Prove that, if $A$ is a square matrix,
$A$ has a group inverse if and only if its index is 1
Definitions:
A matrix $G$ is said to be group inverse of $A$ if it satisfies, $$AGA=A$$ $$GAG=G$$ $$AG=GA$$
Index $1$, means that rank($A$)=rank($A^2$).
Work done:
One way is easy, since, $\rho(A)=\rho( AGA)=\rho(GAA)\leq \rho(A^2)\leq\rho(A)$ where $\rho$ represents rank.
And for the otherway round, if $rank(A)=rank(A^2)$ then their column spaces are same. But further how to proceed??
Any hint will be helpful.
I will express this in terms of endomorphisms of a finite dimensional vector space $V$. Assume $f\colon V\to V$ is such that $\dim\operatorname{im}(f)=\dim\operatorname{im}(f^2)$. Since $\operatorname{im}(f^2)\subseteq\operatorname{im}(f)$ and the dimensions are finite and equal, we have $\operatorname{im}(f)=\operatorname{im}(f^2)$. We know that $f$ restricts to an automorphism $\varphi\colon\operatorname{im}(f)\to\operatorname{im}(f)$, since it take $\operatorname{im}(f)$ surjectively to $\operatorname{im}(f^2)=\operatorname{im}(f)$ and surjective endomorphisms on finite dimensional vector spaces are automorphisms.
Now take any complement $U$ of $\operatorname{im}(f)$, that is $V = \operatorname{im}(f) \oplus U$, and let \begin{align*} g\colon V = \operatorname{im}(f) \oplus U &\longrightarrow V, \\ (v,w) &\longmapsto \varphi^{-1}(v). \end{align*}
Now we can verify that \begin{align*} f\circ g\circ f &= f,\\ g\circ f\circ g &= g,\\ f\circ g &= g\circ f. \end{align*}