TITLE: Schur's Lemma in Group Theory
QUESTION [17 upvotes]: The analogue of well celebrated Schur's Lemma in group theory will be 
"If $G$ is a finite simple group, and $\phi$ is a non-identity homomorphism from $G$ to $G$, then $\phi$ is an isomorphism". 
The proof follows exactly same lines as it is in representation theory.
I would like to ask the following question, which is a converse of Schur's lemma in some sense.
Question 1: If $G$ is a finite group, $|G|>2$, such that every non-identity homomorphism $\phi\colon G\rightarrow G$ is an automorphism, should $G$ necessarily simple?

The progress towards the solution I made is the following: if $G$ satisfies the hypothesis in the question and $G$ is non-abelian, then $G$ is not solvable: if $G$ is non-abelian, solvable, then it has non-identity abelian quotient $G^{ab}=G/[G,G]$. Let $p$ be a prime divisor of $|G^{ab}|$ (hence $p$ also divides $|G|$). Then there is a surjective homomorphism from $G^{ab}$ to $C_p$ (cyclic group of order $p$). We can embedd $C_p$ in $G$. Then, consider the composition of the homomorphisms:
$G\rightarrow G^{ab} \rightarrow C_p\rightarrow G$. It is a non-identity homomorphism from $G$ to $G$ which is clearly not an automorphism. 
Note that if $G$ is abelian, which is not cyclic of prime order, then we can easily find non-identity homomorphisms $G\rightarrow G$ which are not automorphisms.

I would like to ask a general question also (remove the condition of finiteness in above question).
Question 2: If $G$ is an infinite group, such that every non-identity homomorphism $\phi\colon G\rightarrow G$ is an automorphism, should $G$ necessarily simple?

REPLY [5 votes]: By request, posting a counterexample for question 2 as an answer.
The group of rational numbers $\mathbb{Q}$ under addition is clearly not simple. But the endomorphism ring $\mathrm{End}(\mathbb{Q})$ is isomorphic to $\mathbb{Q}$ itself, and is a division ring, which means that all non-zero endomorphisms are automorphisms.