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Let G be a group with subgroup H and let $\Omega$ be the set of right cosets of H in G.

Show that if G is a group with a subgroup of index n then G has a normal subgroup with index dividing n!

Attempt:

There are n-distinct right-cosets of H in G. These distinct right cosets are a collection in $\Omega.$

So $\left | \Omega \right |=n$

Also, by definition of the index of the subgroup H in G and using Lagrange's theorem

We get $\left | G:H \right |=\left | H \right | \mid \left | G \right |=n.$

Also, note that the order of the symmetric group of the finite set $\Omega$ of order n is $\left | Sym\left ( \Omega \right ) \right |=n!$ Note that $ker\left ( \varphi \right ) \triangleleft G$

At this point I am stuck. Looking at the solution, it mentioned that the index of the $ker\left ( \varphi \right )$ is $\left | \left ( G \right )\varphi \right |$ which I do not quite understand. Is this a consequence or some theorem or properties?

Could someone kindly explain? Thanks in advance.

Mathematicing
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1 Answers1

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Let $\phi:G\to S\left(\Omega\right)$ where $\phi\left(g\right)$ is prescribed by $Hk\mapsto Hkg^{-1}$.

It can be shown that $\phi$ is a grouphomomorphism.

If $N$ denotes its kernel then $G/N$ and $\phi\left(G\right)\leq S\left(\Omega\right)$ are isomorphic.

Then $\left[G:N\right]=\left|\phi\left(G\right)\right|$ and $\left|\phi\left(G\right)\right|$ divides $\left|S\left(\Omega\right)\right|=n!$.

drhab
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  • Yes $\phi$ is a homomorphism. And yes $\phi(G)$ is a subgroup of $S(\Omega)$ but for this to be true, the image of G $\phi(G)$ must be a group by Cauchy's theorem and of course is a group by definition of group homomorphism. But I don't see how these relates to the order of the image G being equivalent to the index of the $ker(\phi)$ in G. – Mathematicing May 04 '16 at 09:56
  • Do you agree that $G/N$ and $\phi(G)$ are isomorphic? First isomorphism theorem. – drhab May 04 '16 at 10:00
  • Yes they are isomorphic by the first isomorphism theorem. However, the first isomorphism theorem says the image of $\phi$ is a subgroup of some subgroup. While ker$(\phi)$ is the normal subgroup of the group G. I'm not sure how these relates to the order of the image and the index. – Mathematicing May 04 '16 at 10:03
  • Because they are isomorphic they have the same cardinality. The cardinality of $G/N$ is the index of $N$. The cardinality of the other (equal to the index of $N$) will divide $n!$ because it is a subgroup of a group that has order $n!$. – drhab May 04 '16 at 10:08
  • What is unclear to you? i) $|G/N|=[G:N]$. ii) $|G/N|=|\phi(G)|$ iii)$|\phi(G)|$ divides $n!$ iv) conclusion: $[G:N]$ divides $n!$ – drhab May 04 '16 at 10:25
  • I get it. I forgot the cardinality are the same if the isomorphism holds – Mathematicing May 04 '16 at 10:29