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.
Best Answer
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 $H$.