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 $\psi:B \to A$ such that $\psi \circ \phi=\id_A$.

Proof.

(a) Prove that $\phi$ is a group homomorphism.

For any integers $m, n \in A$, we have
\begin{align*}
\phi(m+n)&=2(m+n)\\
&=2m+2n\\
&=\phi(m)+\phi(n).
\end{align*}
Thus, the map $\phi$ is a group homomorphism.

(b) Prove that $\phi$ is injective.

Suppose that we have
\[\phi(m)=\phi(n)\] for some integers $m, n\in A$.
This yields that we have $2m=2n$, and hence $m=n$.
So $\phi$ is injective.

Since $\phi$ is a group homomorphism, we can also prove the injectivity by showing that $\ker(\phi)=\{0\}$.
(For this, see the post “A Group Homomorphism is Injective if and only if the Kernel is Trivial“.)

Suppose that we have
\[\phi(m)=0.\] Then we have $2m=0$, and hence $m=0$.
It follows that the group homomorphism $\phi$ is injective.

(c) Prove that there does not exist a group homomorphism $\psi:B \to A$ such that $\psi \circ \phi=\id_A$.

Seeking a contradiction, assume that there exists a group homomorphism $\psi:B \to A$ such that $\psi \circ \phi =\id_A$.
Then we compute
\begin{align*}
&1=\id_A(1)=\psi \circ \phi(1)\\
&=\psi(2)=\psi(1+1)\\
&=\psi(1)+\psi(1) && \text{since $\psi$ is a group homomorphism}\\
&=2\psi(1).
\end{align*}
It yields that
\[\psi(1)=\frac{1}{2}.\] However note that $\psi(1)$ is an element in $A$, thus $\psi(1)$ is an integer.
Hence we got a contradiction, and we conclude that there is no such $\psi$.

The post Injective Group Homomorphism that does not have Inverse Homomorphism first appeared on Problems in Mathematics.

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 $G$ to a group $H$ is a map $f:G \to H$ such that we have
\[f(gg’)=f(g)f(g’)\] for any elements $g, g\in G$.

If the group operations for groups $G$ and $H$ are written additively, then a group homomorphism $f:G\to H$ is a map such that
\[f(g+g’)=f(g)+f(g’)\] for any elements $g, g’ \in G$.

Here is a hint for the problem.
For any integer $n$, write it as
\[n=1+1+\cdots+1\] and compute $f(n)$ using the property of a homomorphism.

Proof.

Let us put $a:=f(1)\in \Z$. Then for any integer $n$, writing
\[n=1+1+\cdots+1,\] we have
\begin{align*}
f(n)&=f(1+1+\cdots+1)\\
&=f(1)+f(1)+\cdots+f(1) \quad \text{ since } f \text{ is a homomorphism}\\
&=a+a+\cdots+a\\
&=an.
\end{align*}
Thus we have $f(n)=an$ with $a=f(1)\in \Z$ as required.

The post A Homomorphism from the Additive Group of Integers to Itself first appeared on Problems in Mathematics.