I just did this problem:

"Prove that if $G$ is a finite group and $H$ is a proper normal subgroup of largest order, then $G/H$ is simple."

And I am currently working on this problem:

"Suppose that $p$ is the smallest prime that divides $|G|$. Show that any subgroup of index $p$ in $G$ is normal in $G$."

If $p$ is the smallest prime that divides $|G|$, then 'proper normal subgroup of largest order' and 'normal subgroup of index $p$' are equivalent, aren't they? Based on this, I've been trying to start with the quotient, which is of order $p$ and must be simple, and work backward. I don't know that this is necessarily a good approach.

I'm not making too much progress toward a proof, and any hints (not answers) would be much appreciated.

  • $\begingroup$ @ArturoMagidin I read "order" instead of "index". Sorry. $\endgroup$ – Alex Becker Feb 22 '12 at 18:08
  • $\begingroup$ @Arturo: thanks. totally not equivalent. $\endgroup$ – Jack Schmidt Feb 22 '12 at 18:12
  • $\begingroup$ @ArturoMagidin Can I find you in the chat? $\endgroup$ – Pedro Tamaroff Feb 22 '12 at 18:14
  • $\begingroup$ @Peter: I don't usually go to chat, and I'm about to go to lunch, sorry. $\endgroup$ – Arturo Magidin Feb 22 '12 at 18:21
  • $\begingroup$ @ArturoMagidin Ok. $\endgroup$ – Pedro Tamaroff Feb 22 '12 at 18:23

A normal subgroup of prime index is maximal, but it need not be "of largest order." For example, in the cyclic group of order $6$ generated by $x$, $\langle x^3\rangle$ is of prime index (namely, index $3$), but is not of largest order (the largest order for a proper normal subgroup is $3$, given by $\langle x^2\rangle$). The mistake here is that even though a normal subgroup of prime index cannot be properly contained in a proper subgroup, it can have smaller size than a subgroup that does not contain it.

Also: you cannot assume that you can do a quotient modulo a subgroup of index $p$ unless you first prove that the subgroup is normal. So using the quotient would be circular.

As for the proof of this standard problem: consider the action of $G$ on the left cosets of $H$ given by $g\cdot xH =gxH$. This induces a group homomorphism from $G$ to $S_{G/H}$, the permutation group of the cosets, with kernel contained in $H$.

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