Let $G$ and $H$ be groups and let $\phi: G \to H$ be a group homomorphism.
Suppose that $f:G\to H$ is bijective.
Then there exists a map $\psi:H\to G$ such that
\[\psi \circ \phi=\id_G \text{ and } \phi \circ \psi=\id_H.\]
Then prove that $\psi:H \to G$ is also a group homomorphism.

Isomorphism Criterion of Semidirect Product of Groups
Let $A$, $B$ be groups. Let $\phi:B \to \Aut(A)$ be a group homomorphism.
The semidirect product $A \rtimes_{\phi} B$ with respect to $\phi$ is a group whose underlying set is $A \times B$ with group operation
\[(a_1, b_1)\cdot (a_2, b_2)=(a_1\phi(b_1)(a_2), b_1b_2),\]
where $a_i […]

Injective Group Homomorphism that does not have Inverse Homomorphism
Let $A=B=\Z$ be the additive group of integers.
Define a map $\phi: A\to B$ by sending $n$ to $2n$ for any integer $n\in A$.
(a) Prove that $\phi$ is a group homomorphism.
(b) Prove that $\phi$ is injective.
(c) Prove that there does not exist a group homomorphism $\psi:B […]

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 […]

Multiplicative Groups of Real Numbers and Complex Numbers are not Isomorphic
Let $\R^{\times}=\R\setminus \{0\}$ be the multiplicative group of real numbers.
Let $\C^{\times}=\C\setminus \{0\}$ be the multiplicative group of complex numbers.
Then show that $\R^{\times}$ and $\C^{\times}$ are not isomorphic as groups.
Recall.
Let $G$ and $K$ […]

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 […]

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 $f$ […]