Prove/disprove: Let $G$ a simple infinite group $\implies$ $G$ doesn’t have a proper subgroup of finite index.

group-homomorphismgroup-theorynormal-subgroupssolution-verification

I'm writing this question for proof verification and also since I didn't find the already existing answers very clear:

If $G$ is an infinite simple group then any proper subgroup of $G$ has infinite index.

An infinite simple group has no proper subgroup of finite index.

Here's my proof:

By contradiction assume exists $H \le G$ of finite index $\implies\; \exists: 1 \lt m \in \mathbf N$, such that $[G:H]={|G| \over |H|}=m$.

Take X to be the group of left cosets of $H$ in $G$: $\;G/H$.

Let us observe the action $G \mapsto G/H$ by multiplication from the left implementing we have $g.x = gx, \forall g \in G, \forall \;aH \in G/H$.

In particular let us observe the induced homomorphism $\phi: G \to S_m$.

Denote we have: $g \in Ker\phi \iff \phi(g) = Id_{G/H} \iff g \in H$, thus implementing we have Ker$\phi \neq \{e\}$.

From the first Iso theorem we have Ker$\phi \lhd G$ and since we saw Ker$\phi$ is a proper subgroup of $G$ we have contradiction to the assumption G is simple.

Best Answer

Your solution is going in the correct direction, but I found four important issues, with (3) and (4) being the main ones:

  1. You write $gx$, but this is not defined (I understand what you mean, but this is still an issue).
  2. I am uncomfortable with the line "In particular, let us observe the induced homomorphism $\phi:G\to S_m$"; tell me why it exists, rather than state that it does!
  3. Your proof "shows" that $H=\ker\phi$, which is incorrect as $H$ may not be normal in $G$. (This is the line starting "Denote we have".) If we dig into your proof, we have: $g\in\ker\phi$ if and only if $gaH=aH$ for all $aH\in G/H$, which is equivalent to $a^{-1}ga\in H$ for all $a\in G$. Therefore, $\ker\phi$ is the largest normal subgroup contained in $H$. But, as just noted, $H$ may not be normal.
  4. At the end you state that $\ker\phi$ is proper, but this is never proven.

Below, I have copied-and-pasted your solution, and then edited it to address the issues discussed, and also to make some bits slightly clearer.

Assume, seeking a contradiction, that $G$ is simple but that there exists $H \lneq G$ of finite index, i.e. $1\neq[G:H]=m<\infty$.

Let us observe the action $G \mapsto G/H$ by multiplication from the left implementing we have $g.aH = gaH, \forall g \in G, \forall \;aH \in G/H$.

This action permutes the cosets $G/H$, so we have an induced homomorphism $\phi: G \to \operatorname{Sym}(G/H)\cong S_m$. (We covered this in Theorem ???, Section ??? of the notes.)

The kernel is a proper subgroup: As $H\lneq G$, there exists some $g\in G\setminus H$. The action of $g$ on $G/H$ is non-trivial, and so $g\not\in\ker\phi$.

The kernel is non-trivial: As $\operatorname{im}(\phi)$ is finite, the kernel has finite index in the infinite group $G$, and hence also contained infinitely many elements. In particular, it is non-trivial.

From the First Isomorphism Theorem, we have $\ker\phi \lhd G$ and since we saw $\ker\phi$ is a proper, non-trivial subgroup of $G$ we have contradiction to the assumption G is simple.

Let me finish with proof-structural point: it is not contradiction but contrapositive which you are using here. That is, you are proving that if $G$ has a finite index subgroup then it is not simple. Phrasing this as a contradiction is not incorrect, but is slightly clunky.

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