(c) If K is a subgroup of G, then p(K) is a subgroup of H. Given: K < G (d) If K' is a subgroup of H, then p-'(K') is a subgroup of G.

Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter4: More On Groups
Section4.7: Direct Sums (optional)
Problem 24E
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I did a,b.......need help with c &d for abstract algebra.

 

1. Let p: G → H be a group homomorphism. Show:
(a) p(1g) = 1µ;
Let p: G → H
p(1ç):
= 1µ, where 1gand 1µare the identities of G, H respectively
P(16) = p(1g1c)
= p(1g)9(1ç) (p is a homomorphism)
p(1c)¯'p(1g) = q(1g)--»(1c)q(1c)|
1H = 1µ 9(1g)
1H = p(1g)
: p(1g) = 1µ
(*)
..
%3D
... ... ... ...
(b) o(a-1) = [p(a)]-1, va e G;
p(g-1) = 9(g)-1 vg E G
p(16) = 9(gg¬1) = p(g)p(g¬1)
Since p(1g) = 1µ .
- (*)
..... .... ... u. ..
= 1µ = p(g)q(g¬1)
= p(g¬1)1µ = q(g)p(g¬1)ø(g¬1)
= p(g-1) = 9(g)-1
(c) If K is a subgroup of G, then p(K) is a subgroup of H.
Given: K < G
(d) If K' is a subgroup of H, then p-1(K') is a subgroup of G.
Transcribed Image Text:1. Let p: G → H be a group homomorphism. Show: (a) p(1g) = 1µ; Let p: G → H p(1ç): = 1µ, where 1gand 1µare the identities of G, H respectively P(16) = p(1g1c) = p(1g)9(1ç) (p is a homomorphism) p(1c)¯'p(1g) = q(1g)--»(1c)q(1c)| 1H = 1µ 9(1g) 1H = p(1g) : p(1g) = 1µ (*) .. %3D ... ... ... ... (b) o(a-1) = [p(a)]-1, va e G; p(g-1) = 9(g)-1 vg E G p(16) = 9(gg¬1) = p(g)p(g¬1) Since p(1g) = 1µ . - (*) ..... .... ... u. .. = 1µ = p(g)q(g¬1) = p(g¬1)1µ = q(g)p(g¬1)ø(g¬1) = p(g-1) = 9(g)-1 (c) If K is a subgroup of G, then p(K) is a subgroup of H. Given: K < G (d) If K' is a subgroup of H, then p-1(K') is a subgroup of G.
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