Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
@ArturoMagidin I read "order" instead of "index". Sorry. – Alex Becker Feb 22 '12 at 18:08
@Arturo: thanks. totally not equivalent. – Jack Schmidt Feb 22 '12 at 18:12
@ArturoMagidin Can I find you in the chat? – Pedro Tamaroff Feb 22 '12 at 18:14
@Peter: I don't usually go to chat, and I'm about to go to lunch, sorry. – Arturo Magidin Feb 22 '12 at 18:21
@ArturoMagidin Ok. – Pedro Tamaroff Feb 22 '12 at 18:23
up vote 4 down vote accepted

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 $H$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.