Injective Group Homomorphism that does not have Inverse Homomorphism
Problem 443
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 $\psi \circ \phi=\id_A$.
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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 $\psi$.
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