Question

Let f: G--> H be a surjective homomorphism of groups. Prove that H is abelian iff the commutator subgroup G' is contained in the kernel of f.
Please, help

Answer 1

where are you stuck?

Answer 2

I am stuck at commutator subgroup

Answer 3

ok...what about the commutator subgroup are you stuck on?

Answer 4

by definition, G'={aba^-b^- | a, b in G}, if it is in kernel of f, then f (G') = e_H

Answer 5

and f(G') = f(aba^-b^-) = f(a) f(b) f(a^-) f(b^-) = e_H
I don't see how to force H is abelian from this

Answer 6

\[f(a^{-1})=f(a)^{-1}\]

Answer 7

also f is surjective

Answer 8

yes, but if H is abelian, then we can change their location to get e_H. We don't have it yet.

Answer 9

all we can do is f(ab) f(ba)^- = e_H

Answer 10

which direction are you trying to prove right now

Answer 11

I 've just start, either way is ok. Now, I do if commutator subgroup G' in the kernel, then H is abelian.

Answer 12

ok

so we have \[f(aba^{-1}b^{-1})=ker(f)\]

right

Answer 13

not equal

Answer 14

element of

Answer 15

yes

Answer 16

\[f(aba^{-1}b^{-1})\in ker(f)\]

Answer 17

yes,

Answer 18

oh, no

Answer 19

lol

Answer 20

aba^- b^- in ker f, not f(aba^-b^-) in kernel f

Answer 21

aba^-1b^-1 is is in the ker

Answer 22

yes, :)

Answer 23

either way...


\[f(aba^{-1}b^{-1})=e_H\]


ok :)

Answer 24

yes,

Answer 25

now break it up as

\[f(a)f(b)f(a^{-1})f(b^{-1})=e_{H}\]

Answer 26

ok?

Answer 27

yes

Answer 28

then

\[f(a)f(b)f(a)^{-1}f(b)^{-1}=e_{H}\]

Answer 29

ok?

Answer 30

ok

Answer 31

then \[f(a)f(b)=f(b)^{-1}f(a)^{-1}\]

Answer 32

yes

Answer 33

that last step was a typo (copied too much)

\[f(a)f(b)f(a)^{-1}f(b)^{-1}=e_{H}\]

implies

\[f(a)f(b)=f(b)f(a)\]

Answer 34

\[f(a)f(b)f(a)^{-1}f(b)^{-1}=e_{H}\]

\[f(a)f(b)f(a)^{-1}f(b)^{-1}f(b)f(a)=e_{H}f(b)f(a)\]

Answer 35

yes, I see it

Answer 36

that shows H is abelian, lalala, right?

Answer 37

yes because f is surjective

Answer 38

how it relate to surjective?

Answer 39

every two elements of H can be written as f(a) or f(b)

Answer 40

for any \[\alpha,\beta\in H\]
there exists \[a,b\in G\] such that \[f(a)=\alpha\] and \[f(b)=\beta\]

Answer 41

and that is true because f is surjective

Answer 42

backward, :) for any f(a) , f(b) in H, there exists a, b in G. That's the definition of surjective, right?

Answer 43

lol, yes, my bad,

Answer 44

I see it

Answer 45

that is what I said...for any elements in H there are elements in G that are mapped to the elements in H

Answer 46

The converse part is backward of what we did, right? (no, you did, not me , lol)

Answer 47

pretty much

Answer 48

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Answer 49

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Answer 50

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Answer 51

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Answer 52

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Answer 53

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Answer 54

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Answer 55

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Answer 56

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Answer 57

:)

Answer 58

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Answer 59

no problem ... later