Re: Algebra with isomorphism and group.


"mina_world" <mina_world@xxxxxxxxxxx> wrote in message
news:fhhasu$q87$1@xxxxxxxxxxxxxxxxxxx
Hello sir~

G is a group.

f : G -> f[G] is isomorphism.

then,
Can I say that f[G] is group unconditionally ?

In fact, it's trivial.
(f[G], .) is a binary algebraic structure.
so, It's closed basically.
Since (G, *) ~ (f[G] , .),
f(e) in f[G] for e in G.
f(g^-1) in f[G] for g in G.

so, f[G] is a group trivially.


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