NoteNextra-origin/content/Math4302/Math4302_L23.md

Math4302 Modern Algebra (Lecture 23)

Group

Group acting on a set

Theorem for the orbit of a set with prime power group

Suppose X is a $G$-set, and |G|=p^n where p is prime, then |X_G|\equiv |X|\mod p.

Where 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 p, where p is a prime, divides |G|, then G has a subgroup of order p. (equivalently, g has an element of order p)

This does not hold when p is not prime.

Consider A_4 with order 12, and A_4 has no subgroup of order 6.

Corollary: Center of prime power group is non-trivial

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

Proof

Let G act on G via conjugation, then g\cdot h=ghg^{-1}. This makes G to a $G$-set.

Apply the theorem, the set of elements with trivial orbit is; Let X=G, then 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)|\equiv |G|\mod p.

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

Proposition: Prime square group is abelian

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

Proof

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

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

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

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

Classification of small order

Let G be a group

• |G|=1
  • G=\{e\}
• |G|=2
  • G\simeq\mathbb{Z}_2 (prime order)
• |G|=3
  • G\simeq\mathbb{Z}_3 (prime order)
• |G|=4
  • G\simeq\mathbb{Z}_2\times \mathbb{Z}_2
  • G\simeq\mathbb{Z}_4
• |G|=5
  • G\simeq\mathbb{Z}_5 (prime order)
• |G|=6
  • G\simeq S_3
  • G\simeq\mathbb{Z}_3\times \mathbb{Z}_2\simeq \mathbb{Z}_6

Proof

|G| has an element of order 2, namely b, and an element of order 3, namely a.

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

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

If ab=ba, then G is abelian, then G\simeq \mathbb{Z}_2\times \mathbb{Z}_3.

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

• |G|=7
  • G\simeq\mathbb{Z}_7 (prime order)
• |G|=8
  • G\simeq\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2
  • G\simeq\mathbb{Z}_4\times \mathbb{Z}_2
  • G\simeq\mathbb{Z}_8
  • G\simeq D_4
  • G\simeq quaternion group \{e,i,j,k,-1,-i,-j,-k\} where i^2=j^2=k^2=-1, (-1)^2=1. ij=l, jk=i, ki=j, ji=-k, kj=-i, ik=-j.
• |G|=9
  • G\simeq\mathbb{Z}_3\times \mathbb{Z}_3
  • G\simeq\mathbb{Z}_9 (apply the corollary, 9=3^2, these are all the possible cases)
• |G|=10
  • G\simeq\mathbb{Z}_5\times \mathbb{Z}_2\simeq \mathbb{Z}_{10}
  • G\simeq D_5
• |G|=11
  • G\simeq\mathbb{Z}_11 (prime order)
• |G|=12
  • G\simeq\mathbb{Z}_3\times \mathbb{Z}_4
  • G\simeq\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_3
  • A_4
  • D_6\simeq S_3\times \mathbb{Z}_2
  • ??? One more
• |G|=13
  • G\simeq\mathbb{Z}_{13} (prime order)
• |G|=14
  • G\simeq\mathbb{Z}_2\times \mathbb{Z}_7
  • G\simeq D_7

Lemma for group of order 2p where p is prime

If p is prime, p\neq 2, and |G|=2p, then G is either abelian \simeq \mathbb{Z}_2\times \mathbb{Z}_p or G\simeq D_p

Proof

We know G has an element of order 2, namely b, and an element of order p, namely a.

So e,a,a^2,\dots ,a^{p-1},ba,ba^2,\dots,ba^{p-1} are distinct elements of G.

Consider ab, if ab=ba, then G is abelian, then G\simeq \mathbb{Z}_2\times \mathbb{Z}_p.

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

ab cannot be inverse of other elements, if ab=ba^t, where 2\leq t\leq p-2, then bab=a^t, then (bab)^t=a^{t^2}, then ba^tb=a^{t^2}, therefore a=a^{t^2}, then a^{t^2-1}=e, so p|(t^2-1), therefore p|t-1 or p|t+1.

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