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Let ##G′## be a group and let ##\phi## be a homomorphism from ##G## to ##G′##. Assume that ##G## is simple, that ##|G| \neq 2##, and that ##G′## has a normal subgroup ##N## of index 2. Show that ##\phi (G) \subset N##.

And last year somebody gave me the following solution but which I am not quite understand, especially the last two parts:

(a) Let ##\phi## be homomorphism ##\phi : G \to G'## and ##\sigma## be the canonical homomorphis ##\sigma : G′ \to G′/N##, such that

##G \to G' \to G'/N##.

(b) And let the composition of two mappings ##\psi = \sigma \circ \phi ##.

(c) Since ##|G′/N|=2## and ##|G| \neq 2, \ \psi ## cannot be injective. So ##ker (\psi)=G##.

(d) In other words, ##\psi (G) = \{eN\} = N##.

(e) This implies ##\sigma ( \phi (G)) = N ##, which means ##\phi (G) \subset N##.

##G \to G' \to G'/N##.

(c) Since ##|G′/N|=2## and ##|G| \neq 2, \ \psi ## cannot be injective. So ##ker (\psi)=G##.

(d) In other words, ##\psi (G) = \{eN\} = N##.

(e) This implies ##\sigma ( \phi (G)) = N ##, which means ##\phi (G) \subset N##.

Since the responder may no longer be reached after so many months have passed, I am therefore posting questions here, hoping that some of you would help me with the last two parts:

(1) I understand up to (c), but how do you go from (c) to (d)?

(2) And also, how do you go from (d) to (e)?

(2) And also, how do you go from (d) to (e)?

Thank you very much for your time and help.