Re: Algebra with isomorphism and group.



mina_world wrote:
"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, my question derive from proof of Cayley's theorem.
Namely,
Every group is isomorphic to a group of permutations.
pf)
Let G be a group.
Let f : G -> S_G by f(x) = g_x for all x in G.
Let g_x : G -> G by g_x(g) = xg for all g in G.

so, g_x is a permutation of G. (easy to show)

Since f is homomorphic and 1-1 and onto,
G ~ f(G) (isomorphic)

Can I say that f(G) is (sub)group automatically ?

Yes, in general if G and H are groups and f: G -> H is a homomorphism,
then the image f(G) of f is a subgroup of H. The proof is
straightforward.

(Also, the kernel K of f is a normal subgroup of G and G/K is
isomorphic to f(G).)

Derek Holt.
.

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