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Math4302Modern Algebra (Lecture 23)

Math4302 Modern Algebra (Lecture 23)

Group

Group acting on a set

Theorem for the orbit of a set with prime power group

Suppose XX is a GG-set, and G=pn|G|=p^n where pp is prime, then XGXmodp|X_G|\equiv |X|\mod p.

Where XG={xXgx=x for all gG}={xXorbit of x is trivial}X_G=\{x\in X|g\cdot x=x\text{ for all }g\in G\}=\{x\in X|\text{orbit of }x\text{ is trivial}\}

Corollary: Cauchy’s theorem

If pp, where pp is a prime, divides G|G|, then GG has a subgroup of order pp. (equivalently, gg has an element of order pp)

This does not hold when pp is not prime.

Consider A4A_4 with order 1212, and A4A_4 has no subgroup of order 66.

Corollary: Center of prime power group is non-trivial

If G=pm|G|=p^m, then Z(G)Z(G) is non-trivial. (Z(G){e}Z(G)\neq \{e\})

Proof

Let GG act on GG via conjugation, then gh=ghg1g\cdot h=ghg^{-1}. This makes GG to a GG-set.

Apply the theorem, the set of elements with trivial orbit is; Let X=GX=G, then XG={hGgh=h for all gG}={hGghg1=h for all gG}=Z(G)X_G=\{h\in G|g\cdot h=h\text{ for all }g\in G\}=\{h\in G|ghg^{-1}=h\text{ for all }g\in G\}=Z(G).

Therefore Z(G)Gmodp|Z(G)|\equiv |G|\mod p.

So pp divides Z(G)|Z(G)|, so Z(G)1|Z(G)|\neq 1, therefore Z(G)Z(G) is non-trivial.

Proposition: Prime square group is abelian

If G=p2|G|=p^2, where pp is a prime, then GG is abelian.

Proof

Since Z(G)Z(G) is a subgroup of GG, Z(G)|Z(G)| divides p2p^2 so Z(G)=1,p|Z(G)|=1, p or p2p^2.

By corollary center of prime power group is non-trivial, Z(G)1Z(G)\neq 1.

If Z(G)=p|Z(G)|=p. If Z(G)=p|Z(G)|=p, then consider the group G/Z(G)G/Z(G) (Note that Z(G)GZ(G)\trianglelefteq G). We have G/Z(G)=p|G/Z(G)|=p so G/Z(G)G/Z(G) is cyclic (by problem 13.39), therefore GG is abelian.

If Z(G)=p2|Z(G)|=p^2, then GG is abelian.

Classification of small order

Let GG be a group

  • G=1|G|=1
    • G={e}G=\{e\}
  • G=2|G|=2
    • GZ2G\simeq\mathbb{Z}_2 (prime order)
  • G=3|G|=3
    • GZ3G\simeq\mathbb{Z}_3 (prime order)
  • G=4|G|=4
    • GZ2×Z2G\simeq\mathbb{Z}_2\times \mathbb{Z}_2
    • GZ4G\simeq\mathbb{Z}_4
  • G=5|G|=5
    • GZ5G\simeq\mathbb{Z}_5 (prime order)
  • G=6|G|=6
    • GS3G\simeq S_3
    • GZ3×Z2Z6G\simeq\mathbb{Z}_3\times \mathbb{Z}_2\simeq \mathbb{Z}_6

Proof

G|G| has an element of order 22, namely bb, and an element of order 33, namely aa.

So e,a,a2,b,ba,ba2e,a,a^2,b,ba,ba^2 are distinct.

Therefore, there are only two possibilities for value of abab. (a,a2a,a^2 are inverse of each other, bb is inverse of itself.)

If ab=baab=ba, then GG is abelian, then GZ2×Z3G\simeq \mathbb{Z}_2\times \mathbb{Z}_3.

If ab=ba2ab=ba^2, then GS3G\simeq S_3.

  • G=7|G|=7
    • GZ7G\simeq\mathbb{Z}_7 (prime order)
  • G=8|G|=8
    • GZ2×Z2×Z2G\simeq\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2
    • GZ4×Z2G\simeq\mathbb{Z}_4\times \mathbb{Z}_2
    • GZ8G\simeq\mathbb{Z}_8
    • GD4G\simeq D_4
    • GG\simeq quaternion group {e,i,j,k,1,i,j,k}\{e,i,j,k,-1,-i,-j,-k\} where i2=j2=k2=1i^2=j^2=k^2=-1, (1)2=1(-1)^2=1. ij=lij=l, jk=ijk=i, ki=jki=j, ji=kji=-k, kj=ikj=-i, ik=jik=-j.
  • G=9|G|=9
    • GZ3×Z3G\simeq\mathbb{Z}_3\times \mathbb{Z}_3
    • GZ9G\simeq\mathbb{Z}_9 (apply the corollary, 9=329=3^2, these are all the possible cases)
  • G=10|G|=10
    • GZ5×Z2Z10G\simeq\mathbb{Z}_5\times \mathbb{Z}_2\simeq \mathbb{Z}_{10}
    • GD5G\simeq D_5
  • G=11|G|=11
    • GZ11G\simeq\mathbb{Z}_11 (prime order)
  • G=12|G|=12
    • GZ3×Z4G\simeq\mathbb{Z}_3\times \mathbb{Z}_4
    • GZ2×Z2×Z3G\simeq\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_3
    • A4A_4
    • D6S3×Z2D_6\simeq S_3\times \mathbb{Z}_2
    • ??? One more
  • G=13|G|=13
    • GZ13G\simeq\mathbb{Z}_{13} (prime order)
  • G=14|G|=14
    • GZ2×Z7G\simeq\mathbb{Z}_2\times \mathbb{Z}_7
    • GD7G\simeq D_7

Lemma for group of order 2p2p where pp is prime

If pp is prime, p2p\neq 2, and G=2p|G|=2p, then GG is either abelian Z2×Zp\simeq \mathbb{Z}_2\times \mathbb{Z}_p or GDpG\simeq D_p

Proof

We know GG has an element of order 2, namely bb, and an element of order pp, namely aa.

So e,a,a2,,ap1,ba,ba2,,bap1e,a,a^2,\dots ,a^{p-1},ba,ba^2,\dots,ba^{p-1} are distinct elements of GG.

Consider abab, if ab=baab=ba, then GG is abelian, then GZ2×ZpG\simeq \mathbb{Z}_2\times \mathbb{Z}_p.

If ab=bap1ab=ba^{p-1}, then GDpG\simeq D_p.

abab cannot be inverse of other elements, if ab=batab=ba^t, where 2tp22\leq t\leq p-2, then bab=atbab=a^t, then (bab)t=at2(bab)^t=a^{t^2}, then batb=at2ba^tb=a^{t^2}, therefore a=at2a=a^{t^2}, then at21=ea^{t^2-1}=e, so p(t21)p|(t^2-1), therefore pt1p|t-1 or pt+1p|t+1.

This is not possible since 2tp22\leq t\leq p-2.

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