Let $G$ be an abelian group, and let $f\colon G\to G\colon g\mapsto g^2$. Assume $G$ has no elements of order 4. Show that $f_{|\,f[G]}$ is an isomorphism.
We're looking at $$f_{|\,f[G]}\colon f[G]\to G\colon g\mapsto g^2. $$ So let $a\in f[G]$. We can write $a=x^2$ for some $x\in G$. We want to show that $a\neq e$, which is equivalent to showing that $x\neq e$.
I see what goes wrong if $G$ has an element of order 4, let's call it (again) $x$. Then we have $x^2\neq e$, but $(x^2)^2=e$, so the kernel isn't just $\{e\}$.
I have also checked that we don't have to worry about elements with odd order or with order 2.
So what's left checking is elements of even order. Assume $\operatorname{ord}(x)=2m$, where $m\neq 2$ is even. Then $x^{m}\neq e$, and since $m$ is even, we may as well write $x^{m}=x^{2k}$. Now $x^{k}\in G$, and $y=(x^{k+1})^2\in f[G]$. While we do have that $y^2=e$, we've also found an element, $x^{k}$, of order 4. Now assume $m\neq 1$ is odd. Then we can write $m=2k+1$. However, I can't follow the same logic as above, because I would have to deal with $k+1/2$.
Any ideas on how to proceed?
We are looking at: $$f \upharpoonright_{f[G]} : f[G] \to \color{red}{f[G]}$$
We are trying to prove that this is an isomorphism.
Note that this fails for infinite groups, the simplest example being $G=\Bbb Z$.
So $G$ is finite. Then, so is $f[G]$. So, it remains to show that $f \upharpoonright_{f[G]}$ is injective.
It suffices to show that $\ker f \upharpoonright_{f[G]} = \{e\}$.
Let $g \in \ker f \upharpoonright_{f[G]}$, i.e. $g^2 = e$. Since $g \in f[G]$, write $g = h^2$, so $h^4 = e$. Since the order of $h$ is not $4$, it is at most $2$, i.e. $h^2 = e$, i.e. $g = e$. $\square$