- #1
A.Magnus
- 138
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I am working on myself on a problem looks like this:
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) \subseteq N##.
I have been asking around and here is what I got so far:
(1) Since ##G'## is simple group, ##G'## does not have any non-trivial normal subgroup.
(2) Since ##ker(\phi)## is a normal subgroup of ##G##, therefore it is either ##ker (\phi) = G## or ##ker (\phi) = \{1\}.##
(3) In the first case, ##ker (\phi) = G## means ##\phi (G) = 1 ##, and therefore ##\phi (G) \subseteq N## and we are done.
(4) In the second case, ##ker (\phi) = \{1\}## means that ##\phi## is injective and therefore ##\phi (G) \cong G.## This reduces the problem into proving that ##(G) \subseteq N##.
(5) ...
And after that I am totally stuck, would appreciate any friendly and line-by-line befitting a person still in learning stage. Thank you for your time and help. Happy holidays.
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) \subseteq N##.
I have been asking around and here is what I got so far:
(1) Since ##G'## is simple group, ##G'## does not have any non-trivial normal subgroup.
(2) Since ##ker(\phi)## is a normal subgroup of ##G##, therefore it is either ##ker (\phi) = G## or ##ker (\phi) = \{1\}.##
(3) In the first case, ##ker (\phi) = G## means ##\phi (G) = 1 ##, and therefore ##\phi (G) \subseteq N## and we are done.
(4) In the second case, ##ker (\phi) = \{1\}## means that ##\phi## is injective and therefore ##\phi (G) \cong G.## This reduces the problem into proving that ##(G) \subseteq N##.
(5) ...
And after that I am totally stuck, would appreciate any friendly and line-by-line befitting a person still in learning stage. Thank you for your time and help. Happy holidays.