Math 31- Second Exam Theorems

A subgroup H of a group G is called a normal subgroup of G if aH=Ha for all a in G

A subgroup H of G is normal in G if and only if xHx^(-1) is contained in H for all x in G

(a) Every subgroup of an Abelian group is Normal
(b) The center of a group is always normal
(c) The alternating group An of even permutations is a normal subgroup of Sn

If a subgroup H is normal, then the set of left (or right) cosets of H in G is itself a group (called the factor or quotient group)

The set G/H= { aH : a<G } is a group under the operation (aH)(bH)=abH

Let G be a group and Z(G) be the center. If G/Z(G) is cyclic, then G is Abelian

...Inn(G)

Let G be a finite Abelian group and let p be a prime that divides the order of G. Then G has an element of order p

...and write G=H*K if H and K are normal subgroups of G and G=HK and H^K={e}

Let H1,...,Hn be a finite collection of normal subgroups of G. We say that G is the internal direct product of H1,...,Hn and write G=H1*...*Hn, if
(1) G=H1H2...Hn= {h1h2...hn: hi<Hi}
(2) (H1H2...Hi)^H(i+1)={e} for i=1,2,...,n-1

...is isomorphic to Zp^2 or the external direct product of Zp and Zp

...G is isomorphic to the external direct product of those subgroups

...then G is abelian

...a mapping from G to G' that preserves the group operation; that is R(ab)=R(a)R(b) for all a, b in G

...is the set from group G that maps to the identity in G'

(1) R carries the identity of G to the identity of G'
(2) R(g^n)=R(g)^n for all n in Z
(3) If |g| is finite, then |R(g)| divides |g|

(4) Ker R is a subgroup of G
(5) R(a)=R(b) IFF a(Ker R)=b(Ker R)
(6) If R(g)=g', then R^-1(g')={ x<G | R(x)=g'}= g(Ker R)

(1) R(H) is a subgroup of G'
(2) If H is cyclic, R(H) is cyclic
(3) if H is abelian, R(H) is abelian

(4) If H is normal in G, then R(H) is normal in G'
(5) If |Ker R|= n, then R is an n to 1 mapping from G onto R(G)
(6) if |H|=n, then |R(H)| divides n

(7) If K' is a subgroup of G', then R^(-1) [K'] is a subgroup of G
(8) If K' is a normal subgroup of G', then R^(-1) [K'] is a normal subgroup of G
(9) If R is onto and Ker R= {e}, then R is an isomorphism

Let R be a homomorphism from G to G'. Then the mapping G/Ker R to R(G) given by gKer R ---> R(g) is an isomorphism

If R is a homomorphism from a finite group G to G', then |R(G)| divides |G| and |G'|

...the kernel of a homomorphism. In particular, a normal subgroup N is the kernel of the mapping g ---> gN from G to G/N

Every finite Abelian group is a direct product of cyclic groups of prime-power order. Moreover, the number of terms in the product and the orders of the cyclic groups are uniquely determined by the group

...then G has a subgroup of order m