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.

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

Suppose that $G$ is an abelian group. We prove that $f:G \to G, a \mapsto a^2$ is a group homomorphism.
Let $a, b$ be arbitrary elements in $G$. Then we have
\begin{align*}
f(ab)&=(ab)^2 && (\text{ by definition of } f)\\
&=(ab)(ab)\\
&=a^2 b^2 && (\text{ since $G$ is abelian})\\
&=f(a)f(b) && (\text{ by definition of } f).
\end{align*}
Therefore, we obtain $f(ab)=f(a)f(b)$ for any $a, b\in G$.
Hence $f$ is a group homomorphism from $G$ to $G$.

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

Suppose that $f:G\to G$ is a group homomorphism. We prove that $G$ is an abelian group.
Let $a, b\in G$. We want to prove that $ab=ba$.

Since $f$ is a group homomorphism, we have
\[f(ab)=f(a)f(b).\]
As a result we have
\[(ab)^2=a^2b^2,\]
or equivalently
\[abab=aabb.\]
Multiplying this by $a^{-1}$ on the left and by $b^{-1}$ on the right, we obtain
\[ab=ba.\]
Since $a$ and $b$ are arbitrary, this implies that $G$ is an abelian group.

A Group Homomorphism and an Abelian Group
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.
Proof.
$(\implies)$ If $G$ is an abelian group, then $f$ […]

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 […]

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 […]

Group Homomorphism, Conjugate, Center, and Abelian group
Let $G$ be a group. We fix an element $x$ of $G$ and define a map
\[ \Psi_x: G\to G\]
by mapping $g\in G$ to $xgx^{-1} \in G$.
Then prove the followings.
(a) The map $\Psi_x$ is a group homomorphism.
(b) The map $\Psi_x=\id$ if and only if $x\in Z(G)$, where $Z(G)$ is the […]

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 […]

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 […]

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

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