Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $H$ be a subgroup of $G$ with $[G:H]=n$ and $G$ finite. Let $G$ act on left cosets of $H$ by left multiplication. Let $N$ be the kernel of the action. Show that $[G:N]$ divides $n!$.

Ok so I have a theorem that says that if $G$ is simple then it embeds into $S_n$. So if $G$ is simple than I'm done because $G$ divides $n!$ and thus $N$ divides $n!$ and hence $[G:N]$ divides $n!$.

Now I also know that $N$ is the largest normal subgroup of $G$ contained in $H$. Thus if $G$ isn't simple then it contains some non-trivial normal subgroup $K$ and thus $K\cap H$ is normal in $H$ and by the 2nd Iso Theorem $H/K\cap H\cong KH/H$...

As you can see I'm not sure where to go from here. I was hoping to prove that $K\cap H$ wasn't trivial and thus that $N$ couldn't be trivial since it must contain $K\cap H$, but I'm not sure I can, and even if I could I'm not sure that would help me in showing that $[G:N]$ is small enough to divide $n!$.

Can anyone give some guidance? Thanks.

share|improve this question
Ronald Weasley! –  KUSH Nov 28 '12 at 0:32

1 Answer 1

up vote 3 down vote accepted

Let $S_{G:H}$ be the set of permutations of $G:H$. Consider $\phi:(G:N)\rightarrow S_{G:H}$ that sends $xN$ to $T_x$ (Where $T_x$ is the permutation of $G:H$ that sends $aH$ to $xaH$ ). It is easy to verify that $\phi$ is an injective group homomorphism. Thus, $G:N$ is isomorphic to a subgrop of $S_{G:H}$. Using Lagrange's theorem, we deduce that $|G:N|$ divides $|S_{G:H}|=n!$

share|improve this answer
I'm assuming by isomorphism you mean embedding (injective hom). I've been completely unsuccessful proving that $\phi$ is an embedding. For instance for injectivity: I assume $gaH=faH$ for all $a\in G$ and I need to prove that $gN = fN$, but the most I can prove is that $gf^{-1}\in H$ when what I need to show is that $gf^{-1}\in N$. –  Blood Pudding Nov 26 '12 at 1:02
OH yes. I will edit it –  Amr Nov 26 '12 at 1:05
any ideas for proving injectivity? –  Blood Pudding Nov 26 '12 at 1:24
Let $\phi(x_1)=\phi(x_2)$, therefore $T_{x_1}=T_{x_2}$. Hence for all a in G $x_1aH=x_2aH$. Thus $x^{-1}_2x_1aH=aH$ therefore $x^{-1}_2x_1\in N$. Hence $x_1N=x_2N$ –  Amr Nov 26 '12 at 1:37
You should define $\phi$ on $G$ and use the homomorphism theorem in order to avoid redundant computations. –  Martin Brandenburg Nov 26 '12 at 1:52

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.