{ is abelian}\\
&=f(g)f(h).
\end{align*}
This implies that the map $f$ is a group homomorphism.

$(\impliedby)$ If $f$ is a homomorphism, then $G$ is an abelian group.

Now we suppose that the map $f: G \to G$ is a group homomorphism.
Then for any $g, h \in G$, we have
\[f(gh)=f(g)f(h) \tag{*}\]
since $f$ is a group homomorphism.
The left hand side of (*) is
\[f(gh)=(gh)^{-1}=h^{-1}g^{-1}.\]
Thus we obtain from (*) that
\[h^{-1}g^{-1}=g^{-1}h^{-1}.\]
Taking the inverse of both sides, we have
\[gh=hg\]
for any $g, h \in G$.
It follows that $G$ is an abelian group.

Group Homomorphism Sends the Inverse Element to the Inverse Element
Let $G, G'$ be groups. Let $\phi:G\to G'$ be a group homomorphism.
Then prove that for any element $g\in G$, we have
\[\phi(g^{-1})=\phi(g)^{-1}.\]
Definition (Group homomorphism).
A map $\phi:G\to G'$ is called a group homomorphism […]

Pullback 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 $G_1 \times G_2$.
[…]

A Group is Abelian if and only if Squaring is a Group Homomorphism
Let $G$ be a group and define a map $f:G\to G$ by $f(a)=a^2$ for each $a\in G$.
Then prove that $G$ is an abelian group if and only if the map $f$ is a group homomorphism.
Proof.
$(\implies)$ If $G$ is an abelian group, then $f$ is a homomorphism.
Suppose that […]

Abelian Groups and Surjective Group Homomorphism
Let $G, G'$ be groups. Suppose that we have a surjective group homomorphism $f:G\to G'$.
Show that if $G$ is an abelian group, then so is $G'$.
Definitions.
Recall the relevant definitions.
A group homomorphism $f:G\to G'$ is a map from $G$ to $G'$ […]

A Homomorphism from the Additive Group of Integers to Itself
Let $\Z$ be the additive group of integers. Let $f: \Z \to \Z$ be a group homomorphism.
Then show that there exists an integer $a$ such that
\[f(n)=an\]
for any integer $n$.
Hint.
Let us first recall the definition of a group homomorphism.
A group homomorphism from a […]

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 […]

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 […]

Normal 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 […]

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[…] Another problem about the relation between an abelian group and a group homomorphism is: A group homomorphism and an abelian group. […]