no.16,please show step by step and give me an easy mode, thank you Ch. 2 84 G ha
ID: 3184172 • Letter: N
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no.16,please show step by step and give me an easy mode, thank you
Ch. 2 84 G has an element is a prime and pm. Groups 15. If G is an abelian group and if G has an element of order m order n, where m and n are relatively prime, prove that G has a of order mn. 16. Let G be an abelian group of order p"m, where p is a prime an Let P = {a E G| a'* = e for some k depending on a}. Prove that: (a) P is a subgroup of G. (b) GIP has no elements of order p. (c) |P| =p". 17. Let G be an abelian group of order mn, where m and n are relative prime. Let M = {a E Ga" = e}. Prove that: (a) M is a subgroup of G. (b) GIM has no element, x, other than the identity element, such that x" = unit element of G/M. 18. Let G be an abelian group (possibly infinite) and let the set T = {a E Ga" = e, m >l depending on a}. Prove that: (a) T is a subgroup of G. (b) GIT has no element—other than its identity element—of finite order. 7. THE HOMOMORPHISM THEOREMS Let G be a group and p a homomorphism of G onto G'. If K is the kernel of =p, then K is a normal subgroup of G, hence we can form GIK. It is fairly nat- ural to expect that there should be a very close relationship between G' and GIK. The First Homomorphism Theorem, which we are about to prove. pells out this relationship in exact detail. But first let's look back at some of the examples of factor groups in section 6 to see explicitly what the relationship mentioned above might be. Let G = {T...la + 0, b real} and let G' be the group of nonz under multiplication. From the product rule of these T tea T..Te4 = Tac.ad + b, we determined that the mare: G - by (Ta.b) = a is a homomorphism of God with {71.|b real. On the nthouExplanation / Answer
a)
Identity element e clearly belongs to P
Let, a belong to P then for some k
a^{p^k}=e
Hence, a^{-p^k}=e=(a^{-1})^{p^k}=e
Hence inverse of a is in P
Hence, P is a subgroup
b)
Let, [x] be an element in G/P which is of order p so that x is in G
x cannot be in P otherwise its order would be 0
[x]^p=e means
[x]^p=P
ie x^p lies in P
So, x^p=a for some a in P
a is in P hence, a^{p^k}=e for some k
HEnce, (x^p)^{p^k}=x^{p^{k+1})=e
Hence, x is in P which is a contradiction
Hence not element in G/P is of order P
c)
Let order of P be p^k, k<n
Hence, G/P is of order , mp^{n-k} ie multiple of prime ,p
Hence, G/P must have an element of order which is not true as we proved in part (b)
HEnce, G/P has order p^n
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