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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 ∣XG∣≡∣X∣mod  p|X_G|\equiv |X|\mod p.

Where XG={x∈X∣g⋅x=x for all g∈G}={x∈X∣orbit 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 g⋅h=ghg−1g\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={h∈G∣g⋅h=h for all g∈G}={h∈G∣ghg−1=h for all g∈G}=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)∣≡∣G∣mod  p|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
    • G≃Z2G\simeq\mathbb{Z}_2 (prime order)
  • ∣G∣=3|G|=3
    • G≃Z3G\simeq\mathbb{Z}_3 (prime order)
  • ∣G∣=4|G|=4
    • G≃Z2×Z2G\simeq\mathbb{Z}_2\times \mathbb{Z}_2
    • G≃Z4G\simeq\mathbb{Z}_4
  • ∣G∣=5|G|=5
    • G≃Z5G\simeq\mathbb{Z}_5 (prime order)
  • ∣G∣=6|G|=6
    • G≃S3G\simeq S_3
    • G≃Z3×Z2≃Z6G\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 G≃Z2×Z3G\simeq \mathbb{Z}_2\times \mathbb{Z}_3.

If ab=ba2ab=ba^2, then G≃S3G\simeq S_3.

  • ∣G∣=7|G|=7
    • G≃Z7G\simeq\mathbb{Z}_7 (prime order)
  • ∣G∣=8|G|=8
    • G≃Z2×Z2×Z2G\simeq\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2
    • G≃Z4×Z2G\simeq\mathbb{Z}_4\times \mathbb{Z}_2
    • G≃Z8G\simeq\mathbb{Z}_8
    • G≃D4G\simeq D_4
    • G≃G\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
    • G≃Z3×Z3G\simeq\mathbb{Z}_3\times \mathbb{Z}_3
    • G≃Z9G\simeq\mathbb{Z}_9 (apply the corollary, 9=329=3^2, these are all the possible cases)
  • ∣G∣=10|G|=10
    • G≃Z5×Z2≃Z10G\simeq\mathbb{Z}_5\times \mathbb{Z}_2\simeq \mathbb{Z}_{10}
    • G≃D5G\simeq D_5
  • ∣G∣=11|G|=11
    • G≃Z11G\simeq\mathbb{Z}_11 (prime order)
  • ∣G∣=12|G|=12
    • G≃Z3×Z4G\simeq\mathbb{Z}_3\times \mathbb{Z}_4
    • G≃Z2×Z2×Z3G\simeq\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_3
    • A4A_4
    • D6≃S3×Z2D_6\simeq S_3\times \mathbb{Z}_2
    • ??? One more
  • ∣G∣=13|G|=13
    • G≃Z13G\simeq\mathbb{Z}_{13} (prime order)
  • ∣G∣=14|G|=14
    • G≃Z2×Z7G\simeq\mathbb{Z}_2\times \mathbb{Z}_7
    • G≃D7G\simeq D_7

Lemma for group of order 2p2p where pp is prime

If pp is prime, p≠2p\neq 2, and ∣G∣=2p|G|=2p, then GG is either abelian ≃Z2×Zp\simeq \mathbb{Z}_2\times \mathbb{Z}_p or G≃DpG\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,…,ap−1,ba,ba2,…,bap−1e,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 G≃Z2×ZpG\simeq \mathbb{Z}_2\times \mathbb{Z}_p.

If ab=bap−1ab=ba^{p-1}, then G≃DpG\simeq D_p.

abab cannot be inverse of other elements, if ab=batab=ba^t, where 2≤t≤p−22\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 at2−1=ea^{t^2-1}=e, so p∣(t2−1)p|(t^2-1), therefore p∣t−1p|t-1 or p∣t+1p|t+1.

This is not possible since 2≤t≤p−22\leq t\leq p-2.

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