Abstract Nonsense

Crushing one theorem at a time

A Clever Proof of a Common Fact

Point of Post: In this post we give a new proof that if G is a finite group H is a subgroup G whose index is the smallest prime dividing |G| then that subgroup is normal.

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It is a commonly used theorem in finite group theory that if G is a finite group and H\leqslant G such that \left[G:H\right]=p is the smallest prime dividing |G| then H\unlhd G. We have already seen a proof of this fact by considering the homomorphism G\to S_p which is the induced map from G acting on G/H by left multiplication, and proving that \ker(G\to S_p)=H. We now give an even shorter (and the just mentioned proof is already short) proof of this fact using double cosets.

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The Theorem

We recall that a double coset HgK for subgroups H,K\leqslant G is an orbit of the action of H\times K on G given by (h,k)\cdot g=hgk^{-1}. We also recall that \#(HgK)=|H|\left[H:H\cap gKg^{-1}\right]. With this we can easily prove the theorem

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Theorem: Let G be a finite group and p the smallest prime dividing |G|. Then, if H\leqslant G is such that \left[G:H\right]=p then H\unlhd G.

Proof: We know that G decomposes as

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G=H\sqcup Hg_1H\sqcup\cdots\sqcup Hg_mH

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where g_1,\cdots,g_m. We see then that that

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p=1+\left[H:H\cap g_1Hg_1^{-1}\right]+\cdots+\left[H:H\cap g_mHg_m^{-1}\right]

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Now, we know then that \left[H:H\cap g_k Hg_k^{-1}\right]<p and since it’s a divisor of |G| we evidently mus have that \left[H:H\cap g_k Hg_k^{-1}\right]=1 and so H=g_kHg_k^{-1}. Since we can evidently choose g_k arbritrarily in G-H  (since it is a representative for one of the double cosets) the conclusion follows. \blacksquare

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1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2009






September 16, 2011 - Posted by | Algebra, Group Theory | , , , , , , , ,

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