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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?

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    $\begingroup$ You mean nonzero, not non-identity. $\endgroup$ – Ink May 21 '13 at 3:42
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    $\begingroup$ For question 2 here is an easy counterexample: the group $\mathbb{Q}$ of all rational numbers under addition. $\endgroup$ – Dan Shved May 21 '13 at 5:54
  • $\begingroup$ @Dan Shved: Great! for infinite groups, I thought only up to $\mathbb{Z}$, and hence posted the question. Nice answer. I would like to vote it as "Answer" than comment; please write it in "Answer". $\endgroup$ – RDK May 21 '13 at 7:37
  • $\begingroup$ @Brian: Here I like to write "non-identity" than "non-trivial"; since by by "non-identity element" of a group has meaning in its name, I tought "non-identity function" will be too meaningful for reader. So "non-identity homomorphism", means a homomorphism which is different from $g\mapsto 1$ $\forall g\in G$. $\endgroup$ – RDK May 21 '13 at 7:41
  • $\begingroup$ @RDK: but I would interpret "non-identity homomorphism" as a homomorphism which is not the identity map $g \mapsto g$. $\endgroup$ – Qiaochu Yuan May 21 '13 at 18:25
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Nice question! The answer is no. The smallest potential counterexample works.

Let $G$ be the binary icosahedral group. This is a perfect group of order $120$ (in fact the smallest nontrivial such group which is not simple). Its only nontrivial normal subgroup is its center $\pm 1$, hence its only nontrivial quotient is the icosahedral group $A_5$, which cannot occur as a subgroup of $G$, since any subgroup of $G$ of order $60$ is necessarily normal. Hence $G$ is not simple but every nonzero homomorphism $G \to G$ is an isomorphism.

Dan Shved gives an example for your second question in the comments.

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  • $\begingroup$ Nice answer! I was already starting to type exactly the same counterexample, but you beat me to it :) $\endgroup$ – Dan Shved May 21 '13 at 6:15
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    $\begingroup$ Incidentally, if @Landscape is reading this, $G$ is isomorphic to $\text{SL}_2(\mathbb{F}_5)$. Unfortunately, $\text{SL}_2(\mathbb{F}_3)$ is too small to be a counterexample; in fact it's solvable. $\endgroup$ – Qiaochu Yuan May 21 '13 at 7:09
  • $\begingroup$ Also, $G = \operatorname{SL}_2(F)$ is an example for any finite field $F$ other than $\mathbb{F}_2$ and $\mathbb{F}_3$. The reason is the same, the only nontrivial proper subgroup of $G$ is $Z(G) = \{I, -I\}$. $\endgroup$ – Mikko Korhonen May 21 '13 at 10:12
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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.

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  • $\begingroup$ Very nice. Thanks!! $\endgroup$ – RDK May 21 '13 at 7:45

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