MHB Sorry, I'm not sure what you're asking for. Could you clarify?

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Here is this week's POTW:

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Why can't an infinite group that has a subgroup of finite index $> 1$ be simple?

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Hello again MHB community,

I've made a correction to the problem statement. The term "finite index subgroup" has been replaced by "subgroup of finite index $> 1$". This is important, for there are infinite simple groups. For example, given any infinite field $F$, the projective special linear group $\operatorname{PSL}_2(F)$ is simple. Sorry for the typo.

 

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Congratulations to the following members for their correct solutions:

1. caffeinemachine

2. Deveno

3. johngHere is a solution by Deveno:

Spoiler

So assume $G$ is infinite, and has a proper subgroup of finite index.

Letting $G$ act on the finite set of (left) cosets $G/H$ by left multiplication, we obtain a homomorphism:

$\phi:G \to S_n$, where $n$ is the index of $H$ in $G$.

The kernel of $\phi$ cannot be trivial, since $G$ is infinite, and $S_n$ is not (thus there can be no injection of $G$ into $S_n$ whatsoever, all the more so a homomorphic injection).

$\phi$ cannot be trivial (that is, the kernel cannot be all of $G$) since $H \neq G$, so that there exists at least one coset $gH \neq H$, thus $\phi(g) \neq \text{id}$-so $g$ is not in the kernel.

Thus $\text{ker }\phi$ is a non-trivial proper normal subgroup of $G$; that is, $G$ is not simple.

Here's another solution by johng:

Spoiler

Suppose the infinite group $G$ has a subgroup $H$ of finite index $n >1$.The number of conjugates of $H$ in $G$ is the index of the normalizer of $H$ in $G$, $[G:N(H)]$ again a finite integer since $N(H)$ contains $H$. Now if two subgroups $H$ and $K$ have finite index, the index of $H\cap K$ is also finite (this index is easily shown to be less than or equal to the product of the indices of $H$ and $K$). So the intersection of a finite number of subgroups, each of finite index, has finite index. Consider $\operatorname{Core}(H)$, the largest normal subgroup of G contained in H; it is the intersection of all conjugates of $H$. So $\operatorname{Core}(H)$ has finite index and hence is a proper normal subgroup.