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CAF123
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Homework Statement
Let ##\theta : G \mapsto H## be a group homomorphism.
A) Show that ##\theta## is injective ##\iff## ##\text{Ker}\theta = \left\{e\right\}##
B) If ##\theta## is injective, show that ##G \cong I am \theta ≤ H##.
The Attempt at a Solution
A)The right implication is fine. Briefly, if ##\theta## is a homomorphism, then the identity in G is always sent to the identity in H. Since ##\theta## is injective, no other element maps there. I am not so sure about the left implication, but what I tried was:
Consider ##\theta (g_1) = h_1## and ##\theta (g_2) = h_1. ## Want to show that ##g_1 = g_2## using the fact that ##\text{Ker}\theta = ##identity.
Then ##\theta (g_1 e_G) = \theta (g_1) e_H = h_1 = \theta(g_2)e_H## so ##\theta(g_1) = \theta(g_2) (1)##. But I realize this is a trivial result. From here I want to show that ##g_1 = g_2##. We were given a hint to use inverses more explicitly, but I can't see it right now.
B)I think I have this one, I just would like someone to check over it to make sure I haven't overlooked anything. So prove that ##G \cong I am \theta##
This means ##G## is isomorphic to ##Im \theta## and so we have a bijective homomorphism. It is given that ##\theta## is injective, so we need only show surjectivity.
But simply, by defintion of ##Im \theta##, it is surjective.
Now prove ##Im \theta \leq H##, so use test for subgroup.
i)Non empty: we are dealing with a homomorphism, so the identity in g is always mapped to identity in H, so the image is never nonempty.
ii)Closure: Let ##h_1, h_2 \in I am \theta##. For some ##g_1, g_2 \in G, g_1 \neq g_2##, we have ##\theta(g_1) = h_1 ## and ##\theta(g_2) = h_2##. Then ##\theta(g_1)\theta(g_2) = h_1 h_2 = \theta(g_1 g_2). ## We found such an element that maps to ##h_1 h_2## so closure is satisfied.
iii) Let ##\ell = h_1 h_2 \Rightarrow \ell h_2^{-1} = h_1 \Rightarrow \ell h_2^{-1} h_1^{-1} = e \Rightarrow \ell (h_1 h_2)^{-1} = e.## So we found an element such that when it is smashed with another element we get ##e \in I am \theta##, so closed under inverse.
Is this okay? If so, it is really just the left implication of A) that I need a hint for.
Many thanks